Free math worksheets

- Use upper function minus lower function. - The vertical distance between the two horizontal lines is constant. - Integrate that constant width across the given x-interval.

1. The upper function is \(5\), and the lower function is \(2\). 2. Therefore, \(A=\int_0^4(5-2)\,\text{d}x=\int_0^4 3\,\text{d}x=12\).

\(12\) square units

- Read or solve for where the graph crosses the x-axis. - Use the intercepts as the limits of integration. - The shaded area is above the x-axis, so the definite integral is positive.

1. Solve \(g(x)=0\): \(-0.5(x-2)^2+4.5=0\), so \((x-2)^2=9\). Thus, \(x=-1\) or \(x=5\). 2. The bounded region extends from \(x=-1\) to \(x=5\). 3. A numerical evaluation gives \(\int_{-1}^{5}\left[-0.5(x-2)^2+4.5\right] \,\text{d}x\approx18\). 4. An exact check is \(\left[-\frac16(x-2)^3+4.5x\right]_{-1}^{5}=18\).

The x-intercepts are \((-1,0)\) and \((5,0)\). The area is approximately \(18\) square units.

- Substitute \(x=0\) to find the y-intercept. - Solve \(h(x)=0\) and use the positive solution. - Use the intercepts as the limits of the area integral.

1. The y-intercept is found from \(h(0)=3.6\), so it is \((0,3.6)\). For the positive x-intercept, solve \(-0.1x^2+3.6=0\). Then \(x^2=36\), so \(x=6\). The intercept is \((6,0)\). 2. The region lies under \(h\) from \(x=0\) to \(x=6\). Its area is \(\int_0^6(-0.1x^2+3.6)\,\text{d}x\). 3. Evaluating gives \(\left[-\frac{x^3}{30}+3.6x\right]_0^6=-7.2+21.6=14.4\).

1. The intercepts are \((0,3.6)\) and \((6,0)\). 2. The area is \(14.40\) square units.

- Use the graph to identify the upper function. - Form an upper-minus-lower difference before integrating. - Pay close attention to signs when subtracting the quadratic expression. - Integrate the difference over \([0,3]\).

1. The graph shows that \(f(x)>g(x)\) throughout \([0,3]\). 2. Form the difference: \(f(x)-g(x)=(2x+5)-(x^2-2x+1)=-x^2+4x+4\). 3. The area is \(\int_0^3(-x^2+4x+4)\,\text{d}x\). 4. An antiderivative is \(-\frac13x^3+2x^2+4x\). 5. Evaluate: \(\left[-\frac13x^3+2x^2+4x\right]_0^3=-9+18+12=21\).

The area is \(21\) square units.

- Use the power rule to find an antiderivative. - Use the endpoints \(-1\) and \(2\) as the limits of integration. - Apply the Fundamental Theorem of Calculus. - Be careful with signs when substituting the negative endpoint.

1. Since \(f(x)>0\), the area equals the definite integral \(\int_{-1}^{2}(x^2+2)\,dx\). 2. An antiderivative is \(F(x)=\frac{x^3}{3}+2x\). 3. Evaluate: \(F(2)=\frac{8}{3}+4=\frac{20}{3}\) and \(F(-1)=-\frac{1}{3}-2=-\frac{7}{3}\). 4. Therefore, \(A=F(2)-F(-1)=\frac{20}{3}+\frac{7}{3}=9\).

The area is \(9\) square units.

- Verify that the function is nonnegative on the interval. - Use the power rule to find an antiderivative. - Apply upper bound minus lower bound. - Keep \(a\) and \(b\) symbolic in part 2.

1. Since \(g(x)\ge 0\) for \(x\ge 0\), the area is the definite integral. An antiderivative is \(G(x)=x^3+x^2\). 2. On \([1, 2]\), \(A=G(2)-G(1)=12-2=10\). 3. On \([a, b]\), \(A=G(b)-G(a)=b^3+b^2-a^3-a^2\).

1) \(A=10\) square units 2) \(A=b^3+b^2-a^3-a^2\)

- Use the x-intercepts of the displayed parabola as the integration limits. - Check that the graph stays above the \(x\)-axis between the intercepts. - Find an antiderivative with the power rule. - Apply the Fundamental Theorem of Calculus.

1. Factor: \(-x^2+2x+3=-(x-3)(x+1)\), so the zeros are \(x=-1\) and \(x=3\). 2. The function is nonnegative between the zeros, so \(A=\int_{-1}^{3}(-x^2+2x+3)\,dx\). 3. An antiderivative is \(F(x)=-\frac{x^3}{3}+x^2+3x\). 4. Evaluate: \(F(3)=9\) and \(F(-1)=-\frac{5}{3}\). Thus, \(A=9-\left(-\frac{5}{3}\right)=\frac{32}{3}\).

The area is \(\frac{32}{3}\approx10.67\) square units.

- Factor out \(x\) to locate the two boundary points of the shaded region. - Notice that the graph touches, rather than crosses, the \(x\)-axis at the second zero. - Verify that the function is nonnegative between the zeros. - Integrate with the power rule.

1. Factor: \(f(x)=x(x-2)^2\). The relevant zeros are \(x=0\) and \(x=2\). 2. The function is nonnegative on \([0,2]\), so \(A=\int_0^2(x^3-4x^2+4x)\,dx\). 3. An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{4x^3}{3}+2x^2\). 4. Evaluate: \(A=F(2)-F(0)=4-\frac{32}{3}+8=\frac{4}{3}\).

The area is \(\frac{4}{3}\approx1.33\) square units.

- Use the graph and each formula to locate the two x-intercepts that bound the region. - Split the area integral where the function definition changes at \(x=1\). - Both pieces lie above the \(x\)-axis inside the bounded region. - Add the two definite integrals.

1. For \(x\le1\), solve \(4-x^2=0\). The relevant zero is \(x=-2\). 2. For \(x>1\), solve \(4.5-1.5x=0\). The zero is \(x=3\). 3. Split the integral at the definition change \(x=1\): \(A=\int_{-2}^{1}(4-x^2)\,dx+\int_1^3(4.5-1.5x)\,dx\). 4. The first integral is \(9\), and the second is \(3\). 5. Therefore, \(A=9+3=12\).

The area is \(12\) square units.

- Set the two formulas equal to find the horizontal bounds. - Use the graph to decide which function supplies the upper boundary. - Area between curves is represented by upper function minus lower function. - The problem asks only for the setup, so do not find an antiderivative.

1. Set the functions equal: \(x^2-4x+5=-x^2+8x-11\). This gives \(x^2-6x+8=(x-2)(x-4)=0\), so the intersections occur at \(x=2\) and \(x=4\). 2. From the graph, \(g\) is above \(f\) on \([2,4]\). 3. Therefore, an integral for the enclosed area is \(A=\int_2^4\left[g(x)-f(x)\right] \,dx=\int_2^4(-2x^2+12x-16)\,dx\).

a) \(x=2\) and \(x=4\) b) \(A=\int_2^4\left[g(x)-f(x)\right] \,dx=\int_2^4(-2x^2+12x-16)\,dx\)

- Factor the polynomial to find all x-intercepts. - Decide whether the graph crosses or only touches the x-axis at \(x=0\). - Use even symmetry to calculate the area on only half of the interval.

1. Factor \(f(x)=2x^2(4-x^2)\). The x-intercepts are \(-2\), \(0\), and \(2\). The graph touches the x-axis at \(x=0\) and is nonnegative on \([-2,2]\). 2. Because \(f\) is even, the total area is \(A=2\int_0^2(8x^2-2x^4)\,dx\). 3. Therefore, \(A=2\left[\frac{8x^3}{3}-\frac{2x^5}{5}\right]_0^2=\frac{256}{15}\) square units.

The total area is \(\frac{256}{15}\) square units.

- Use the x-intercept to separate the two triangular regions. - Area below the x-axis contributes negatively to the definite integral. - Total geometric area adds the positive magnitudes of both triangles.

1. The graph crosses the x-axis at \(x=3\). It lies below the axis on \([0,3]\) and above it on \([3,5]\). 2. The region below the axis is a right triangle with base \(3\) and height \(3\), so its area is \(\frac{1}{2}\cdot3\cdot3=4.5\). The region above the axis is a right triangle with base \(2\) and height \(2\), so its area is \(\frac{1}{2}\cdot2\cdot2=2\). 3. For a), the definite integral is the net signed area: \(\int_0^5 f(x)\,dx=2-4.5=-2.5\). 4. For b), total area adds the magnitudes: \(A=2+4.5=6.5\) square units.

a) \(\int_0^5 f(x)\,dx=-2.5\) b) The total area is \(6.5\) square units.

- Determine whether the graph is above or below the \(x\)-axis on the interval. - Treat \(e\) as a constant when finding an antiderivative. - Use a definite integral to calculate the area once the sign is known.

1. The equation \(e^x-e=0\) has the solution \(x=1\). For \(x\in[1,2]\), \(e^x\geq e\), so \(h(x)\geq 0\). Therefore, the area equals the definite integral. 2. An antiderivative is \(H(x)=e^x-ex\). 3. Thus, \(A=\int_{1}^{2}(e^x-e)\,dx=[e^x-ex]_{1}^{2}\) \(=(e^2-2e)-(e-e)=e^2-2e\) square units.

The area is \(e^2-2e\) square units.

- Integrate the two terms separately. - Pay attention to the sign when finding an antiderivative of \(e^{-x}\). - Keep the constant factor attached to the exponential term.

1. On \([0,2]\), both \(3e^{-x}\) and \(x\) are nonnegative, so \(f(x)>0\). The area equals the definite integral. 2. An antiderivative is \(F(x)=-3e^{-x}+\frac{1}{2}x^2\). 3. Therefore, \(A=F(2)-F(0)\) \(=\left(-3e^{-2}+2\right)-(-3)\) \(=5-3e^{-2}\approx 4.594\) square units.

The area is \(5-3e^{-2}\approx 4.594\) square units.

- Relate the area under a positive graph to a definite integral. - Use the logarithmic antiderivative for a reciprocal linear expression. - Simplify a difference of logarithms with a logarithm property.

1. The function is positive on \([0,3]\), so the area is \(A=\int_{0}^{3}\frac{2}{x+3}\,dx\). 2. An antiderivative is \(F(x)=2\ln|x+3|\). 3. Apply the fundamental theorem of calculus: \(A=2\ln(6)-2\ln(3)\). 4. Using \(\ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right)\), \(A=2\ln\left(\frac{6}{3}\right)=2\ln(2)\) square units.

\(2\ln(2)\) square units

- Use the range of the sine function to find the minimum value of \(f_k\). - Separate the integral into the sine term and the constant term. - Think about the signed area of one complete sine period. - Compare the graph with its horizontal midline.

1. Since \(-1\le\sin(2x)\le1\), \(k\le f_k(x)\le3k\). Because \(k>0\), the graph stays above the x-axis. 2. The area is \(A=\int_0^\pi\left(k\sin(2x)+2k\right)\,dx\). 3. The sine term completes one full period on \([0,\pi]\), so its signed integral is \(0\). Therefore, \(A=2k\cdot\pi=2\pi k\) square units. 4. Geometrically, the portions of the sine wave above and below the midline \(y=2k\) have equal signed areas. They cancel, leaving the area of the rectangle with width \(\pi\) and height \(2k\).

a) \(f_k(x)\ge k>0\), so the graph stays above the x-axis. b) \(A=2\pi k\) square units. The oscillating term has net area \(0\), leaving the \(2k\)-by-\(\pi\) rectangle.

- Use point symmetry about \(W(1,0)\) to compare the two shaded regions. - Keep the sign of each definite integral separate from the positive geometric area. - Use interval additivity over \([-1,3]\). - Add the magnitudes for total area.

1. Point symmetry about \(W(1,0)\) maps \(A_1\) on \([-1,1]\) onto the congruent region \(A_2\) on \([1,3]\). Therefore, \(A_2\) also has area \(2\) square units. 2. Since \(A_1\) lies above the x-axis, \(\int_{-1}^{1}f(x)\,dx=2\). 3. Since \(A_2\) lies below the x-axis, \(\int_{1}^{3}f(x)\,dx=-2\). 4. By interval additivity, \(\int_{-1}^{3}f(x)\,dx=2+(-2)=0\). 5. The total geometric area is \(2+2=4\) square units.

a) The area of \(A_2\) is \(2\) square units; \(\int_{-1}^{1}f(x)\,dx=2\), and \(\int_{1}^{3}f(x)\,dx=-2\). b) \(\int_{-1}^{3}f(x)\,dx=0\) c) The total area is \(4\) square units.

- Differentiate the logarithm using the chain rule. - Read the two vertical integration bounds from the shaded region. - Because the graph is above the x-axis, the definite integral equals the area. - Use a logarithm quotient to simplify the exact result.

1. Differentiate \(G\): \(G'(x)=4\frac{e^x}{e^x+2}=g(x)\). 2. The function is positive, and the vertical boundaries are \(x=0\) and \(x=2\). Therefore, \(A=\int_0^2g(x)\,dx\). 3. Using the given antiderivative, \(A=4\ln(e^2+2)-4\ln3=4\ln\left(\frac{e^2+2}{3}\right)\approx4.564\) square units.

a) \(G'(x)=g(x)\) b) \(4\ln\left(\frac{e^2+2}{3}\right)\approx4.564\) square units

- Use the shaded interval as the integration bounds. - Check that the sine graph stays above the x-axis. - Recall an antiderivative of \(\sin x\).

1. Since \(\sin x\ge0\) on \([0,\pi]\), the area is \(A=\int_0^\pi\sin x\,dx\). 2. An antiderivative is \(-\cos x\). 3. Therefore, \(A=[-\cos x]_0^\pi=2\) square units.

\(2\) square units

- Use \(x=0\) and \(x=2\) as the integration limits. - Treat \(a\) as a constant when integrating \(ax^3\). - Set the resulting area expression equal to \(10\).

1. The region lies on \([0,2]\), so \(A(a)=\int_0^2 ax^3\,\mathrm{d}x\). 2. Treating \(a\) as a constant gives \(A(a)=\left[\frac{a}{4}x^4\right]_0^2=4a\). 3. Set \(4a=10\). Therefore, \(a=2.5\).

\(a=2.5\)

- The graph stays above the x-axis on the stated interval. - Use \(2\sin x\) as an antiderivative of \(2\cos x\). - Evaluate the exact sine values at the two endpoints.

1. The graph is nonnegative on the interval, so \(A=\int_{-\pi/2}^{\pi/2}2\cos x\,\mathrm{d}x\). 2. An antiderivative is \(2\sin x\). 3. Therefore, \(A=\left[2\sin x\right]_{-\pi/2}^{\pi/2}=2-(-2)=4\).

\(4\) square units

- Find the two x-intercepts to determine the integration limits. - The function is negative between the intercepts. - Negate the signed integral so that the geometric area is positive.

1. The x-intercepts satisfy \(\frac{1}{2}x^2-2=0\), so they are \(x=-2\) and \(x=2\). 2. The graph lies below the x-axis between the intercepts. Therefore, \(A=-\int_{-2}^{2}\left(\frac{1}{2}x^2-2\right)\,\mathrm{d}x\). 3. An antiderivative is \(F(x)=\frac{1}{6}x^3-2x\). The signed integral is \(-\frac{16}{3}\), so \(A=\frac{16}{3}\).

\(\frac{16}{3}\) square units, or approximately \(5.33\) square units

- Compare \(h(x)\) with \(f(x-2)\). - Track how the interval endpoints change under the translation. - A horizontal translation preserves every vertical slice of the region.

1. Since \(h(x)=f(x-2)\), the graph of \(h\) is the graph of \(f\) translated \(2\) units to the right. 2. The interval \([-1,1]\) is translated to \([1,3]\). A horizontal translation does not change the widths or heights of the region. 3. Therefore, the area for \(h\) is also \(\frac{4}{3}\) square units.

\(\frac{4}{3}\) square units; the horizontal translation does not change area.

- Factor the function to find all three zeros. - Check that \(p(-x)=-p(x)\). - Find the area of one region and use the origin symmetry.

1. Factor \(p(x)=x(x-2)(x+2)\), so the zeros are \(x=-2\), \(x=0\), and \(x=2\). 2. Since \(p\) is odd, the two regions have equal areas. It is enough to find the magnitude of the signed area on \([0,2]\) and double it. 3. An antiderivative is \(P(x)=\frac{1}{4}x^4-2x^2\), and \(\int_0^2p(x)\,\mathrm{d}x=-4\). 4. Therefore, the total area is \(2\cdot4=8\) square units.

\(8\) square units

- Distinguish signed integral from geometric area. - Use the odd symmetry of \(x^3\). - On an interval crossing the x-axis, use absolute value or split the integral. - Form an equation for the unknown endpoint.

1. Choose an interval symmetric about the origin, \([-c, c]\). Since \(x^3\) is odd, the two geometric regions have equal area. 2. The total area is \(2\int_0^c x^3\,dx=2\left[\frac{x^4}{4}\right]_0^c=\frac{c^4}{2}\). 3. Set \(\frac{c^4}{2}=8\). Then \(c^4=16\), so \(c=2\). 4. One valid interval is \([-2, 2]\), and the area can be written as \(\int_{-2}^{2}|x^3|\,dx\).

One possible interval is \([-2, 2]\), with area \(\int_{-2}^{2}|x^3|\,dx=8\).

- First determine which graph is above the other on each interval. - Use the integral of the upper function minus the lower function. - Recall antiderivatives for rational and exponential functions. - Use logarithm properties to simplify the result in part a).

1. For a), \(f(0)=2>g(0)=1\), so \(f\) is above \(g\) throughout the interval. The area is \(\int_0^4\left(\frac{4}{x+2}-1+0.5x\right)\,dx\). An antiderivative is \(4\ln(x+2)-x+0.25x^2\). Therefore, \(A=\left[4\ln(x+2)-x+0.25x^2\right]_0^4=4\ln(6)-4\ln(2)=4\ln(3)\approx4.394\). 2. For b), \(f(0)=2>g(0)=1\), so \(f\) is above \(g\). The area is \(\int_0^1(e^x+1-e^{-x})\,dx\). An antiderivative is \(e^x+x+e^{-x}\). Thus, \(A=\left[e^x+x+e^{-x}\right]_0^1=e+e^{-1}-1\approx2.086\).

a) \(A=4\ln(3)\approx4.394\) b) \(A=e+e^{-1}-1\approx2.086\)

- Find the x-intercepts before setting the limits. - A substitution involving \(\sqrt{x}\) simplifies the equation for the zeros. - Rewrite the square root as a power before integrating. - The shaded graph is below the x-axis, so reverse the sign of the definite integral.

1. Find the zeros by substituting \(u=\sqrt{x}\). The equation \(x-4\sqrt{x}+3=0\) becomes \(u^2-4u+3=0\), so \(u=1\) or \(u=3\). Therefore, the zeros are \(x=1\) and \(x=9\). 2. An antiderivative is \(F(x)=\frac12x^2-\frac83x^{3/2}+3x\). 3. On \([1,9]\), the function is below the x-axis. The signed integral is \(\left[F(x)\right]_1^9=-\frac{16}{3}\). 4. The geometric area is the absolute value, so \(A=\frac{16}{3}\).

The area is \(\frac{16}{3}\approx5.33\) square units.

- Express the area as a definite integral. - Rewrite the square root as a fractional power. - Treat \(k\) as a constant while integrating. - Set the area formula equal to \(32\).

1. Because \(f_k(x)\ge 0\) on \([0, 4]\), \(A(k)=\int_0^4k\sqrt{x}\,dx\). 2. Use the power rule: \(A(k)=\left[\frac{2k}{3}x^{3/2}\right]_0^4=\frac{2k}{3}\cdot 8=\frac{16k}{3}\). 3. Set the area equal to \(32\): \(\frac{16k}{3}=32\), so \(k=6\).

1) \(A(k)=\frac{16k}{3}\) 2) \(k=6\)

- Express the area as a definite integral. - Account for the linear exponent when integrating. - Factor the common expression in the area equation. - Isolate the parameter.

1. Since \(g_a(x)>0\), \(A(a)=\int_0^2ae^{0.5x}\,dx\). 2. An antiderivative is \(2ae^{0.5x}\), so \(A(a)=[2ae^{0.5x}]_0^2=2ae-2a=2a(e-1)\). 3. Set \(2a(e-1)=4(e-1)\). Since \(e-1\ne 0\), divide to obtain \(a=2\).

1) \(A(a)=2a(e-1)\) 2) \(a=2\)

- Determine which function is above the other. - Express the area between the graphs as a definite integral. - Recall the area formula for a triangle. - Set the two area expressions equal and use \(k>0\).

1. Since \(f(x)-g(x)=\frac{1}{2}x^2+4>0\), \(f\) is above \(g\). 2. The area function is \(A(k)=\int_0^k\left(\frac{1}{2}x^2+4\right)\,dx=\left[\frac{x^3}{6}+4x\right]_0^k=\frac{k^3}{6}+4k\). 3. The triangle's area is \(\frac{1}{2}\cdot k\cdot10=5k\). 4. Set the areas equal: \(\frac{k^3}{6}+4k=5k\). Because \(k>0\), divide by \(k\) to obtain \(\frac{k^2}{6}=1\). Thus, \(k=\sqrt{6}\).

a) \(A(k)=\frac{k^3}{6}+4k\) b) \(k=\sqrt{6}\)

- Use symmetry about the \(y\)-axis to simplify the integral. - Integrate the difference between the upper and lower functions. - Write the rectangle's area in terms of \(t\). - Use the fact that \(t>0\) when solving the equation.

1. The difference is \(f(x)-g(x)=x^2+3>0\), so \(f\) is above \(g\). 2. Because the difference is even, \(A(t)=\int_{-t}^{t}(x^2+3)\,dx=2\int_0^t(x^2+3)\,dx\). Therefore, \(A(t)=2\left[\frac{x^3}{3}+3x\right]_0^t=\frac{2}{3}t^3+6t\). 3. The rectangle's area is \(8t\). Set the areas equal: \(\frac{2}{3}t^3+6t=8t\). 4. Since \(t>0\), divide by \(t\): \(\frac{2}{3}t^2+6=8\). Hence, \(t^2=3\), so \(t=\sqrt{3}\).

a) \(A(t)=\frac{2}{3}t^3+6t\) b) \(t=\sqrt{3}\)

- Write the equation of the tangent line at \(x=e\). - Find where the curve and the tangent line meet the x-axis. - View the shaded region as a larger triangle-like area minus the area under \(\ln(x)\). - Use integration by parts for \(\int\ln(x)\,\text{d}x\).

1. Since \(f(e)=1\) and \(f'(x)=\frac{1}{x}\), the tangent slope at \(x=e\) is \(\frac{1}{e}\). The tangent line is \(y-1=\frac{1}{e}(x-e)\), or \(y=\frac{x}{e}\). 2. The graph of \(f\) meets the x-axis at \(x=1\), and the tangent line meets it at \(x=0\). 3. Compute the area under the tangent line from \(0\) to \(e\), then subtract the area under \(f\) from \(1\) to \(e\): \(\int_0^e\frac{x}{e}\,\text{d}x-\int_1^e\ln(x)\,\text{d}x\). The first integral is \(\frac{e}{2}\). Using the antiderivative \(x\ln(x)-x\), the second integral is \(1\). 4. Therefore, \(A=\frac{e}{2}-1\approx0.359\).

The area is \(\frac{e}{2}-1\approx0.359\) square units.

- Find where the curve meets the \(x\)-axis and the line \(g\). - Use symmetry to calculate only one of the two regions. - Split one region into a rectangle and a curved part. - For the curved part, integrate \(0-f(x)\).

1. Substitute \(x=1\): \(f(1)=2-\frac{18}{1^2}=-16\). Therefore, \((1, -16)\) lies on both graphs. 2. For the right-hand region, the part from \(x=0\) to \(x=1\) is a rectangle between \(y=0\) and \(y=-16\), with area \(16\). 3. From \(x=1\) to \(x=3\), the curve lies below the \(x\)-axis. This part has area \(\int_1^3\left(0-f(x)\right)\,dx=\int_1^3\left(\frac{18}{x^2}-2\right)\,dx\). 4. An antiderivative is \(-\frac{18}{x}-2x\). Thus, \(\left[-\frac{18}{x}-2x\right]_1^3=(-12)-(-20)=8\). 5. The right-hand region has area \(16+8=24\). By symmetry, the combined area is \(2\cdot24=48\).

a) \(f(1)=-16\), so \(x=1\) is an intersection \(x\)-coordinate. b) The combined area is \(48\) square units.

- Find the x-intercepts to determine the bounds. - Check where the graph lies above the x-axis. - Express the area as a definite integral. - Set the area formula equal to \(36\).

1. Factor to find the x-intercepts: \(f_k(x)=x(k-x)\), so the intercepts are \(x=0\) and \(x=k\). The function is nonnegative between them. 2. Therefore, \(A(k)=\int_0^k(kx-x^2)\,dx\) \(=\left[\frac{kx^2}{2}-\frac{x^3}{3}\right]_0^k=\frac{k^3}{6}\). 3. Set \(\frac{k^3}{6}=36\). Then \(k^3=216\), so \(k=6\).

1) \(A(k)=\frac{k^3}{6}\) 2) \(k=6\)

- Find all x-intercepts. - Use the odd symmetry to compare the two regions. - Add geometric areas, not signed integrals. - Set the area formula equal to \(16\).

1. Factor: \(g_a(x)=ax(a^2-x^2)\), so the x-intercepts are \(-a\), \(0\), and \(a\). 2. The function is odd, so the two regions have equal area. On \([0, a]\), the function is nonnegative. Thus, \(A(a)=2\int_0^a(a^3x-ax^3)\,dx\). 3. Evaluate: \(A(a)=2\left[\frac{a^3x^2}{2}-\frac{ax^4}{4}\right]_0^a\) \(=2\left(\frac{a^5}{2}-\frac{a^5}{4}\right)=\frac{a^5}{2}\). 4. Set \(\frac{a^5}{2}=16\). Then \(a^5=32\), so \(a=2\).

1) \(A(a)=\frac{a^5}{2}\) 2) \(a=2\)

- Find all intersection points by setting the functions equal. - Three intersection points can create two bounded regions. - Check which graph is above the other on each interval. - Add the absolute values of the signed integrals.

1. Set the functions equal: \(x^3-2x=x^2\), so \(x(x^2-x-2)=x(x-2)(x+1)=0\). The intersection \(x\)-values are \(-1, 0,\) and \(2\). 2. Let \(h(x)=f(x)-g(x)=x^3-x^2-2x\). An antiderivative is \(H(x)=\frac{x^4}{4}-\frac{x^3}{3}-x^2\). 3. On \([-1, 0]\), \(f\) is above \(g\), and \(\int_{-1}^{0}h(x)\,dx=\frac{5}{12}\). 4. On \([0, 2]\), \(g\) is above \(f\), and \(\int_0^2h(x)\,dx=-\frac{8}{3}\), so this region has area \(\frac{8}{3}\). 5. The total area is \(\frac{5}{12}+\frac{8}{3}=\frac{37}{12}\).

The total area is \(\frac{37}{12}\approx3.083\) square units.

- Use the point-slope form of the tangent line at \(x=a\). - The \(y\)-axis makes \(x=0\) one integration boundary. - Determine which graph is above the other on \([0, a]\). - Integrate the difference using the power rule.

1. Since \(f'(x)=3x^2\), the tangent slope at \(x=a\) is \(3a^2\). The tangent line is \(t(x)=3a^2(x-a)+a^3=3a^2x-2a^3\). 2. On \([0, a]\), the cubic lies above its tangent line. Thus, \(A=\int_0^a\left[x^3-(3a^2x-2a^3)\right]\,dx\). 3. An antiderivative is \(\frac{x^4}{4}-\frac{3}{2}a^2x^2+2a^3x\). 4. Evaluating from \(0\) to \(a\) gives \(A=\frac{a^4}{4}-\frac{3}{2}a^4+2a^4=\frac{3}{4}a^4\).

\(A=\frac{3}{4}a^4\)

- Write the tangent line using the function value and derivative at \(x=1\). - Identify the interval bounded by the \(y\)-axis and the point of tangency. - Integrate the curve minus the tangent line. - Leave \(e\) in the exact answer.

1. Since \(f(1)=e\) and \(f'(1)=e\), the tangent line is \(t(x)=e(x-1)+e=ex\). 2. The region extends from the \(y\)-axis, \(x=0\), to the point of tangency, \(x=1\). 3. Because \(e^x\ge ex\) on this interval, \(A=\int_0^1(e^x-ex)\,dx\). 4. Evaluate: \(\left[e^x-\frac{e}{2}x^2\right]_0^1=\left(e-\frac{e}{2}\right)-1=\frac{e}{2}-1\).

The exact area is \(\frac{e}{2}-1\) square units.

- Use the known \(y\)-intercept to write \(g(x)=mx+b\). - Express the area between the line and the curve as a definite integral. - Use the given area to solve for the slope. - Use the stated graph order when setting up the area integral.

1. Since \(f(0)=18\), write the line as \(g(x)=mx+18\). 2. The area under \(f\) on \([0, 2]\) is \(\int_0^2\frac{18}{(x+1)^2}\,dx=\left[-\frac{18}{x+1}\right]_0^2=12\). 3. Since \(g\) lies above \(f\) on \([0,2]\), the enclosed area is \(\int_0^2(g(x)-f(x))\,dx=(2m+36)-12=2m+24\). 4. Set the area equal to \(10\): \(2m+24=10\), so \(m=-7\). 5. Therefore, \(g(x)=-7x+18\).

\(g(x)=-7x+18\)

- Find the intersections of \(y=x^2\) and \(y=k^2\). - Determine which graph is above the other. - Use symmetry about the \(y\)-axis. - Integrate with the power rule.

1. The intersections satisfy \(x^2=k^2\), so \(x=-k\) and \(x=k\). 2. The line is above the parabola between the intersections. Thus, \(A(k)=\int_{-k}^{k}(k^2-x^2)\,dx\). 3. Evaluate: \(\left[k^2x-\frac{x^3}{3}\right]_{-k}^{k}=\frac{4}{3}k^3\). 4. For an area of \(36\), solve \(\frac{4}{3}k^3=36\). Then \(k^3=27\), and because \(k>0\), \(k=3\).

a) \(A(k)=\frac{4}{3}k^3\) b) \(k=3\)

- Find where \(ae^{-x}=1\). - Use \(a>1\) to determine which graph is above the other. - Integrate the difference between the curve and \(y=1\). - Simplify using \(e^{-\ln(a)}=\frac{1}{a}\).

1. Find the intersection with \(y=1\): \(ae^{-x}=1\), so \(e^{-x}=\frac{1}{a}\) and \(x=\ln(a)\). 2. Since \(a>1\), the curve is above \(y=1\) on \([0, \ln(a)]\). 3. Therefore, \(A(a)=\int_0^{\ln(a)}(ae^{-x}-1)\,dx\). 4. An antiderivative is \(-ae^{-x}-x\). Evaluating gives \(A(a)=\left[-ae^{-x}-x\right]_0^{\ln(a)}=-1-\ln(a)+a\). 5. Thus, \(A(a)=a-\ln(a)-1\).

\(A(a)=a-\ln(a)-1\)

- Use the power rule for antiderivatives. - Apply \(F(1)=0\) to determine the constant. - Represent the area with a definite integral from \(1\) to \(k\). - Solve the resulting equation for \(k>1\).

1. Integrate \(f(x)=6x^{-4}\): \(F(x)=-2x^{-3}+C=-\frac{2}{x^3}+C\). 2. Use \(F(1)=0\): \(-2+C=0\), so \(C=2\). Thus, \(F(x)=2-\frac{2}{x^3}\). 3. The area from \(x=1\) to \(x=k\) is \(\int_1^k\frac{6}{x^4}\,dx=\left[-\frac{2}{x^3}\right]_1^k=2-\frac{2}{k^3}\). 4. Set this equal to \(1.75=\frac{7}{4}\): \(2-\frac{2}{k^3}=\frac{7}{4}\). Then \(\frac{2}{k^3}=\frac{1}{4}\), so \(k^3=8\) and \(k=2\).

a) \(F(x)=2-\frac{2}{x^3}\) b) \(k=2\)

- Rewrite cube roots using rational exponents. - Differentiate \(G\) to verify the antiderivative. - Use the Fundamental Theorem of Calculus for the area. - Use \(a>1\) when solving the power equation.

1. Rewrite \(G(x)=2x^{2/3}\). Then \(G'(x)=2\left(\frac{2}{3}\right)x^{-1/3}=\frac{4}{3\sqrt[3]{x}}=g(x)\). Therefore, \(G\) is an antiderivative of \(g\). 2. By the Fundamental Theorem of Calculus, \(\int_1^a g(x)\,dx=G(a)-G(1)=2a^{2/3}-2\). 3. Set the area equal to \(6\): \(2a^{2/3}-2=6\), so \(a^{2/3}=4\). 4. Raise both sides to the \(\frac{3}{2}\) power. Since \(a>1\), \(a=4^{3/2}=8\).

a) \(G'(x)=g(x)\), so \(G\) is an antiderivative of \(g\). b) \(a=8\)

- Find the zeros that determine the integration interval. - Rewrite the square root as a power before integrating. - Treat \(k\) as a constant during integration. - Set the area expression equal to \(13.5\).

1. Solve \(k\sqrt{x}-x=0\): \(\sqrt{x}(k-\sqrt{x})=0\), so the zeros are \(x=0\) and \(x=k^2\). 2. The function is nonnegative on \([0, k^2]\), so \(A=\int_0^{k^2}(k\sqrt{x}-x)\,dx\). 3. Evaluate: \(\left[\frac{2}{3}kx^{3/2}-\frac{x^2}{2}\right]_0^{k^2} =\frac{2}{3}k^4-\frac{1}{2}k^4=\frac{1}{6}k^4\). 4. Set \(\frac{1}{6}k^4=13.5\). Then \(k^4=81\), and because \(k>0\), \(k=3\).

\(k=3\)

- Factor the function to find its zeros. - Check whether the function has symmetry. - Account for the region below the \(x\)-axis as positive area. - Set the total area expression equal to \(8\).

1. Factor: \(g_a(x)=x(a^2-x^2)\), so the zeros are \(x=-a, 0,\) and \(a\). 2. The function is odd, so the two bounded regions have equal area. On \([0, a]\), \(g_a(x)\ge0\). 3. The combined area is \(A=2\int_0^a(a^2x-x^3)\,dx\). 4. Evaluate: \(A=2\left[\frac{1}{2}a^2x^2-\frac{1}{4}x^4\right]_0^a =2\left(\frac{1}{4}a^4\right)=\frac{1}{2}a^4\). 5. Set \(\frac{1}{2}a^4=8\). Then \(a^4=16\), and because \(a>0\), \(a=2\).

\(a=2\)

- The average rate of change is the slope of the secant line. - Use the two endpoint values to write the secant line. - Use concavity to decide which graph is above the other. - Integrate the vertical difference between the two graphs.

1. The endpoint values are \(f(0)=\frac{1}{2}\) and \(f(2)=\frac{1}{2}e^2-2\). 2. The average rate of change is \(m=\frac{f(2)-f(0)}{2-0}=\frac{\frac{1}{2}e^2-\frac{5}{2}}{2}=\frac{1}{4}e^2-\frac{5}{4}\). 3. The secant line is \(s(x)=\left(\frac{1}{4}e^2-\frac{5}{4}\right)x+\frac{1}{2}\). 4. Since \(f''(x)=\frac{1}{2}e^x>0\), \(f\) is concave up, so its secant line lies above the graph on \([0, 2]\). 5. The area is \(A=\int_0^2\left[s(x)-f(x)\right] \,dx\). Simplifying the integrand gives \(\left(\frac{1}{4}e^2-\frac{1}{4}\right)x+\frac{1}{2}-\frac{1}{2}e^x\). 6. Evaluating the integral gives \(A=1\).

The average rate of change is \(\frac{1}{4}e^2-\frac{5}{4}\). The secant line is \(s(x)=\left(\frac{1}{4}e^2-\frac{5}{4}\right)x+\frac{1}{2}\). The area is \(1\) square unit.

- A definite integral uses signed area, while total area counts every region positively. - Find the zeros of \(f\) that lie in the displayed interval. - Split the interval wherever the graph crosses the \(x\)-axis. - Reverse the sign of any negative integral contribution when finding total area.

1. An antiderivative is \(F(x)=\frac{1}{4}x^4-2x^2\). Therefore, \(\int_{-1}^{3}(x^3-4x)\,dx=F(3)-F(-1)=\frac{9}{4}-\left(-\frac{7}{4}\right)=4\). 2. The zeros are \(x=-2\), \(x=0\), and \(x=2\). The zeros in \([-1,3]\) are \(0\) and \(2\), so split the interval there. On \([-1,0]\), \(\int_{-1}^{0}f(x)\,dx=\frac{7}{4}\). On \([0,2]\), \(\int_{0}^{2}f(x)\,dx=-4\), so this region has area \(4\). On \([2,3]\), \(\int_{2}^{3}f(x)\,dx=\frac{25}{4}\). Thus, the total area is \(\frac{7}{4}+4+\frac{25}{4}=12\) square units.

1) \(\int_{-1}^{3}(x^3-4x)\,dx=4\) 2) The total area is \(12\) square units.

- Use the factored form to find every zero in the displayed interval. - The graph changes sign at each of the three zeros. - Evaluate the definite integral on each resulting subinterval. - Add the absolute values of the four signed integrals.

1. Expand: \(f(x)=x^3-x^2-4x+4\). An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{x^3}{3}-2x^2+4x\). 2. From the factored form, the zeros are \(x=-2\), \(x=1\), and \(x=2\). Split the interval at these points. 3. The signed integrals are \(\int_{-3}^{-2}f(x)\,dx=-\frac{103}{12}\), \(\int_{-2}^{1}f(x)\,dx=\frac{45}{4}\), \(\int_{1}^{2}f(x)\,dx=-\frac{7}{12}\), and \(\int_{2}^{3}f(x)\,dx=\frac{47}{12}\). 4. Add their absolute values: \(A=\frac{103}{12}+\frac{135}{12}+\frac{7}{12}+\frac{47}{12}=\frac{292}{12}=\frac{73}{3}\).

The total area is \(\frac{73}{3}\approx24.33\) square units.

- Find which zeros of \(f\) lie in the displayed interval. - Split the interval at the zero inside \([-1,3]\). - Use the graph to determine where \(f\) is below and above the \(x\)-axis. - Add positive geometric areas rather than signed integrals.

1. Solve \(\frac{1}{2}x^2-2=0\). The zeros are \(x=-2\) and \(x=2\), but only \(x=2\) lies inside \([-1,3]\). 2. Split the interval at \(x=2\). An antiderivative is \(F(x)=\frac{x^3}{6}-2x\). 3. On \([-1,2]\), the function is below the axis: \(\int_{-1}^{2}f(x)\,dx=-\frac{9}{2}\). 4. On \([2,3]\), the function is above the axis: \(\int_{2}^{3}f(x)\,dx=\frac{7}{6}\). 5. The total area is \(\frac{9}{2}+\frac{7}{6}=\frac{27}{6}+\frac{7}{6}=\frac{17}{3}\).

The total area is \(\frac{17}{3}\approx5.67\) square units.

- Use the factored form to locate every sign change in the displayed interval. - A single definite integral gives net signed area, not necessarily total area. - Evaluate the integral separately on the three regions. - Add the magnitudes of the three contributions.

1. Factor: \(f(x)=x(x-1)(x-2)\), so the graph changes sign at \(x=1\) and \(x=2\) inside the interval. 2. The student found the signed integral. The positive and negative contributions partly cancel, so that value is not the total geometric area. 3. An antiderivative is \(F(x)=\frac{x^4}{4}-x^3+x^2\). The signed integrals on the three regions are \(\int_0^1f(x)\,dx=\frac{1}{4}\), \(\int_1^2f(x)\,dx=-\frac{1}{4}\), and \(\int_2^{2.5}f(x)\,dx=\frac{25}{64}\). 4. Add their absolute values: \(A=\frac{1}{4}+\frac{1}{4}+\frac{25}{64}=\frac{57}{64}=0.890625\).

The student calculated signed area, so the first two regions canceled. The total area is \(\frac{57}{64}=0.890625\) square unit.

- Find where the two displayed graphs intersect in the interval. - Determine which graph is above the other on each side of the interior intersection. - Split the area integral where the upper graph changes. - Integrate upper function minus lower function on each subinterval.

1. Set the functions equal: \(x^2-1=1-x\), so \(x^2+x-2=(x+2)(x-1)=0\). The intersections occur at \(x=-2\) and \(x=1\). 2. On \([-2,1]\), \(g\) is above \(f\). On \([1,2]\), \(f\) is above \(g\). 3. Therefore, \(A=\int_{-2}^{1}(g(x)-f(x))\,dx+\int_1^2(f(x)-g(x))\,dx\). 4. The first integral is \(\frac{9}{2}\), and the second is \(\frac{11}{6}\). 5. Thus, \(A=\frac{9}{2}+\frac{11}{6}=\frac{19}{3}\).

The total area is \(\frac{19}{3}\approx6.33\) square units.

- Compare integrating a difference with integrating its absolute value. - Find where sine and cosine intersect on \([0,\pi]\). - Use the graph to identify the upper function on each side of the intersection. - Split the absolute-value integral where \(f(x)-g(x)\) changes sign.

1. \(I_1=\int_{0}^{\pi}(\sin(x)-\cos(x))\,dx=[-\cos(x)-\sin(x)]_{0}^{\pi}=2\). 2. The graphs intersect where \(\sin(x)=\cos(x)\). On \([0,\pi]\), this occurs at \(x=\frac{\pi}{4}\). 3. On \(\left[0,\frac{\pi}{4}\right]\), \(g(x)>f(x)\), and on \(\left[\frac{\pi}{4},\pi\right]\), \(f(x)>g(x)\). Therefore, \(I_2=\int_{0}^{\frac{\pi}{4}}(\cos(x)-\sin(x))\,dx+\int_{\frac{\pi}{4}}^{\pi}(\sin(x)-\cos(x))\,dx\). 4. Evaluating gives \(I_2=[\sin(x)+\cos(x)]_{0}^{\frac{\pi}{4}}+[-\cos(x)-\sin(x)]_{\frac{\pi}{4}}^{\pi}\) \(=(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}\approx2.83\). 5. The first integral is the signed difference between the two regions. The second integral adds both regions as positive quantities, so it is the total area between the graphs.

1) \(I_1=2\) 2) \(I_2=2\sqrt{2}\approx2.83\) 3) \(I_1\) is the net signed difference, while \(I_2\) is the total geometric area between the graphs.

- Factor the function to find its zeros. - The order of the zero locations depends on the sign of \(k\). - Use an absolute value for geometric area. - A fourth-power equation can have two real solutions.

1. Factor: \(f_k(x)=x^2(k-x)\). If \(k\ne0\), the zeros are \(x=0\), with multiplicity \(2\), and \(x=k\). If \(k=0\), then \(f_0(x)=-x^3\), so \(x=0\) is a triple zero. 2. For \(k\ne0\), the geometric area between the two distinct zeros is \(A=\left|\int_0^k(kx^2-x^3)\,dx\right|\). 3. Evaluate: \(\int_0^k(kx^2-x^3)\,dx =\left[\frac{kx^3}{3}-\frac{x^4}{4}\right]_0^k =\frac{k^4}{12}\). Because \(k^4\ge0\), \(A=\frac{k^4}{12}\). 4. Set \(\frac{k^4}{12}=6.75=\frac{27}{4}\). Then \(k^4=81\), so \(k=3\) or \(k=-3\).

1. If \(k\ne0\), \(x=0\) is a double zero and \(x=k\) is the other zero. If \(k=0\), \(x=0\) is a triple zero. 2. \(k=3\) or \(k=-3\)

- Determine when \(x^2-k=0\) has two distinct real solutions. - Track the vertex \((0, -k)\) as \(k\) changes. - Use symmetry to simplify the area integral. - Solve the resulting equation involving \(k^{3/2}\).

1. The graph is an upward-opening parabola with vertex \((0, -k)\). It has two zeros \(x=\pm\sqrt{k}\) only when \(k>0\), so a bounded region is enclosed only for \(k>0\). For \(k=0\), the graph only touches the axis. For \(k<0\), the graph lies above the axis. 2. For \(k>0\), the area is \(A=\int_{-\sqrt{k}}^{\sqrt{k}}(k-x^2)\,dx\). 3. By symmetry, \(A=2\int_0^{\sqrt{k}}(k-x^2)\,dx =2\left[kx-\frac{x^3}{3}\right]_0^{\sqrt{k}} =\frac{4}{3}k^{3/2}\). 4. Set \(\frac{4}{3}k^{3/2}=\frac{32}{3}\). Then \(k^{3/2}=8\), so \(k=4\).

1. A bounded region is enclosed only when \(k>0\). 2. \(k=4\)

- Use the displayed endpoints and the formula change at \(x=0\) to organize the area. - Integrate the square-root piece with the power rule. - Account for the factor \(0.5\) when integrating the cosine piece. - Add the two positive area contributions.

1. The square-root piece is zero at \(x=-4\). The cosine piece is zero when \(0.5x=\frac{\pi}{2}\), so \(x=\pi\). 2. Split the integral at \(x=0\): \(A=\int_{-4}^{0}\sqrt{x+4}\,dx+\int_0^\pi2\cos(0.5x)\,dx\). 3. The first integral is \(\left[\frac{2}{3}(x+4)^{3/2}\right]_{-4}^{0}=\frac{16}{3}\). 4. The second integral is \(\left[4\sin(0.5x)\right]_0^\pi=4\). 5. Therefore, \(A=\frac{16}{3}+4=\frac{28}{3}\).

The area is \(\frac{28}{3}\approx9.33\) square units.

- Find the two intersection values shown at the ends of the shaded region. - Determine which parabola is above the other between the intersections. - Subtract the lower function from the upper function. - Integrate the difference over the bounded interval.

1. Set the functions equal: \(-x^2+6x-5=x^2-4x+3\). This simplifies to \(x^2-5x+4=0\), so the intersections occur at \(x=1\) and \(x=4\). 2. On \([1,4]\), \(f\) is above \(g\). Their difference is \(f(x)-g(x)=-2x^2+10x-8\). 3. The area is \(\int_1^4(-2x^2+10x-8)\,dx\). 4. An antiderivative is \(-\frac{2}{3}x^3+5x^2-8x\). 5. Evaluating gives \(\left[-\frac{2}{3}x^3+5x^2-8x\right]_1^4=9\).

The area is \(9\) square units.

- Find all three intersections visible in the displayed interval. - Three intersections create two separate bounded regions. - The upper graph changes at the middle intersection. - Add the absolute values of the two signed integrals.

1. Set the functions equal: \(x^3-4x^2+3x=x-x^2\). This simplifies to \(x^3-3x^2+2x=x(x-1)(x-2)=0\). The intersection \(x\)-values are \(0\), \(1\), and \(2\). 2. The difference is \(d(x)=f(x)-g(x)=x^3-3x^2+2x\), with antiderivative \(D(x)=\frac{x^4}{4}-x^3+x^2\). 3. On \([0,1]\), \(\int_0^1d(x)\,dx=\frac{1}{4}\). 4. On \([1,2]\), \(\int_1^2d(x)\,dx=-\frac{1}{4}\). 5. Add the absolute values: \(A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\).

The combined area is \(\frac{1}{2}\) square unit.

- Find the intersection points in terms of \(k\). - Determine which parabola is above the other. - Use symmetry to simplify the integral. - Solve the resulting equation for \(k>0\).

1. Set the functions equal: \(x^2-k=k-x^2\), so \(x^2=k\). The intersections occur at \(x=\pm\sqrt{k}\). 2. On the bounded interval, \(g\) is above \(f\), and \(g(x)-f(x)=2k-2x^2\). 3. The area is \(A=\int_{-\sqrt{k}}^{\sqrt{k}}(2k-2x^2)\,dx\). 4. By symmetry, \(A=2\int_0^{\sqrt{k}}(2k-2x^2)\,dx=\frac{8}{3}k^{3/2}\). 5. Set \(\frac{8}{3}k^{3/2}=72\). Then \(k^{3/2}=27\), so \(k=9\).

\(k=9\)

- Factor the polynomial to find the three x-intercepts. - Evaluate the signed integral between each pair of consecutive intercepts. - Convert a negative signed integral into positive geometric area. - Compare the two magnitudes before adding them.

1. Factor by grouping: \(f(x)=(x^2-1)(x-3)=(x+1)(x-1)(x-3)\). The zeros are \(-1\), \(1\), and \(3\). 2. An antiderivative is \(F(x)=\frac{x^4}{4}-x^3-\frac{x^2}{2}+3x\). 3. On \([-1,1]\), \(\int_{-1}^{1}f(x)\,dx=4\), so the first region has area \(4\). 4. On \([1,3]\), \(\int_1^3f(x)\,dx=-4\), so the second region also has area \(4\). 5. Therefore, the regions have equal areas and their combined area is \(4+4=8\).

Each region has area \(4\) square units, so the combined area is \(8\) square units.

- Find all zeros and identify the repeated zero visible at the center. - Use symmetry about the \(y\)-axis to halve the calculation. - The graph is below the \(x\)-axis on both bounded subintervals. - Use the magnitude of the signed integral.

1. Factor: \(f(x)=\frac{1}{2}x^2(x^2-4)\). The zeros are \(x=-2\), \(x=0\) with multiplicity \(2\), and \(x=2\). 2. The function is even and lies below the \(x\)-axis on \((-2,2)\), except at \(x=0\). 3. Therefore, \(A=2\left|\int_0^2\left(\frac{1}{2}x^4-2x^2\right)\,dx\right|\). 4. An antiderivative is \(\frac{x^5}{10}-\frac{2x^3}{3}\). The integral from \(0\) to \(2\) is \(-\frac{32}{15}\). 5. Thus, \(A=2\left(\frac{32}{15}\right)=\frac{64}{15}\).

The total area is \(\frac{64}{15}\approx4.27\) square units.

- Differentiate \(f\) to find the tangent slope at \(x=-2\). - The point of tangency is a repeated solution of \(f(x)=t(x)\). - Find the other intersection visible at the right end of the shaded region. - Integrate the tangent line minus the curve.

1. Since \(f'(x)=\frac{3}{2}x^2-2\), the tangent slope at \(x=-2\) is \(4\). The tangent line is \(t(x)=4(x+2)=4x+8\). 2. Solve \(f(x)=t(x)\): \(\frac{1}{2}x^3-2x=4x+8\), or \(x^3-12x-16=0\). The point of tangency \(x=-2\) is a double root, and the other intersection is \(x=4\). 3. On \([-2,4]\), the tangent line is above the curve. Thus, \(A=\int_{-2}^{4}\left[(4x+8)-\left(\frac{1}{2}x^3-2x\right)\right] \,dx\). 4. Evaluate: \(\left[-\frac{x^4}{8}+3x^2+8x\right]_{-2}^{4}=48-(-6)=54\).

The area is \(54\) square units.

- Differentiate \(f\) to verify that the tangent at \(P\) is horizontal. - The tangency value \(x=4\) is a repeated solution of \(f(x)=8\). - Find the other intersection at the left end of the shaded region. - Integrate the horizontal line minus the curve.

1. Since \(f'(x)=-\frac{3}{4}x^2+3x\), \(f'(4)=0\). The tangent line is therefore \(t(x)=8\). 2. Solve \(f(x)=8\): \(-\frac{1}{4}x^3+\frac{3}{2}x^2=8\). The point of tangency \(x=4\) is a double root, and the other intersection is \(x=-2\). 3. On \([-2,4]\), the tangent line is above the curve. Thus, \(A=\int_{-2}^{4}\left(8+\frac{1}{4}x^3-\frac{3}{2}x^2\right)\,dx\). 4. Evaluate: \(\left[8x+\frac{x^4}{16}-\frac{x^3}{2}\right]_{-2}^{4}=16-(-11)=27\).

The area is \(27\) square units.

- Differentiate \(f\) to write the tangent line at \(x=4\). - Locate the x-intercept of the parabola and the x-intercept of the tangent line. - The lower boundary changes at \(x=3\), as shown in the graph. - Subtracting the triangle under the tangent line from the area under the parabola gives a shorter calculation.

1. Since \(f(4)=4\) and \(f'(x)=2x-4\), the tangent slope at \(x=4\) is \(4\). The tangent line is \(t(x)=4x-12\). 2. The parabola touches the \(x\)-axis at \(x=2\), and the tangent line crosses the \(x\)-axis at \(x=3\). 3. From \(x=2\) to \(x=3\), the region lies between the parabola and the \(x\)-axis. From \(x=3\) to \(x=4\), it lies between the parabola and the tangent line. 4. Equivalently, subtract the triangle under the tangent line from the area under the parabola: \(A=\int_2^4(x^2-4x+4)\,dx-\frac{1}{2}\cdot1\cdot4\). 5. The integral is \(\frac{8}{3}\), so \(A=\frac{8}{3}-2=\frac{2}{3}\).

The area is \(\frac{2}{3}\) square unit.

- Differentiate \(f\) to find the tangent line at \(x=2\). - Locate the relevant x-intercepts of the tangent line and the parabola. - The lower boundary changes at \(x=1\), as shown in the graph. - Subtract the area under the parabola from the triangle under the tangent line.

1. Since \(f(2)=3\) and \(f'(x)=-2x+6\), the tangent slope is \(f'(2)=2\). Thus, the tangent line is \(t(x)=2x-1\). 2. The tangent line crosses the \(x\)-axis at \(x=\frac{1}{2}\). The parabola has zeros at \(x=1\) and \(x=5\); the relevant zero is \(x=1\). 3. The region lies below the tangent line. From \(x=\frac{1}{2}\) to \(x=1\), the lower boundary is the \(x\)-axis. From \(x=1\) to \(x=2\), it is the parabola. 4. The triangle under the tangent line from \(x=\frac{1}{2}\) to \(x=2\) has area \(\frac{1}{2}\left(\frac{3}{2}\right)(3)=\frac{9}{4}\). 5. The area under the parabola from \(1\) to \(2\) is \(\int_1^2(-x^2+6x-5)\,dx=\frac{5}{3}\). 6. Therefore, \(A=\frac{9}{4}-\frac{5}{3}=\frac{7}{12}\).

The area is \(\frac{7}{12}\) square unit.

- Factor \(h\) to determine its sign on \([0,3]\). - Since the graph is below the axis, integrate \(-h(x)\) for geometric area. - Pull the positive constant \(k\) outside the definite integral. - Set the scaled area equal to \(108\).

1. Factor: \(h(x)=-x(x-3)^2\), so \(h(x)\le0\) on \([0,3]\). 2. The area for \(h\) is \(A_1=\int_0^3[-h(x)]\,dx=\int_0^3(x^3-6x^2+9x)\,dx=\frac{27}{4}\). 3. Multiplying every function value by \(k>0\) multiplies every vertical slice of the region by \(k\). Equivalently, \(A_k=\int_0^3[-kh(x)]\,dx=k\int_0^3[-h(x)]\,dx=kA_1\). 4. Therefore, \(A_k=\frac{27}{4}k\). Set \(\frac{27}{4}k=108\), which gives \(k=16\).

a) \(\frac{27}{4}\) square units b) Vertical scaling by \(k\) multiplies the area by \(k\). c) \(k=16\)

- Differentiate \(f\) to find the tangent slope at the origin. - Factor \(f\) to locate the second intersection with the tangent line. - The graph lies below its horizontal tangent on the shaded interval. - Integrate the tangent line minus the curve.

1. Since \(f(0)=0\) and \(f'(0)=0\), the tangent line at the origin is \(y=0\). 2. Solve \(f(x)=0\): \(x^4-4x^3=x^3(x-4)=0\). The intersections are \(x=0\) and \(x=4\). 3. The graph lies below the \(x\)-axis on \((0,4)\). Therefore, \(A=\int_0^4(0-f(x))\,dx=\int_0^4(-x^4+4x^3)\,dx\). 4. An antiderivative is \(-\frac{x^5}{5}+x^4\). 5. Evaluate: \(A=\left[-\frac{x^5}{5}+x^4\right]_0^4=\frac{256}{5}=51.2\).

The area is \(\frac{256}{5}=51.2\) square units.

- Find the intersections in terms of \(c\). - Use symmetry about the \(y\)-axis. - Integrate the line minus the parabola. - Solve the resulting equation involving \(c^{3/2}\).

1. The intersections satisfy \(\frac{1}{4}x^2=c\), so \(x=\pm2\sqrt{c}\). 2. By symmetry, the area is \(A=2\int_0^{2\sqrt{c}}\left(c-\frac{1}{4}x^2\right)\,dx\). 3. Evaluate: \(A=2\left[cx-\frac{x^3}{12}\right]_0^{2\sqrt{c}} =\frac{8}{3}c^{3/2}\). 4. Set \(\frac{8}{3}c^{3/2}=\frac{64}{3}\). Then \(c^{3/2}=8\), so \(c=4\).

\(c=4\)

- Find the intersections in terms of \(k\). - Use symmetry about the \(y\)-axis. - Integrate the line minus the parabola. - Solve the area equation for \(k>0\).

1. The intersections satisfy \(kx^2=4\), so \(x=\pm\frac{2}{\sqrt{k}}\). 2. The line is above the parabola between the intersections. By symmetry, \(A=2\int_0^{2/\sqrt{k}}(4-kx^2)\,dx\). 3. Evaluate: \(A=2\left[4x-\frac{kx^3}{3}\right]_0^{2/\sqrt{k}} =\frac{32}{3\sqrt{k}}\). 4. Set \(\frac{32}{3\sqrt{k}}=16\). Then \(3\sqrt{k}=2\), so \(k=\frac{4}{9}\).

\(k=\frac{4}{9}\)

- Find the full shaded area first. - A \(1{:}3\) split places one fourth of the total area on the left. - Represent the left area with an integral whose upper limit is unknown. - Solve the resulting exponential equation.

1. The total area is \(\int_0^{\ln(9)}e^x\,dx=\left[e^x\right]_0^{\ln(9)}=9-1=8\). 2. A \(1{:}3\) ratio means the left part is \(\frac{1}{4}\) of the total, so its area is \(2\). 3. Let the dividing line be \(x=k\). Then \(\int_0^k e^x\,dx=2\). 4. Thus, \(e^k-1=2\), so \(e^k=3\) and \(k=\ln(3)\). 5. The dividing line is \(x=\ln(3)\).

\(x=\ln(3)\)

- Compare the graph heights on the left and right of the proposed line. - Find the total shaded area over \([1,4]\). - A \(2{:}1\) split places two thirds of the total area on the left. - Use an integral with upper limit \(k\) and solve the resulting rational equation.

1. Equal or proportional widths do not produce equal or proportional areas when the function height changes. Since \(f\) decreases, the left portion has greater average height than the right portion. 2. The total area is \(\int_1^4x^{-2}\,dx=\left[-\frac{1}{x}\right]_1^4=\frac{3}{4}\). 3. A \(2{:}1\) ratio means the left part is \(\frac{2}{3}\) of the total, so its area must be \(\frac{1}{2}\). 4. Therefore, \(\int_1^k x^{-2}\,dx=\frac{1}{2}\). Evaluating gives \(1-\frac{1}{k}=\frac{1}{2}\), so \(k=2\).

The width argument ignores the changing height of the graph. The required dividing line is \(x=2\).

- A line through the origin has equation \(y=mx\). - Find the intersections in terms of \(m\). - Use an absolute value because the order of the intersections depends on \(m\). - Solve the resulting absolute-value equation.

1. Set the functions equal: \(x^2-4x=mx\), so \(x[x-(m+4)]=0\). The intersections occur at \(x=0\) and \(x=m+4\). 2. The signed integral of \(g-f\) between the intersections is \(\int_0^{m+4}\left[(m+4)x-x^2\right]\,dx\). 3. Evaluating gives \(\frac{(m+4)^3}{6}\). Therefore, the geometric area is \(\left|\frac{(m+4)^3}{6}\right|\). 4. Set this equal to \(36\): \(|m+4|^3=216\), so \(|m+4|=6\). 5. Thus, \(m=2\) or \(m=-10\).

\(m=2\) or \(m=-10\)

- Write the line as \(y=mx\). - Find the second intersection in terms of \(m\). - Integrate the curve minus the line. - Apply exponent rules carefully at the upper limit.

1. Solve \(\sqrt{x}=mx\). The intersections occur at \(x=0\) and \(x=\frac{1}{m^2}\). 2. On this interval, \(\sqrt{x}\) is above \(mx\). Thus, \(A=\int_0^{1/m^2}(\sqrt{x}-mx)\,dx\). 3. Evaluate: \(A=\left[\frac{2}{3}x^{3/2}-\frac{m}{2}x^2\right]_0^{1/m^2} =\frac{2}{3m^3}-\frac{1}{2m^3} =\frac{1}{6m^3}\). 4. Set \(\frac{1}{6m^3}=\frac{4}{3}\). Then \(m^3=\frac{1}{8}\), so \(m=\frac{1}{2}\).

\(m=\frac{1}{2}\)

- Use the form \(ax^2+c\) for a parabola symmetric about the \(y\)-axis. - Symmetry gives the second zero. - Express the enclosed area as a definite integral. - Use the area condition to determine the coefficient.

1. Because the axis of symmetry is the \(y\)-axis, write \(f(x)=ax^2+c\). 2. Since \(f(3)=0\), \(9a+c=0\), so \(c=-9a\). By symmetry, the other zero is \(x=-3\). 3. The enclosed area is \(\int_{-3}^{3}(ax^2-9a)\,dx=-36a\). 4. Set \(-36a=24\). Then \(a=-\frac{2}{3}\). 5. Therefore, \(c=-9a=6\), and \(f(x)=-\frac{2}{3}x^2+6\).

\(f(x)=-\frac{2}{3}x^2+6\)

- Use the two visible factors to identify the integration interval. - Express geometric area using the absolute value of the signed integral. - The area condition determines \(|a|\), not its sign. - Use the second derivative at \(x=2\) to select the correct sign.

1. The distinct zeros are \(x=0\) and \(x=3\). 2. The signed integral between the zeros is \(\int_0^3 ax^2(x-3)\,dx=-\frac{27}{4}a\). Therefore, the geometric area is \(\frac{27}{4}|a|\). 3. Set \(\frac{27}{4}|a|=6.75=\frac{27}{4}\). Thus, \(|a|=1\). 4. The second derivative is \(f_a''(x)=6a(x-1)\). For \(x=2\) to be a local minimum, \(f_a''(2)=6a>0\). Therefore, \(a=1\).

\(a=1\)

- An origin-symmetric cubic contains only odd powers. - Use the zero at \(x=3\) to relate the coefficients. - Integrate over the first-quadrant interval. - Use the required sign of the function to choose the coefficient's sign.

1. An origin-symmetric cubic has the form \(f(x)=ax^3+bx\). 2. Since \(f(3)=0\), \(27a+3b=0\), so \(b=-9a\). Thus, \(f(x)=ax^3-9ax\). 3. The first-quadrant region extends from \(x=0\) to \(x=3\). Its signed integral is \(\int_0^3(ax^3-9ax)\,dx=-\frac{81}{4}a\). 4. For the graph to be above the \(x\)-axis on \((0, 3)\), \(a<0\). Set \(-\frac{81}{4}a=\frac{9}{2}\). Then \(a=-\frac{2}{9}\). 5. Therefore, \(b=2\), and \(f(x)=-\frac{2}{9}x^3+2x\).

\(f(x)=-\frac{2}{9}x^3+2x\)

- Factor the function to find the positive zero. - Use the coordinate axes to identify the integration interval. - Integrate the function from \(0\) to the positive zero. - Use \(k>0\) when solving the fourth-power equation.

1. Factor: \(f(x)=(x-k)^2(x+k)\). The relevant positive zero is \(x=k\), and \(f(0)=k^3\). 2. The graph is above the \(x\)-axis on \([0, k]\), so \(A=\int_0^k(x^3-kx^2-k^2x+k^3)\,dx\). 3. Evaluate: \(A=\left[\frac{x^4}{4}-\frac{kx^3}{3}-\frac{k^2x^2}{2}+k^3x\right]_0^k =\frac{5}{12}k^4\). 4. Set \(\frac{5}{12}k^4=\frac{20}{3}\). Then \(k^4=16\). Since \(k>0\), \(k=2\).

\(k=2\)

- Factor \(f_a\) to locate the positive intercept. - Differentiate the function to write the tangent line at that intercept. - The \(y\)-axis makes \(x=0\) the left boundary of the region. - Simplify the vertical difference before integrating.

1. Factor: \(f_a(x)=x(x-a)\). The positive zero is \(x=a\), so \(P=(a,0)\). 2. Since \(f_a'(x)=2x-a\), the tangent slope at \(P\) is \(a\). The tangent line is \(t_a(x)=a(x-a)=ax-a^2\). 3. On \([0,a]\), \(f_a(x)-t_a(x)=x^2-2ax+a^2=(x-a)^2\ge0\). 4. The enclosed area is \(A=\int_0^a(x-a)^2\,dx=\frac{a^3}{3}\). 5. Set \(\frac{a^3}{3}=9\). Then \(a^3=27\), so \(a=3\).

a) \(P=(a,0)\) b) \(a=3\)

- Solve \(g_k(x)=0\) for the positive intercept. - Differentiate the square-root term to find the tangent slope at \(x=k^2\). - A tangent to this concave-down graph lies above it on the enclosed interval. - Integrate from the \(y\)-axis to the point of tangency.

1. Solve \(k\sqrt{x}-x=0\): \(\sqrt{x}(k-\sqrt{x})=0\). The positive zero is \(x_S=k^2\). 2. Since \(g_k'(x)=\frac{k}{2\sqrt{x}}-1\), the slope at \(x=k^2\) is \(-\frac{1}{2}\). The tangent line is \(t_k(x)=-\frac{1}{2}(x-k^2)=-\frac{1}{2}x+\frac{1}{2}k^2\). 3. On \([0,k^2]\), the tangent line is above the graph. Their difference is \(\frac{1}{2}x+\frac{1}{2}k^2-k\sqrt{x}\). 4. Integrating gives \(A=\int_0^{k^2}\left(\frac{1}{2}x+\frac{1}{2}k^2-k\sqrt{x}\right)\,dx=\frac{1}{12}k^4\). 5. Set \(\frac{1}{12}k^4=\frac{4}{3}\). Then \(k^4=16\), so \(k=2\).

a) \(x_S=k^2\) b) \(k=2\)

- Match the endpoint \((2,4)\) on \(y=x^2\) with its reflected point \((4,2)\) on \(y=\sqrt{x}\). - Both displayed rectangles have area \(8\). - The two reflected areas are complementary portions of the rectangle. - Compute the simpler polynomial integral and subtract it from \(8\).

1. The function \(g(x)=\sqrt{x}\) is the inverse of \(f(x)=x^2\) for \(x\ge0\), so their graphs and the corresponding regions are reflections across \(y=x\). 2. The point \((2,4)\) lies on \(f\), and the point \((4,2)\) lies on \(g\). Each corresponding rectangle has area \(2\cdot4=8\). 3. The reflected regions fill the rectangle, so \(\int_0^4\sqrt{x}\,dx=8-\int_0^2x^2\,dx\). 4. Since \(\int_0^2x^2\,dx=\left[\frac{1}{3}x^3\right]_0^2=\frac{8}{3}\), the desired value is \(8-\frac{8}{3}=\frac{16}{3}\).

\(\int_0^4\sqrt{x}\,dx=\frac{16}{3}\)

- Use the two supplied intersection values as the integration limits. - Subtract the cubic from the horizontal tangent line. - Treat \(k\) as a constant while integrating with respect to \(x\). - Use \(k>0\) when solving the fourth-power equation.

1. The tangent line is above the curve between the two intersections, so \(A=\int_{-k/3}^{2k/3}\left(\frac{4}{27}k^3-kx^2+x^3\right)\,dx\). 2. Evaluating the integral gives \(A=\frac{1}{12}k^4\). 3. Set \(\frac{1}{12}k^4=108\). Then \(k^4=1296\). Since \(k>0\), \(k=6\).

\(k=6\)

- Use the supplied factorization to identify both intersections. - Determine the sign of \(g_k-t_k\) between the intersections. - Use the symmetry of the integration interval to eliminate odd terms. - Solve the resulting fifth-power equation for \(k>0\).

1. From the supplied factorization, the two intersection values are \(x=-\frac{k}{2}\) and \(x=\frac{k}{2}\). 2. Between these values, \(x+\frac{k}{2}>0\) and \(x-\frac{k}{2}<0\). Therefore, \(g_k(x)-t_k(x)<0\), so the line is above the curve. 3. The area is \(A=\int_{-k/2}^{k/2}[t_k(x)-g_k(x)]\,dx\). The odd-power terms cancel over the symmetric interval, giving \(A=\frac{1}{20}k^5\). 4. Set \(\frac{1}{20}k^5=1.6\). Then \(k^5=32\), so \(k=2\).

\(k=2\)

- Find all three intersections visible in the graph. - Three intersections create two bounded regions. - The upper graph changes at the middle intersection. - Add the magnitudes of the two signed integrals.

1. Set the functions equal: \(x^3-6x^2+9x=x\), so \(x^3-6x^2+8x=x(x-2)(x-4)=0\). The intersection \(x\)-values are \(0\), \(2\), and \(4\). 2. The difference is \(d(x)=f(x)-g(x)=x^3-6x^2+8x\), with antiderivative \(D(x)=\frac{x^4}{4}-2x^3+4x^2\). 3. On \([0,2]\), \(\int_0^2d(x)\,dx=4\). 4. On \([2,4]\), \(\int_2^4d(x)\,dx=-4\). 5. Therefore, the total area is \(|4|+|-4|=8\).

The total area is \(8\) square units.

- Substitute \(u=x^2\) to find the four x-intercepts. - Use symmetry about the \(y\)-axis to reduce the calculation. - On the right half, split the integral at \(x=1\). - Add the magnitudes of the negative and positive contributions before doubling.

1. Let \(u=x^2\). Then \(-u^2+5u-4=0\), or \((u-1)(u-4)=0\). Thus, \(x^2=1\) or \(x^2=4\), giving zeros \(x=-2\), \(x=-1\), \(x=1\), and \(x=2\). 2. The function is even, so compute the area on \([0,2]\) and double it. Split at \(x=1\). 3. An antiderivative is \(F(x)=-\frac{1}{5}x^5+\frac{5}{3}x^3-4x\). 4. On \([0,1]\), \(\int_0^1f(x)\,dx=-\frac{38}{15}\). On \([1,2]\), \(\int_1^2f(x)\,dx=\frac{22}{15}\). 5. Therefore, \(A=2\left(\frac{38}{15}+\frac{22}{15}\right)=8\) square units.

The total area is \(8\) square units.

- Factor the intersection equation to locate all three crossing points. - Three intersections create two bounded regions of very different sizes. - The upper graph changes at \(x=2\). - Add the magnitudes of the two signed integrals.

1. Set the functions equal: \(x^3-2x^2-5x+6=x^2-x-6\). The resulting equation is \(x^3-3x^2-4x+12=(x+2)(x-2)(x-3)=0\). The intersection \(x\)-values are \(-2\), \(2\), and \(3\). 2. Let \(d(x)=f(x)-g(x)=x^3-3x^2-4x+12\). An antiderivative is \(D(x)=\frac{x^4}{4}-x^3-2x^2+12x\). 3. On \([-2,2]\), \(\int_{-2}^{2}d(x)\,dx=32\). 4. On \([2,3]\), \(\int_2^3d(x)\,dx=-\frac{3}{4}\). 5. The total area is \(32+\frac{3}{4}=\frac{131}{4}=32.75\).

The total area is \(\frac{131}{4}=32.75\) square units.

- Factor the function to find the interior sign change. - Compare \(\left|\int_0^3f(x)\,dx\right|\) with the sum of the absolute values on the two subintervals. - Split the calculation at \(x=1\). - Add both geometric regions as positive quantities.

1. Factor: \(f(x)=x(x-1)(x-3)\). The graph changes sign at the interior zero \(x=1\). 2. Taking the absolute value of one integral over the entire interval still allows the two regions to cancel before the absolute value is applied. 3. An antiderivative is \(F(x)=\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2}\). Between \(0\) and \(1\), \(\int_0^1f(x)\,dx=\frac{5}{12}\). Between \(1\) and \(3\), \(\int_1^3f(x)\,dx=-\frac{8}{3}\). 4. Adding the geometric areas gives \(A=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}\).

The student took the absolute value after cancellation had already occurred. The correct total area is \(\frac{37}{12}\approx3.083\) square units.

- Find half of the full shaded area first. - If the dividing line stays below the curve, the area beneath it is a triangle. - Express the triangle's height at \(x=9\) in terms of \(m\). - After solving, verify that the line does not cross the curve before \(x=9\).

1. The total area under the curve is \(\int_0^9\sqrt{x}\,dx=\left[\frac{2}{3}x^{3/2}\right]_0^9=18\). Each half must have area \(9\). 2. If the line remains below the curve on \([0,9]\), the lower part is a triangle with base \(9\) and height \(9m\). Its area is \(\frac{1}{2}(9)(9m)=\frac{81}{2}m\). 3. Set \(\frac{81}{2}m=9\). Then \(m=\frac{2}{9}\). 4. Check the position of the line: \(\sqrt{x}=\frac{2}{9}x\) has intersections at \(x=0\) and \(x=\frac{81}{4}>9\). Thus, the line stays below the curve on the required interval.

\(m=\frac{2}{9}\)

- The zeros of \(h=f-g\) are the intersections of the original graphs. - The sign of \(h\) identifies which original graph is above the other. - Evaluate the integral of \(h\) on each interval between consecutive zeros. - Compare the magnitudes of the two signed results.

1. The difference is \(h(x)=x^3-9x^2+23x-15=(x-1)(x-3)(x-5)\). Therefore, the graphs of \(f\) and \(g\) intersect at \(x=1\), \(x=3\), and \(x=5\). 2. An antiderivative of \(h\) is \(H(x)=\frac{x^4}{4}-3x^3+\frac{23}{2}x^2-15x\). 3. On \([1,3]\), \(\int_1^3h(x)\,dx=4\). 4. On \([3,5]\), \(\int_3^5h(x)\,dx=-4\). 5. These signed integrals have equal magnitude, so both regions between \(f\) and \(g\) have area \(4\).

Each region has area \(4\) square units.

- Find the intersections in terms of \(k\). - Check the symmetry of the difference function. - Compute one of the two regions. - Double that area and use the given total.

1. Set the functions equal: \(x^3=k^2x\), so \(x(x-k)(x+k)=0\). The intersections occur at \(x=-k\), \(x=0\), and \(x=k\). 2. The difference \(h(x)=x^3-k^2x\) is odd. Therefore, the two regions are symmetric about the origin and have equal area. 3. The area of the right-hand region is \(\left|\int_0^k(x^3-k^2x)\,dx\right|=\left|\frac{k^4}{4}-\frac{k^4}{2}\right|=\frac{k^4}{4}\). 4. The combined area is \(\frac{k^4}{2}\). Set \(\frac{k^4}{2}=8\). Then \(k^4=16\), and since \(k>0\), \(k=2\).

a) Each region has area \(\frac{k^4}{4}\), so the two areas are equal. b) \(k=2\)

- Use derivatives to locate and classify the two local maxima. - Compare their function values to identify the common horizontal tangent. - The tangency points provide both endpoints of the shaded region. - Use symmetry to simplify the area integral.

1. The first derivative is \(f'(x)=-4x^3+4x=-4x(x-1)(x+1)\). The critical points are \(-1\), \(0\), and \(1\). 2. Since \(f''(x)=-12x^2+4\), the points at \(x=\pm1\) are local maxima. Both have function value \(4\). 3. The common tangent line is therefore \(y=4\). 4. Solving \(f(x)=4\) gives \(x^4-2x^2+1=(x^2-1)^2=0\). The bounded region extends from \(x=-1\) to \(x=1\). 5. Its area is \(A=\int_{-1}^{1}\left[4-f(x)\right] \,dx=\int_{-1}^{1}(x^4-2x^2+1)\,dx=\frac{16}{15}\).

The area is \(\frac{16}{15}\approx1.07\) square units.

- Set the cubic equal to the supplied tangent line. - The tangency point is a repeated intersection at \(x=0\). - Use the other intersection as the upper integration limit. - Integrate the line minus the function.

1. Solve \(f(x)=4\): \(x^3-3x^2=0\), so \(x^2(x-3)=0\). The tangent point is \(x=0\), and the other intersection is \(x=3\). 2. On \([0,3]\), the tangent line lies above the graph. Thus, \(A=\int_0^3\left[4-f(x)\right] \,dx=\int_0^3(-x^3+3x^2)\,dx\). 3. Evaluating gives \(A=\left[-\frac{x^4}{4}+x^3\right]_0^3=\frac{27}{4}\).

The area is \(\frac{27}{4}=6.75\) square units.

- Find half of the full shaded area. - For the vertical split, use an integral with unknown upper limit \(k\). - Before modeling the horizontal split, compare \(c\) with the minimum graph height \(1\). - A horizontal strip below the entire graph forms a rectangle.

1. The total area is \(A=\int_0^3\sqrt{x+1}\,dx=\left[\frac{2}{3}(x+1)^{3/2}\right]_0^3=\frac{14}{3}\). 2. For the vertical division, \(\int_0^k\sqrt{x+1}\,dx=\frac{7}{3}\). Thus, \(\frac{2}{3}\left((k+1)^{3/2}-1\right)=\frac{7}{3}\), so \((k+1)^{3/2}=\frac{9}{2}\) and \(k=\left(\frac{9}{2}\right)^{2/3}-1\approx1.726\). 3. The minimum value of \(f\) on the interval is \(f(0)=1\). If \(c<1\), the area below \(y=c\) is a rectangle of area \(3c\). 4. Set \(3c=\frac{7}{3}\). Then \(c=\frac{7}{9}\), which is less than \(1\), so the rectangular-area model is valid.

a) \(A=\frac{14}{3}\) square units b) \(k=\left(\frac{9}{2}\right)^{2/3}-1\approx1.726\) c) \(c=\frac{7}{9}\approx0.778\)

- Use the first and second derivatives to locate the two minima. - Compare their function values to determine the connecting segment. - The graph is above the segment between the two minimum points. - Use symmetry to simplify the integral if desired.

1. Differentiate: \(f'(x)=\frac{3}{4}x^3-3x=\frac{3}{4}x(x^2-4)\). The critical points are \(x=-2\), \(x=0\), and \(x=2\). 2. Since \(f''(x)=\frac{9}{4}x^2-3\) is positive at \(x=\pm2\), the local minima occur at \(x=-2\) and \(x=2\). 3. Both minima have value \(0\), so the connecting segment lies on \(y=0\). 4. The enclosed area is \(A=\int_{-2}^{2}\left(\frac{3}{16}x^4-\frac{3}{2}x^2+3\right)\,dx\) \(=\left[\frac{3}{80}x^5-\frac{1}{2}x^3+3x\right]_{-2}^{2}=\frac{32}{5}=6.4\).

The enclosed area is \(\frac{32}{5}=6.4\) square units.

- Find the triangle's height from its base line to the intersection of the sloping sides. - Compute the lower region as graph minus base line on \([-1,1]\). - Subtract the lower region from the triangle to find the upper region. - Simplify the two areas before forming the ratio.

1. The two sloping sides meet at \((0,\frac{11}{2})\). The triangle has base \(2\) and height \(4\), so its area is \(A_{\triangle}=\frac{1}{2}\cdot2\cdot4=4\). 2. The lower region lies between the graph and the base line \(y=\frac{3}{2}\). Its area is \(A_1=\int_{-1}^{1}\left(g(x)-\frac{3}{2}\right)\,dx\) \(=\int_{-1}^{1}\left(\frac{1}{2}x^4-3x^2+\frac{5}{2}\right)\,dx=\frac{16}{5}\). 3. The upper region has area \(A_2=4-\frac{16}{5}=\frac{4}{5}\). 4. Therefore, \(A_1{:}A_2=\frac{16}{5}{:}\frac{4}{5}=4{:}1\).

The lower-to-upper area ratio is \(4{:}1\).

- Differentiate \(f\) to find the tangent slope at \(x=3\). - Use point-slope form for the tangent line. - The point of tangency is a repeated solution of the intersection equation. - Integrate the tangent line minus the cubic between the two intersections.

1. Since \(f(3)=0\) and \(f'(x)=-3x^2+6x\), the slope at \(x=3\) is \(f'(3)=-9\). Therefore, \(t(x)=-9(x-3)=-9x+27\). 2. Solve \(f(x)=t(x)\): \(-x^3+3x^2=-9x+27\), so \((x-3)^2(x+3)=0\). The graphs meet at \(x=-3\) and \(x=3\). 3. On \([-3,3]\), the tangent line lies above the graph of \(f\). Thus, \(A=\int_{-3}^{3}\left[t(x)-f(x)\right] \,dx=\int_{-3}^{3}(x^3-3x^2-9x+27)\,dx\). 4. Evaluating gives \(A=108\) square units.

a) \(t(x)=-9x+27\) b) \(A=108\) square units

- Set the function equal to the horizontal tangent line to find both boundaries. - The high-multiplicity intersection at \(x=0\) is the tangency point. - Use an absolute value because the second intersection may lie to the left or right. - Solve the resulting fifth-power equation for both real parameter values.

1. The intersections with \(y=2\) satisfy \(x^4+ax^3+2=2\), so \(x^3(x+a)=0\). The distinct intersection values are \(x=0\) and \(x=-a\). 2. The enclosed area is \(A=\left|\int_0^{-a}(x^4+ax^3)\,dx\right|\). 3. Evaluating gives \(A=\frac{|a|^5}{20}\). 4. Set \(\frac{|a|^5}{20}=51.2\). Then \(|a|^5=1024\), so \(|a|=4\). Therefore, \(a=4\) or \(a=-4\).

\(a=4\) or \(a=-4\)

- Find the x-intercepts to determine the bounds. - Check the sign of the function between the intercepts. - Express the area as a definite integral. - Use \(a>0\) when solving the resulting equation.

1. Factor: \(f_a(x)=x^2(a-x)\), so the relevant x-intercepts are \(0\) and \(a\). The function is nonnegative on \([0, a]\). 2. The area is \(\int_0^a(ax^2-x^3)\,dx\) \(=\left[\frac{ax^3}{3}-\frac{x^4}{4}\right]_0^a=\frac{a^4}{12}\). 3. Set \(\frac{a^4}{12}=6.75=\frac{27}{4}\). Then \(a^4=81\). Since \(a>0\), \(a=3\).

\(a=3\)

- Use the graph to notice how the curve touches, rather than crosses, the x-axis at the positive intercept. - Find both x-intercepts in terms of \(k\), and check the sign of the function between them. - Expand the integrand before applying the power rule. - Use the condition \(k>0\) when solving the resulting power equation.

1. The x-intercepts are \(0\) and \(k\), with \(k\) a double intercept. The function is nonnegative on \([0, k]\). 2. The area is \(A=\int_0^k x(x-k)^2\,dx=\int_0^k(x^3-2kx^2+k^2x)\,dx=\frac{k^4}{12}\). 3. Set \(\frac{k^4}{12}=6.75=\frac{27}{4}\). Then \(k^4=81\). Since \(k>0\), \(k=3\).

\(k=3\)

- Use the graph to identify the symmetry and the two x-intercepts. - Solve the intercept equation in terms of \(k\) before setting up the integral. - When evaluating at \(x=k^{3/2}\), apply exponent rules carefully. - Solve the final equation by raising both sides to the reciprocal exponent.

1. Solve \(k-\frac{x^2}{k^2}=0\). The x-intercepts are \(x=\pm k^{3/2}\). 2. By even symmetry, \(A=2\int_0^{k^{3/2}}\left(k-\frac{x^2}{k^2}\right)\,dx\). 3. Evaluate: \(A=2\left[kx-\frac{x^3}{3k^2}\right]_0^{k^{3/2}}=\frac{4}{3}k^{5/2}\). 4. Set \(\frac{4}{3}k^{5/2}=\frac{128}{3}\). Then \(k^{5/2}=32\), so \(k=32^{2/5}=4\).

\(k=4\)

- Find the x-intercepts. - Check the sign on the bounded interval. - Treat \(k\) as a constant while integrating. - Set the area formula equal to \(18\).

1. Factor: \(f_k(x)=x^2\left(1-\frac{x}{k}\right)\), so the relevant x-intercepts are \(0\) and \(k\). The function is nonnegative on \([0, k]\). 2. The area is \(A=\int_0^k\left(x^2-\frac{x^3}{k}\right)\,dx\) \(=\left[\frac{x^3}{3}-\frac{x^4}{4k}\right]_0^k=\frac{k^3}{12}\). 3. Set \(\frac{k^3}{12}=18\). Then \(k^3=216\), so \(k=6\).

\(k=6\)

- Use the example graph to determine whether \(f_a\) is positive or negative between \(0\) and \(a\). - A definite integral gives signed area; geometric area must be nonnegative. - Use even symmetry after correcting the sign. - Factor the function to identify the bounds in terms of \(a\).

1. Factor \(f_a(x)=x^2\left(\frac{x^2}{a^2}-1\right)\). The x-intercepts are \(-a\), \(0\), and \(a\), and \(f_a(x)\le 0\) on \([-a,a]\). 2. The student's integral is negative because the graph lies below the x-axis. Geometric area must be positive, so \(7.2=-2\int_0^a f_a(x)\,dx=2\int_0^a\left(x^2-\frac{x^4}{a^2}\right)\,dx\). 3. Evaluate: \(7.2=2\left[\frac{x^3}{3}-\frac{x^5}{5a^2}\right]_0^a=\frac{4a^3}{15}\). 4. Since \(7.2=\frac{36}{5}\), \(a^3=27\). Therefore, \(a=3\).

The student used signed area, which is negative on \([0,a]\). A correct equation is \(7.2=-2\int_0^a f_a(x)\,dx\), and \(a=3\).

- Find the x-intercepts. - Use even symmetry. - Treat \(k\) as a constant during integration. - Solve the resulting fifth-power equation.

1. Factor: \(g_k(x)=x^2(k^2-x^2)\), so the intercepts are \(-k\), \(0\), and \(k\). The function is nonnegative on \([-k, k]\). 2. By even symmetry, \(A=2\int_0^k(k^2x^2-x^4)\,dx\) \(=2\left[\frac{k^2x^3}{3}-\frac{x^5}{5}\right]_0^k=\frac{4k^5}{15}\). 3. Set \(\frac{4k^5}{15}=64.8=\frac{324}{5}\). Then \(k^5=243\), so \(k=3\).

\(k=3\)

- Use the graph or solve \(x^2-4=0\) to locate the sign change in the interval. - Replace the absolute-value expression with \(4-x^2\) where the graph is below the x-axis. - Split the integral at the zero before evaluating.

1. The expression \(x^2-4\) is zero at \(x=2\) in \([0,3]\). 2. On \([0,2]\), \(x^2-4\le 0\), so \(|x^2-4|=4-x^2\). On \([2,3]\), \(x^2-4\ge 0\), so \(|x^2-4|=x^2-4\). 3. Split the integral: \(\int_0^3|x^2-4|\,dx=\int_0^2(4-x^2)\,dx+\int_2^3(x^2-4)\,dx\). 4. The first integral is \(\left[4x-\frac{x^3}{3}\right]_0^2=\frac{16}{3}\). The second is \(\left[\frac{x^3}{3}-4x\right]_2^3=\frac{7}{3}\). 5. Therefore, \(\int_0^3|x^2-4|\,dx=\frac{23}{3}\).

\(\frac{23}{3}\)

- Check whether the function can be zero or negative on the interval. - Use the antiderivative rule for exponential functions with a constant factor in the exponent. - Apply the fundamental theorem of calculus.

1. Rewrite the function as \(f(x)=(e^x-1)^2+2\). Since this is positive for every real \(x\), the graph lies above the \(x\)-axis on \([0,1]\). 2. An antiderivative is \(F(x)=\frac{1}{2}e^{2x}-2e^x+3x\). 3. Therefore, \(A=F(1)-F(0)\) \(=\left(\frac{1}{2}e^2-2e+3\right)-\left(\frac{1}{2}-2\right)\) \(=\frac{1}{2}e^2-2e+\frac{9}{2}\approx 2.758\) square units.

The area is \(\frac{1}{2}e^2-2e+\frac{9}{2}\approx 2.758\) square units.

- Because \(e^{x/2}>0\), use the factor \(2-x\) to determine the sign. - The signed integral can be evaluated over the full interval without splitting. - Total geometric area requires splitting at the x-intercept and reversing the sign of the negative contribution.

1. The exponential factor is always positive, so the sign of \(f(x)\) is determined by \(2-x\). Thus, \(f(x)\ge 0\) on \([0,2]\) and \(f(x)\le 0\) on \([2,4]\). 2. For b), \(\int_0^4 f(x)\,dx=F(4)-F(0)=0-8=-8\). 3. The positive area is \(\int_0^2 f(x)\,dx=F(2)-F(0)=4e-8\). 4. The signed integral on \([2,4]\) is \(F(4)-F(2)=-4e\), so the geometric area there is \(4e\). 5. Therefore, the total area is \((4e-8)+4e=8e-8\approx13.746\) square units.

a) The graph is above the x-axis on \([0,2]\) and below it on \([2,4]\). b) \(\int_0^4 f(x)\,dx=-8\) c) The total area is \(8e-8\approx13.746\) square units.

- Determine whether the graph lies above or below the \(x\)-axis on the interval. - Use the logarithmic antiderivative for a reciprocal linear expression. - Take the magnitude of a negative definite integral when finding geometric area.

1. An antiderivative is \(G(x)=5\ln|x-7|\). 2. For every \(x\in[2,6]\), \(x-7<0\), so the graph lies below the \(x\)-axis. 3. The definite integral is \(\int_{2}^{6}\frac{5}{x-7}\,dx=[5\ln|x-7|]_{2}^{6}\) \(=5\ln(1)-5\ln(5)=-5\ln(5)\). 4. Because geometric area is positive, \(A=|-5\ln(5)|=5\ln(5)\approx 8.05\) square units.

The area is \(5\ln(5)\approx 8.05\) square units.

- Tangent graphs have the same function value and the same first derivative at the contact point. - Use the y-axis and the tangency point as the horizontal bounds. - Read from the graph which function is above the other on \([0,1]\). - Apply the chain rule carefully to \(e^{1-x}\).

1. At \(x=1\), \(f(1)=g(1)=1\). 2. The derivatives are \(f'(x)=e^{x-1}\) and \(g'(x)=e^{1-x}\), so \(f'(1)=g'(1)=1\). Therefore, the graphs are tangent at \((1,1)\). 3. On \([0,1]\), \(f(x)\ge g(x)\). Thus, \(A=\int_0^1\left(e^{x-1}-2+e^{1-x}\right)\,dx\). 4. Evaluating, \(A=\left[e^{x-1}-2x-e^{1-x}\right]_0^1=e-\frac{1}{e}-2\approx0.3504\) square units.

a) \(f(1)=g(1)=1\) and \(f'(1)=g'(1)=1\), so the graphs are tangent at \((1,1)\). b) \(A=e-\frac{1}{e}-2\approx0.3504\) square units

- Use the product rule to differentiate the proposed antiderivative. - The coordinate axes give the bounds \(x=0\) and \(x=a\). - The function is negative in the enclosed Quadrant IV region, so reverse the sign of its definite integral. - Evaluate the derivative at the x-intercept \(x=a\) for the slope condition.

1. Differentiate \(G_a\): \(G_a'(x)=2e^{0.5x}+\frac{1}{2}(2x-2a-4)e^{0.5x}=(x-a)e^{0.5x}=g_a(x)\). 2. The graph meets the y-axis at \(x=0\) and the x-axis at \(x=a\). Since \(g_a(x)\le0\) on \([0,a]\), \(A(a)=-\int_0^a g_a(x)\,dx=G_a(0)-G_a(a)\). 3. Therefore, \(A(a)=(-2a-4)-(-4e^{0.5a})=4e^{0.5a}-2a-4\). 4. The derivative is \(g_a'(x)=\left(1+\frac{x-a}{2}\right)e^{0.5x}\). At the x-intercept \(x=a\), the slope is \(g_a'(a)=e^{0.5a}\). 5. Set \(e^{0.5a}=e\). Then \(a=2\), and the corresponding area is \(A(2)=4e-8\) square units.

a) \(G_a'(x)=g_a(x)\) b) \(A(a)=4e^{0.5a}-2a-4\) c) \(a=2\), and the area is \(4e-8\) square units.

- Find the \(x\)-intercepts from the factors in the numerator. - Rewrite the rational expression as a sum of powers of \(x\). - Use logarithm properties to simplify the result. - Check that the function is positive between its intercepts.

1. The \(x\)-intercepts are \(x=a\) and \(x=4a\). 2. Rewrite the function: \(f_a(x)=-1+\frac{5a}{x}-\frac{4a^2}{x^2}\). 3. An antiderivative is \(F_a(x)=-x+5a\ln x+\frac{4a^2}{x}\). 4. Therefore, \(A=\int_a^{4a}f_a(x)\,dx =\left[-x+5a\ln x+\frac{4a^2}{x}\right]_a^{4a} =5a\ln4-6a =a(10\ln2-6)\).

The area is \(a(10\ln2-6)\) square units.

- Find where the graph meets each coordinate axis. - Determine the interval on which the graph is below the \(x\)-axis. - Account for the factor \(k\) when integrating \(e^{x/k}\). - Area must be positive even when the signed integral is negative.

1. The \(y\)-intercept is \((0, 1-e)\). The \(x\)-intercept satisfies \(e^{x/k}=e\), so \(x=k\). 2. The function is nonpositive on \([0, k]\). Therefore, \(A=\int_0^k\left(e-e^{x/k}\right)\,dx\). 3. An antiderivative is \(ex-ke^{x/k}\). Thus, \(A=\left[ex-ke^{x/k}\right]_0^k=0-(-k)=k\).

The area is \(k\) square units.

- Find the \(x\)-intercepts. - Check whether the graph is above or below the \(x\)-axis between them. - Use polynomial division before integrating. - Recall the antiderivative of \(\frac{1}{x+a}\).

1. The \(x\)-intercepts are \(x=-3\) and \(x=3\). 2. On \([-3, 3]\), \(f(x)\le0\). Therefore, \(A=\int_{-3}^{3}\frac{9-x^2}{x+4}\,dx\). 3. Polynomial division gives \(\frac{9-x^2}{x+4}=-x+4-\frac{7}{x+4}\). 4. Thus, \(A=\left[-\frac{x^2}{2}+4x-7\ln|x+4|\right]_{-3}^{3} =24-7\ln7\approx10.379\).

The area is \(24-7\ln7\approx10.379\) square units.

- Set the two functions equal to find the bounds. - Use the graph to decide which function is the upper boundary. - Integrate the line and reciprocal function term by term. - Simplify \(\ln4\) as \(2\ln2\).

1. Solve \(\frac{4}{x}=5-x\). This gives \(x^2-5x+4=0\), so the intersections occur at \(x=1\) and \(x=4\). 2. On \([1,4]\), \(g(x)\ge f(x)\). Therefore, \(A=\int_1^4\left(5-x-\frac{4}{x}\right)\,dx\). 3. Evaluating, \(A=\left[5x-\frac{x^2}{2}-4\ln x\right]_1^4=\frac{15}{2}-8\ln2\approx1.955\) square units.

The area is \(\frac{15}{2}-8\ln2\approx1.955\) square units.

- Use the graph to determine the sign between the two consecutive zeros. - Integrate sine and cosine separately. - Evaluate the exact unit-circle values at both bounds. - Set the resulting area expression equal to the given area.

1. An antiderivative is \(G_t(x)=t(-\cos x+\sin x)\). 2. The function is nonpositive between the given zeros, so \(A_t=-\int_{3\pi/4}^{7\pi/4}g_t(x)\,dx\). 3. Evaluating, \(A_t=-\left[-t\sqrt2-t\sqrt2\right]=2\sqrt2\,t\). 4. Set \(2\sqrt2\,t=4\sqrt2\). Then \(t=2\).

a) \(A_t=2\sqrt2\,t\) b) \(t=2\)

- Substitute the two proposed zeros directly into the function. - Use the graph to identify whether the right-hand bounded region is above or below the x-axis. - Recall that an antiderivative of \(\tan x\) is \(-\ln|\cos x|\). - Compare \(f(-x)\) with \(-f(x)\) before doubling the area.

1. Direct substitution gives \(f(0)=0\) and \(f\left(\frac{\pi}{4}\right)=1-1=0\). On \(\left(0,\frac{\pi}{4}\right)\), the graph shows \(f(x)<0\). 2. Therefore, the right-hand area is \(A=\int_0^{\pi/4}\left(\frac{4}{\pi}x-\tan x\right)\,dx\). 3. An antiderivative is \(\frac{2}{\pi}x^2+\ln(\cos x)\). Thus, \(A=\frac{\pi}{8}-\frac{1}{2}\ln2\). 4. Since both \(\tan x\) and \(x\) are odd, \(f\) is odd. The two bounded regions have equal areas, so the total area is \(2A=\frac{\pi}{4}-\ln2\).

a) The consecutive zeros are \(0\) and \(\frac{\pi}{4}\), and \(f(x)<0\) between them. b) \(\frac{\pi}{8}-\frac{1}{2}\ln2\) square units c) \(\frac{\pi}{4}-\ln2\) square units

- Evaluate both \(f(\pi)\) and \(f'(\pi)\) to write the tangent line. - The y-axis and the tangency point give the horizontal bounds. - Use the graph to identify the tangent line as the upper boundary. - Simplify \(t(x)-f(x)\) before integrating.

1. The derivative is \(f'(x)=\cos x-\frac{1}{2}\). Therefore, \(f'(\pi)=-\frac{3}{2}\), and \(f(\pi)=1-\frac{\pi}{2}\). 2. The tangent line is \(t(x)=-\frac{3}{2}(x-\pi)+1-\frac{\pi}{2}=-\frac{3}{2}x+\pi+1\). 3. On \([0,\pi]\), the tangent line lies above the graph. Their difference is \(t(x)-f(x)=\pi-x-\sin x\). 4. Therefore, the area is \(\int_0^\pi(\pi-x-\sin x)\,dx=\frac{\pi^2}{2}-2\approx2.935\) square units.

a) \(t(x)=-\frac{3}{2}x+\pi+1\) b) \(\frac{\pi^2}{2}-2\approx2.935\) square units

- Rewrite \(\sin(2x)\) and factor out \(\sin x\). - A product is zero when at least one factor is zero. - Account for the inner factor \(2\) when integrating \(\sin(2x)\). - Use the graph to check the sign on \([0,\pi]\).

1. Using \(\sin(2x)=2\sin x\cos x\), \(g(x)=\sin x(1+\cos x)\). Thus, the zeros on \([0,2\pi]\) are \(0\), \(\pi\), and \(2\pi\). 2. One antiderivative is \(G(x)=-\cos x-\frac{1}{4}\cos(2x)+C\). 3. Since \(g(x)\ge0\) on \([0,\pi]\), \(A=\int_0^\pi g(x)\,dx=\left[-\cos x-\frac{1}{4}\cos(2x)\right]_0^\pi=2\) square units.

a) \(x=0,\pi,2\pi\) b) \(G(x)=-\cos x-\frac{1}{4}\cos(2x)+C\) c) \(A=2\) square units

- Set the function values equal and factor out \(x\). - Use the graph and odd symmetry to reduce the calculation to one region. - For the exponential term, use a substitution based on \(1-x^2\). - Double the positive one-sided area.

1. Solve \(x=xe^{1-x^2}\): \(x(1-e^{1-x^2})=0\). Thus, \(x=-1,0,1\), giving the points \((-1,-1)\), \((0,0)\), and \((1,1)\). 2. Both functions are odd, so the two enclosed regions have equal area. On \([0,1]\), \(g(x)\ge f(x)\). 3. Hence, \(A=2\int_0^1\left(xe^{1-x^2}-x\right)\,dx\). 4. Since \(\int xe^{1-x^2}\,dx=-\frac{1}{2}e^{1-x^2}\), \(A=2\left[-\frac{1}{2}e^{1-x^2}-\frac{x^2}{2}\right]_0^1=e-2\) square units.

1) \((-1,-1)\), \((0,0)\), and \((1,1)\) 2) \(A=e-2\approx0.718\) square units

- Factor the intersection equation after moving both sides together. - Use the graph to identify the linear function as the upper boundary. - Apply integration by parts to \(xe^x\). - Evaluate the lower endpoint carefully because \((x-1)e^x\) is not zero at \(x=0\).

1. Solve \(e^2x=xe^x\): \(x(e^2-e^x)=0\). Thus, the intersections occur at \(x=0\) and \(x=2\). 2. On \([0,2]\), \(f(x)\ge g(x)\). Therefore, \(A=\int_0^2\left(e^2x-xe^x\right)\,dx\). 3. Integration by parts gives \(\int xe^x\,dx=(x-1)e^x\). 4. Thus, \(A=\left[\frac{e^2x^2}{2}-(x-1)e^x\right]_0^2=e^2-1\) square units.

1) \(x=0\) and \(x=2\) 2) \(A=e^2-1\approx6.389\) square units

- Use the given maximum location as the upper integration bound. - Substitute \(u=ax\) and transform both the integrand and the bounds. - Use integration by parts for \(\int ue^{-u}\,du\). - Check whether the final expression still contains the parameter.

1. The area is \(A=\int_0^{1/a}a^2xe^{-ax}\,dx\). 2. Let \(u=ax\). Then \(x=\frac{u}{a}\), \(dx=\frac{1}{a}\,du\), and the bounds become \(0\) and \(1\). Therefore, \(A=\int_0^1ue^{-u}\,du\). 3. Since \(\int ue^{-u}\,du=-(u+1)e^{-u}\), \(A=\left[-(u+1)e^{-u}\right]_0^1=1-\frac{2}{e}\). 4. The final expression contains no \(a\), so the area is independent of the parameter.

The area is \(1-\frac{2}{e}\approx0.264\) square units, independent of \(a\).

- Compare \(g(-x)\) with \(g(x)\). - Substitute \(u=x^2\) to solve the quartic intercept equation. - Use symmetry to calculate only the right half of the central region and one side region. - Remember to reverse the sign where the graph lies below the x-axis.

1. Since \(g(-x)=g(x)\), the function is even and its graph has y-axis symmetry. 2. Let \(u=x^2\). Then \(u^2-6u+5=0\), so \(u=1\) or \(u=5\). Thus, the x-intercepts are \(x=-\sqrt5,-1,1,\sqrt5\). 3. The central region lies above the x-axis. By symmetry, its area is \(A_C=2\int_0^1(x^4-6x^2+5)\,dx=\frac{32}{5}\) square units. 4. Each side region lies below the x-axis and has area \(A_S=-\int_1^{\sqrt5}(x^4-6x^2+5)\,dx=\frac{16}{5}\) square units. 5. Therefore, the total area is \(A_C+2A_S=\frac{32}{5}+2\left(\frac{16}{5}\right)=\frac{64}{5}\) square units.

a) The graph has y-axis symmetry. The x-intercepts are \((-\sqrt5,0)\), \((-1,0)\), \((1,0)\), and \((\sqrt5,0)\). b) The central area is \(\frac{32}{5}\) square units. c) The total area is \(\frac{64}{5}\) square units.

- Apply the chain rule when differentiating the logarithm. - Use the derivative or the graph to locate the positive maximum. - Evaluate the first area with the supplied antiderivative. - Translate “twice the area” into an equation for a second definite integral.

1. By the chain rule, \(G'(x)=\frac{1}{2}\frac{2x}{x^2+4}=g(x)\). 2. Since \(g'(x)=\frac{4-x^2}{(x^2+4)^2}\), the positive local maximum occurs at \(x=2\). 3. The shaded area is \(A=\int_0^2g(x)\,dx=\left[\frac{1}{2}\ln(x^2+4)\right]_0^2=\frac{1}{2}\ln2\). 4. For c), set \(\int_2^kg(x)\,dx=\ln2\). Then \(\frac{1}{2}\ln\left(\frac{k^2+4}{8}\right)=\ln2\), so \(k^2=28\) and \(k=2\sqrt7\).

a) \(G'(x)=g(x)\) b) \(A=\frac{1}{2}\ln2\approx0.347\) square units c) \(k=2\sqrt7\approx5.292\)

- Factor the function to identify both x-intercepts. - Use the sign of each factor to check that the graph is above the x-axis between the zeros. - Treat \(k\) as a constant while integrating. - Solve the resulting fifth-power equation using \(k>0\).

1. The zeros are \(x=0\) and \(x=k\). Since \(f_k(x)\ge0\) on \([0,k]\), the enclosed area is \(A(k)=\int_0^k x^3(k-x)\,dx\). 2. Therefore, \(A(k)=\int_0^k(kx^3-x^4)\,dx=\left[\frac{kx^4}{4}-\frac{x^5}{5}\right]_0^k=\frac{k^5}{20}\). 3. Set \(\frac{k^5}{20}=51.2\). Then \(k^5=1024\), so \(k=4\).

a) The zeros are \(0\) and \(k\), and \(A(k)=\frac{k^5}{20}\). b) \(k=4\)

- Use the derivative at \(x=0\) for the tangent slope. - Set the function equal to the tangent line to find the second intersection. - The repeated solution \(x=0\) represents tangency. - Use the graph to identify the tangent line as the upper boundary.

1. Differentiate: \(f'(x)=x^2-2x-3\). Since \(f'(0)=-3\), the tangent line is \(t(x)=-3x\). 2. Solve \(f(x)=t(x)\): \(\frac{1}{3}x^3-x^2-3x=-3x\), so \(x^2\left(\frac{x}{3}-1\right)=0\). Besides the tangent point \(x=0\), the other intersection occurs at \(x=3\). 3. Since \(t(3)=-9\), \(Q=(3,-9)\). 4. On \([0,3]\), \(t(x)\ge f(x)\). Therefore, \(A=\int_0^3\left(x^2-\frac{1}{3}x^3\right)\,dx=\left[\frac{x^3}{3}-\frac{x^4}{12}\right]_0^3=\frac{9}{4}\) square units.

a) \(t(x)=-3x\) b) \(Q=(3,-9)\) c) \(A=\frac{9}{4}=2.25\) square units

- Verify an antiderivative by differentiating it once. - The exponential factor is never zero, so use the linear factor for the x-intercept. - The coordinate axes determine the interval of integration. - Apply the Fundamental Theorem of Calculus using the supplied antiderivative.

1. Differentiate the proposed antiderivative: \(F'(x)=-e^x+(4-x)e^x=(3-x)e^x=f(x)\). 2. Since \(e^x>0\), the x-intercept is determined by \(3-x=0\), so \(x=3\). 3. The graph is nonnegative on \([0,3]\). Therefore, the area is \(\int_0^3f(x)\,dx=F(3)-F(0)=e^3-4\approx16.086\) square units.

a) \(F'(x)=f(x)\), so \(F\) is an antiderivative of \(f\). b) The x-intercept is \((3,0)\), and the area is \(e^3-4\approx16.086\) square units.

- Set the numerator equal to zero while checking that the denominator is nonzero. - Differentiate the proposed logarithmic antiderivative using the chain rule. - Use the graph to see that the signed integral is negative on the enclosed interval. - Simplify the endpoint values with logarithm properties.

1. A fraction equals zero when its numerator equals zero and its denominator is nonzero. Since \(e^x-3=0\) gives \(e^x=3\), the zero is \(x=\ln3\). 2. Differentiate \(G\): \(G'(x)=2\cdot\frac{e^x}{e^x+3}-1=\frac{e^x-3}{e^x+3}=g(x)\). 3. The region extends from \(x=0\) to \(x=\ln3\). On this interval, \(g(x)<0\), so \(A=-\int_0^{\ln3}g(x)\,dx=G(0)-G(\ln3)\). 4. Since \(G(0)=2\ln4=\ln16\) and \(G(\ln3)=2\ln6-\ln3=\ln12\), \(A=\ln16-\ln12=\ln\left(\frac{4}{3}\right)\approx0.288\) square units.

a) \(g(\ln3)=0\). b) \(G'(x)=g(x)\), so \(G\) is an antiderivative of \(g\). c) \(A=\ln\left(\frac{4}{3}\right)\approx0.288\) square units

- Find all three intersection values. - Use the shaded regions to identify the upper graph on each subinterval. - Compute each region as a positive upper-minus-lower integral. - Use symmetry to check that the two areas should be equal.

1. Solve \(f(x)=g(x)\): \(x^3-4x=x(x^2-4)=0\). The intersections occur at \(x=-2,0,2\). 2. On \([-2,0]\), \(f(x)\ge g(x)\), and on \([0,2]\), \(g(x)\ge f(x)\). 3. Therefore, \(A=\int_{-2}^{0}(x^3-4x)\,dx+\int_0^2(4x-x^3)\,dx=4+4=8\) square units.

The total area is \(8\) square units.

- Set the two quadratic expressions equal to find the bounds. - Use the shaded graph to identify the upper function. - Simplify \(f(x)-g(x)\) before integrating. - Evaluate the antiderivative at both intersection values.

1. Solve \(f(x)=g(x)\): \(x^2-x-2=0\), so \(x=-1\) and \(x=2\). 2. On \([-1,2]\), \(f(x)\ge g(x)\). 3. Thus, \(A=\int_{-1}^{2}\left(-2x^2+2x+4\right)\,dx=\left[-\frac{2}{3}x^3+x^2+4x\right]_{-1}^{2}=9\) square units.

The enclosed area is \(9\) square units.

- Treat geometric areas as positive numbers and determine the sign each contributes to an integral. - Use interval additivity for the two pieces of \(I_1\). - Combine the two integrals in part c) because they have the same bounds. - In panel c), use upper minus lower to interpret the difference integral.

1. For a), \(f\) is below the x-axis on \([0,1]\) and above it on \([1,4]\). Therefore, \(I_1=-A_1+A_2=A_2-A_1\). 2. For b), \(g\) is below the x-axis on \([0,2]\), so \(I_2=-A_3\). 3. For c), linearity gives \(I_3=\int_0^3(f(x)-g(x))\,dx\). Since \(f(x)\ge g(x)\) on \([0,3]\), \(I_3=A_4\).

a) \(I_1=A_2-A_1\) b) \(I_2=-A_3\) c) \(I_3=A_4\)

- Use the zeros and vertex of the parabola for the triangle vertices. - Apply the triangle area formula using the horizontal base and vertical height. - Integrate the parabola between its two zeros. - Divide the triangle area by the full shaded area.

1. Solve \(4-x^2=0\) to get \(x=\pm2\). The x-intercepts are \((-2,0)\) and \((2,0)\), and the vertex is \((0,4)\). 2. The triangle has base \(4\) and height \(4\), so its area is \(A_T=\frac{1}{2}\cdot4\cdot4=8\) square units. 3. The function is nonnegative on \([-2,2]\), so the enclosed area is \(A=\int_{-2}^{2}(4-x^2)\,dx=\frac{32}{3}\) square units. 4. Therefore, \(\frac{A_T}{A}=\frac{8}{32/3}=\frac{3}{4}=75\%\).

a) The vertices are \((-2,0)\), \((2,0)\), and \((0,4)\); the triangle area is \(8\) square units. b) \(\frac{32}{3}\) square units c) \(75\%\)

- Use vertex form and the point \(P\) to determine the parabola. - Differentiate the parabola to find the tangent slope. - Use the y-axis and the tangency point as the horizontal bounds. - The shaded graph identifies the tangent as the upper boundary.

1. Write \(f(x)=a(x-2)^2+3\). Using \(P(4,1.5)\) gives \(a=-\frac{3}{8}\), so \(f(x)=-\frac{3}{8}(x-2)^2+3\). 2. Since \(f'(x)=-\frac{3}{4}(x-2)\), the tangent slope at \(x=4\) is \(-\frac{3}{2}\). Thus, \(t(x)=-\frac{3}{2}x+\frac{15}{2}\). 3. On \([0,4]\), the tangent lies above the parabola, and \(t(x)-f(x)=\frac{3}{8}(x-4)^2\). 4. Therefore, \(A=\int_0^4\frac{3}{8}(x-4)^2\,dx=8\) square units.

a) \(f(x)=-\frac{3}{8}(x-2)^2+3\), \(t(x)=-\frac{3}{2}x+\frac{15}{2}\) b) \(A=8\) square units

- Identify the three pairwise intersections that form the boundary. - Use the shaded region to locate where the lower boundary changes. - Keep the parabola as the upper boundary in both integrals. - Add the two positive subareas.

1. The relevant intersections occur at \(x=-2\) for \(f\) and \(h\), at \(x=-1\) for \(g\) and \(h\), and at \(x=2\) for \(f\) and \(g\). 2. The lower boundary changes at \(x=-1\). Thus, \(A=\int_{-2}^{-1}\left[f(x)-h(x)\right]\,dx+\int_{-1}^{2}\left[f(x)-g(x)\right]\,dx\). 3. Evaluating, \(A=\frac{19}{6}+\frac{27}{2}=\frac{50}{3}\approx16.667\) square units.

The area is \(\frac{50}{3}\approx16.667\) square units.

- Factor the cubic to find all three zeros. - Use the shaded graph to determine the sign on each subinterval. - Reverse the sign of the integral for the region below the x-axis. - Add the two positive geometric areas.

1. Factor \(f(x)=-\frac{1}{2}x(x-2)(x+1)\). The zeros are \(x=-1,0,2\). 2. The graph is below the x-axis on \([-1,0]\) and above it on \([0,2]\). Therefore, \(A=-\int_{-1}^{0}f(x)\,dx+\int_{0}^{2}f(x)\,dx\). 3. An antiderivative is \(F(x)=-\frac{1}{8}x^4+\frac{1}{6}x^3+\frac{1}{2}x^2\). 4. The first region has area \(\frac{5}{24}\), and the second has area \(\frac{4}{3}=\frac{32}{24}\). 5. Thus, \(A=\frac{37}{24}\) square units.

a) \(A=-\int_{-1}^{0}f(x)\,dx+\int_{0}^{2}f(x)\,dx\) b) \(A=\frac{37}{24}\) square units

- Use the two x-intercepts as the area bounds. - Find the vertex height from the axis of symmetry. - The rectangle area is its fixed width multiplied by the vertex height. - Set the two area expressions equal and use \(a>0\).

1. The x-intercepts are \(0\) and \(a\). Therefore, \(A_1=\int_0^a(ax-x^2)\,dx=\left[\frac{ax^2}{2}-\frac{x^3}{3}\right]_0^a=\frac{a^3}{6}\). 2. The vertex occurs at \(x=\frac{a}{2}\), with height \(f_a\left(\frac{a}{2}\right)=\frac{a^2}{4}\). 3. The rectangle area is \(A_2=4\cdot\frac{a^2}{4}=a^2\). 4. Set the areas equal: \(\frac{a^3}{6}=a^2\). Since \(a>0\), \(a=6\).

a) \(A_1=\frac{a^3}{6}\) b) \(a=6\)

- Set the two functions equal and factor completely. - The upper graph switches at the middle intersection. - Compute each region with a separate positive integral. - Compare the two values, or examine the symmetry of \(f-g\) around \(x=1\).

1. Solve \(f(x)=g(x)\): \(-x(x-1)(x-2)=0\). The intersection points are \((0,0)\), \((1,2)\), and \((2,4)\). 2. On \([0,1]\), \(g(x)\ge f(x)\). The first area is \(A_1=\int_0^1(g-f)\,dx=\frac{1}{4}\). 3. On \([1,2]\), \(f(x)\ge g(x)\). The second area is \(A_2=\int_1^2(f-g)\,dx=\frac{1}{4}\). 4. Therefore, the two regions have equal areas, and the total area is \(A_1+A_2=\frac{1}{2}\) square unit. Their equality also follows from the point symmetry of the difference \(f(x)-g(x)\) about \((1,0)\).

a) \((0,0)\), \((1,2)\), and \((2,4)\) b) Each region has area \(\frac{1}{4}\) square unit, and the total area is \(\frac{1}{2}\) square unit.

- Factor the function using the double-angle identity. - Use the interior zero to split the signed regions. - An integral of zero does not mean that both geometric areas are zero. - Add the magnitudes of the positive and negative contributions.

1. Use \(\sin(2x)=2\sin x\cos x\): \(f(x)=2\sin x(1-\cos x)\). The zeros are \(x=0,\pi,2\pi\). 2. An antiderivative is \(F(x)=-2\cos x+\frac{1}{2}\cos(2x)\). Therefore, \(\int_0^{2\pi}f(x)\,dx=0\); the positive and negative signed areas cancel. 3. On \([0,\pi]\), the signed integral is \(4\). On \([\pi,2\pi]\), it is \(-4\), so that region has geometric area \(4\). 4. The total area is \(4+4=8\) square units.

a) \(x=0,\pi,2\pi\) b) \(\int_0^{2\pi}f(x)\,dx=0\); the signed areas cancel. c) The total area is \(8\) square units.

- Subtract the two consecutive family members before solving for intersections. - Use either graph to find the y-coordinates at the common x-values. - Integrate the difference of the upper and lower graphs on \([0,1]\). - Check whether the final difference or bounds contain \(a\).

1. The difference is \(f_a(x)-f_{a-1}(x)=x-x^2=x(1-x)\). Thus, the intersections occur at \(x=0\) and \(x=1\). 2. The corresponding y-values are \(0\) and \(1\), so the intersection points are \((0,0)\) and \((1,1)\), independent of \(a\). 3. On \([0,1]\), \(x-x^2\ge0\), so \(A=\int_0^1(f_a(x)-f_{a-1}(x))\,dx=\int_0^1(x-x^2)\,dx=\frac{1}{6}\). 4. Both the difference and the bounds are independent of \(a\), so the area is independent of \(a\).

a) The intersection points are \((0,0)\) and \((1,1)\). b) The enclosed area is \(\frac{1}{6}\) square unit, independent of \(a\).

- Read each main region as a triangle. - A region below the x-axis contributes a negative signed integral. - Add the three signed values for the full definite integral. - Add their magnitudes for total geometric area.

1. On \([-4,-1]\), the triangle has base \(3\) and height \(2\), so \(I_1=3\). 2. On \([-1,3]\), the triangle lies below the x-axis and has base \(4\) and height \(2\), so \(I_2=-4\). 3. On \([3,6]\), the triangle has base \(3\) and height \(2\), so \(I_3=3\). 4. Therefore, \(\int_{-4}^{6}f(x)\,dx=3-4+3=2\). The positive area exceeds the magnitude of the negative area by \(2\). 5. The total geometric area is \(3+4+3=10\) square units. It adds all magnitudes instead of subtracting the region below the axis.

a) 1) \(I_1=3\) 2) \(I_2=-4\) 3) \(I_3=3\) b) \(\int_{-4}^{6}f(x)\,dx=2\) c) The total area is \(10\) square units.

- Set the two functions equal and factor the intersection equation. - Use the shaded graph to identify where the upper function changes. - Write one positive upper-minus-lower integral for each region. - Add the two region areas.

1. Solve \(f(x)=g(x)\): \((x^2-4)(x-1)=0\), so \(x=-2,1,2\). 2. On \([-2,1]\), \(f\ge g\), and on \([1,2]\), \(g\ge f\). 3. Therefore, \(A=\int_{-2}^{1}(f-g)\,dx+\int_1^2(g-f)\,dx=\frac{45}{4}+\frac{7}{12}=\frac{71}{6}\) square units.

The total area is \(\frac{71}{6}\approx11.833\) square units.

- Use the derivative to locate where the profile changes from increasing to decreasing. - Evaluate the original function at the maximizing x-coordinate. - Interpret the shaded cross section as a definite integral. - Use integration by parts for the product \(xe^{-0.2x}\).

1. The derivative is \(g'(x)=(10-2x)e^{-0.2x}\). It changes from positive to negative at \(x=5\), so the highest point is \((5,50e^{-1})\approx(5,18.39)\). 2. Integration by parts gives an antiderivative \(G(x)=-50(x+5)e^{-0.2x}\). 3. Therefore, \(\int_0^{15}g(x)\,dx=250-1000e^{-3}\approx200.21\,\text{m}^2\).

a) \((5,50e^{-1})\approx(5,18.39)\) b) \(250-1000e^{-3}\approx200.21\,\text{m}^2\)

- Use the symmetry between the upper and lower edges for the full area. - The angle with the x-axis comes from the magnitude of the tangent slope. - Set the first derivative equal to zero to locate the maximum height. - Include square inches for area and inches for height.

1. By symmetry, the area is \(2\int_0^4f(x)\,dx=\frac{128}{15}\,\text{in}^2\approx8.53\,\text{in}^2\). 2. The derivative is \(f'(x)=\frac{4-3x}{4\sqrt{x}}\), so \(f'(4)=-1\). Therefore, the acute angle satisfies \(\tan\alpha=1\), giving \(\alpha=45^\circ\). 3. Set \(f'(x)=0\), giving \(x=\frac{4}{3}\). Then \(f\left(\frac{4}{3}\right)=\frac{8}{3\sqrt3}\approx1.54\,\text{in}\).

a) \(\frac{128}{15}\,\text{in}^2\approx8.53\,\text{in}^2\) b) \(45^\circ\) c) \(\frac{8}{3\sqrt3}\,\text{in}\approx1.54\,\text{in}\)

- Evaluate the full definite integral before splitting. - Use the x-intercept at \(x=2\) to separate the two regions. - A below-axis region contributes negatively to a definite integral. - Add positive magnitudes for total geometric area.

1. An antiderivative is \(F(x)=\frac{x^3}{3}-4x\). Thus, \(\int_0^3f(x)\,dx=F(3)-F(0)=-3\). 2. The graph crosses the x-axis at \(x=2\). The two geometric areas are \(A_1=\frac{16}{3}\) and \(A_2=\frac{7}{3}\). Therefore, \(A_1+A_2=\frac{23}{3}\approx7.67\) square units. 3. The definite integral is net signed area, so \(A_1\) contributes negatively. Total geometric area adds both positive magnitudes.

1) \(-3\) 2) \(\frac{23}{3}\approx7.67\) square units 3) The integral subtracts the below-axis region, while total area adds both magnitudes.

- Use the right-hand shaded region for the area integral. - Compare \(f(-x)\) with \(-f(x)\). - Reverse the chain rule when integrating the sine function. - Evaluate cosine at \(0\) and \(\pi\).

1. On \([0,2]\), the graph lies above the x-axis, so \(A=\int_0^2 2\sin\left(\frac{\pi}{2}x\right)\,dx\). 2. The function is odd. The signed-area contributions on \([-2,0]\) and \([0,2]\) have equal magnitudes and opposite signs, so they cancel. 3. An antiderivative is \(-\frac{4}{\pi}\cos\left(\frac{\pi}{2}x\right)\). Therefore, \(A=\frac{8}{\pi}\approx2.55\) square units.

a) \(A=\int_0^2 2\sin\left(\frac{\pi}{2}x\right)\,dx\) b) The symmetric signed areas cancel. c) \(A=\frac{8}{\pi}\approx2.55\) square units

- Identify which regions lie above and below the \(x\)-axis. - A definite integral uses signed contributions. - An absolute-value integral counts every region positively. - Compare net signed area with total geometric area.

1. On \([0,6]\), region \(A\) lies below the \(x\)-axis and region \(B\) lies above it. Thus, \(\int_{0}^{6}g(x)\,dx=-A+B=-2.5+5.0=2.5\). 2. On \([0,8]\), all three regions contribute: \(\int_{0}^{8}g(x)\,dx=-A+B-C=-2.5+5.0-1.5=1.0\). 3. The absolute-value integral makes every contribution nonnegative: \(\int_{0}^{8}|g(x)|\,dx=A+B+C=2.5+5.0+1.5=9.0\). 4. Part b) gives net signed area, while part c) gives total geometric area.

a) \(2.5\) b) \(1.0\) c) \(9.0\) d) Part b) subtracts regions below the \(x\)-axis, while part c) adds the magnitudes of all regions.

- Trace each requested boundary from left to right. - Treat the shaded pieces as adjacent parts of a larger region. - Check whether the x-axis divides a requested region into separate components.

1. Region I consists of \(A_1\) and \(A_4\), so its area is \(A_1+A_4\). 2. Region II consists of \(A_2\) on the left and \(A_3+A_4\) on the right, so its area is \(A_2+A_3+A_4\). 3. Region III consists of \(A_1+A_2\) on the left and \(A_3\) on the right, so its area is \(A_1+A_2+A_3\).

I. \(A_1+A_4\) II. \(A_2+A_3+A_4\) III. \(A_1+A_2+A_3\)

- Use the interior x-intercept to split the interval. - The cosine graph is above the axis first and below it second. - Add the positive magnitude of each signed integral.

1. The interior zero is \(x=\frac{\pi}{2}\). 2. The first area is \(A_1=\int_0^{\pi/2}\cos x\,dx=1\). 3. The second signed integral is \(\int_{\pi/2}^{\pi}\cos x\,dx=-1\), so the second geometric area is \(A_2=1\). 4. Therefore, the total area is \(A_1+A_2=2\) square units.

The total area is \(2\) square units.

- Find every zero in the stated interval. - Use the graph to identify the sign on each subinterval. - Reverse the sign of the integral over the below-axis region.

1. Solve \((x-1)^2-1=0\). The zeros in the interval are \(x=0\) and \(x=2\). 2. The graph lies below the x-axis on \([0,2]\) and above it on \([2,3]\). 3. With \(F(x)=\frac{1}{3}(x-1)^3-x\), the first region has area \(\frac{4}{3}\), and the second also has area \(\frac{4}{3}\). 4. Therefore, the total area is \(\frac{8}{3}\approx2.67\) square units.

The total area is \(\frac{8}{3}\approx2.67\) square units.

- Solve for the sign-change point. - Rewrite \(\sqrt{x}\) as a power before integrating. - Add the magnitude of the below-axis contribution.

1. Solve \(2-\sqrt{x}=0\), giving \(x=4\). 2. The graph lies above the x-axis on \([0,4]\) and below it on \([4,9]\). 3. An antiderivative is \(F(x)=2x-\frac{2}{3}x^{3/2}\). Each region has area \(\frac{8}{3}\). 4. Therefore, the total area is \(\frac{16}{3}\approx5.33\) square units.

The total area is \(\frac{16}{3}\approx5.33\) square units.

- Find the x-intercept of the shifted exponential. - Split the interval at the intercept. - Reverse the sign of the integral on the left subinterval.

1. The graph crosses the x-axis at \(x=0\), lies below it on \([-1,0]\), and lies above it on \([0,1]\). 2. An antiderivative is \(F(x)=e^x-x\). 3. The left region has area \(\frac{1}{e}\), and the right region has area \(e-2\). 4. Therefore, the total area is \(e+\frac{1}{e}-2\approx1.086\) square units.

The total area is \(e+\frac{1}{e}-2\approx1.086\) square units.

- Account for the inner factor \(0.5\) when integrating. - Use the horizontal base and vertical height for the triangle. - Divide the triangle area by the full shaded area. - Keep square meters in both area results.

1. The arch area is \(\int_0^{2\pi}2\sin(0.5x)\,dx=8\,\text{m}^2\). 2. The triangle has base \(2\pi\,\text{m}\) and height \(2\,\text{m}\), so its area is \(2\pi\,\text{m}^2\). 3. The covered percentage is \(\frac{2\pi}{8}(100\%)=25\pi\%\approx78.5\%\).

a) \(8\,\text{m}^2\) b) Triangle area: \(2\pi\,\text{m}^2\); percentage covered: \(25\pi\%\approx78.5\%\)

- Think about what equal areas above and below the x-axis imply about a definite integral. - Use signed area rather than adding both areas as positive quantities. - Find an antiderivative of each function. - Evaluate from \(0\) to \(k\), then solve the resulting equation.

1. Equal areas above and below the x-axis mean that the signed area is zero, so \(\int_0^k f(x)\,dx=0\). 2. For part a, an antiderivative is \(F(x)=\frac{1}{3}x^3-x\). Therefore, \(\frac{1}{3}k^3-k=0\), or \(k\left(\frac{1}{3}k^2-1\right)=0\). 3. Since \(k>0\), \(k^2=3\), so \(k=\sqrt{3}\approx1.732\). 4. For part b, an antiderivative is \(F(x)=0.025x^4-0.8x\). Therefore, \(0.025k^4-0.8k=0\), or \(k(0.025k^3-0.8)=0\). 5. Since \(k>0\), \(k^3=32\), so \(k=\sqrt[3]{32}\approx3.175\).

a) \(k=\sqrt{3}\approx1.732\) b) \(k=\sqrt[3]{32}\approx3.175\)

- Match each label to its actual upper and lower boundaries. - Geometric area must be positive. - Check the zeros \(x=2,4\) and the intersection \(x=\sqrt{10}\). - Reverse upper and lower functions after the two graphs cross.

1. (1) is correct because \(f(x)>0\) on \([0,2]\). 2. (2) is correct because \(f(x)\ge g(x)\) on \([2,\sqrt{10}]\). 3. (3) is incorrect because \(g(x)<0\) on \([0,2]\). The correct expression is \(A_4=-\int_0^2g(x)\,dx\). 4. (4) is incorrect because \(g(x)\ge f(x)\) on \([\sqrt{10},4]\). The correct expression is \(A_6=\int_{\sqrt{10}}^4(g(x)-f(x))\,dx\).

(1) Correct. (2) Correct. (3) Incorrect; \(A_4=-\int_0^2g(x)\,dx\). (4) Incorrect; \(A_6=\int_{\sqrt{10}}^4(g(x)-f(x))\,dx\).

- Use the derivative and point \(Q\) to find the tangent equation. - The shaded graph identifies the tangent as the upper boundary. - Simplify the difference before integrating.

1. Since \(f'(x)=-2x+4\), the tangent slope at \(x=3\) is \(-2\). Thus, \(t(x)=-2x+9\). 2. The difference is \(t(x)-f(x)=(x-3)^2\ge0\). 3. Therefore, \(A=\int_1^3(x-3)^2\,dx=\frac{8}{3}\approx2.667\) square units.

The area is \(\frac{8}{3}\approx2.667\) square units.

- Set the two functions equal to find the integration limits. - Check any candidate solutions introduced when you square the equation. - Integrate the upper function minus the lower function.

1. Set the functions equal: \(\sqrt{x+1}=\frac{x+1}{2}\). After checking the solutions introduced by squaring, the intersections are \(x=-1\) and \(x=3\). 2. On \([-1,3]\), \(f(x)\ge g(x)\). 3. Therefore, \(A=\int_{-1}^{3}\left(\sqrt{x+1}-\frac{1}{2}x-\frac{1}{2}\right)\,\mathrm{d}x=\frac{4}{3}\).

\(\frac{4}{3}\) square units, or approximately \(1.333\) square units

- Use the intercepts as both the integration bounds and the endpoints of the rectangle's base. - The vertex height is \(a^2\). - Use symmetry when evaluating the area under the parabola. - Subtract the curved-region area from the rectangle area.

1. The x-intercepts are \(-a\) and \(a\). Therefore, \(A(a)=\int_{-a}^{a}(a^2-x^2)\,\mathrm{d}x=\frac{4}{3}a^3\). 2. The rectangle has width \(2a\) and height \(a^2\), so its area is \(2a^3\). 3. The stated difference gives \(2a^3-\frac{4}{3}a^3=18\). Thus, \(\frac{2}{3}a^3=18\), so \(a=3\).

a) \(A(a)=\frac{4}{3}a^3\) b) \(a=3\)

- Solve the product equation \(x\cos x=0\). - Use integration by parts to find an antiderivative of \(x\cos x\). - Split the total area where the graph crosses the x-axis. - Keep the below-axis contribution negative only for the signed integral.

1. The zeros satisfy \(x=0\) or \(\cos x=0\). On \([0,\pi]\), they are \(x=0\) and \(x=\frac{\pi}{2}\). 2. Integration by parts gives \(\int x\cos x\,\mathrm{d}x=x\sin x+\cos x\). 3. The positive area on \(\left[0,\frac{\pi}{2}\right]\) is \(\frac{\pi}{2}-1\). The region below the axis on \(\left[\frac{\pi}{2},\pi\right]\) has area \(1+\frac{\pi}{2}\). Their total is \(\pi\). 4. The signed integral is \(\int_0^\pi x\cos x\,\mathrm{d}x=-2\). It subtracts the region below the x-axis, while total geometric area adds its magnitude.

a) \(x=0\) and \(x=\frac{\pi}{2}\) b) \(\pi\) square units c) \(\int_0^\pi f(x)\,\mathrm{d}x=-2\); the region below the x-axis contributes negatively to the signed integral.

- Use the sign of \(x-k\) to identify the interval that forms the fourth-quadrant region. - Find an antiderivative by integration by parts or verify the proposed product form. - Negate the signed integral because the graph lies below the x-axis.

1. Since \(e^{0.5x}>0\), the function is negative for \(0\le x<k\) and equals \(0\) at \(x=k\). Therefore, \(A(k)=-\int_0^k(x-k)e^{0.5x}\,\mathrm{d}x\). 2. An antiderivative is \(F(x)=(2x-2k-4)e^{0.5x}\). 3. Thus, \(A(k)=-[F(k)-F(0)]=4e^{k/2}-2k-4\). 4. For \(k=2\), \(A(2)=4e-8\approx2.87\) square units.

a) \(A(k)=4e^{k/2}-2k-4\) b) \(A(2)=4e-8\approx2.87\) square units

- Read two exact points on the line from the grid. - Set the two function formulas equal to find the area bounds. - Integrate the upper graph minus the lower graph.

1. The line passes through \((-2,0)\) and \((0,2)\), so its slope is \(1\) and its y-intercept is \(2\). Thus, \(g(x)=x+2\). 2. Solve \(-x^2+4=x+2\). The intersections occur at \(x=-2\) and \(x=1\). 3. On \([-2,1]\), \(f(x)\ge g(x)\). Therefore, \(A=\int_{-2}^{1}(-x^2-x+2)\,\mathrm{d}x=\frac{9}{2}\).

a) \(g(x)=x+2\) b) \(\frac{9}{2}=4.5\) square units

- Find all x-values where the graphs intersect. - A substitution using \(u=x^2\) simplifies the intersection equation. - Use the symmetry of the two graphs to calculate only one region. - Integrate upper minus lower on \([1,2]\).

1. Set the functions equal: \(\frac{4}{x^2}=5-x^2\). Multiplying by \(x^2\) gives \(x^4-5x^2+4=0\). 2. Let \(u=x^2\). Then \(u^2-5u+4=0\), so \(u=1\) or \(u=4\). Therefore, the intersection x-values are \(-2\), \(-1\), \(1\), and \(2\). 3. Both functions are even, so the two bounded regions have equal area. On \([1,2]\), \(g(x)\) is above \(f(x)\). 4. One region has area \(\int_1^2\left(5-x^2-\frac{4}{x^2}\right)\,\text{d}x\). An antiderivative is \(5x-\frac{x^3}{3}+\frac{4}{x}\), so the area is \(\left[5x-\frac{x^3}{3}+\frac{4}{x}\right]_1^2=\frac{2}{3}\). 5. Doubling gives \(A=2\left(\frac{2}{3}\right)=\frac{4}{3}\).

The total area is \(\frac{4}{3}\) square units.

- Find the tangent line and both x-intercepts. - Use the image to identify where the upper boundary changes. - Split the area at the tangent line’s x-intercept. - Integrate upper minus lower on each interval.

1. Since \(f(1)=-1\) and \(f'(x)=\frac{1}{2\sqrt{x}}\), the tangent slope at \(x=1\) is \(\frac12\). Thus, the tangent line is \(t(x)=\frac12x-\frac32\). 2. The curve meets the x-axis at \(x=4\), and the tangent line meets it at \(x=3\). 3. From \(x=1\) to \(x=3\), the region lies between the tangent line and the curve. From \(x=3\) to \(x=4\), it lies between the x-axis and the curve. 4. Therefore, \(A=\int_1^3\left[\left(\frac12x-\frac32\right)-(\sqrt{x}-2)\right] \,\text{d}x+\int_3^4(2-\sqrt{x})\,\text{d}x=\frac13\).

The area is \(\frac13\) square unit.

- Solve two equations: \(f(x)=0\) and \(f(x)=-6\). - Identify the upper and lower boundary on each part of the interval. - Split the region into a rectangle and a curved part. - Because the curve is below the \(x\)-axis, integrate \(0-f(x)\).

1. Solve \(f(x)=0\): \(\frac{1}{2}x^2-8=0\), so \(x^2=16\) and \(x=\pm4\). 2. Solve \(f(x)=-6\): \(\frac{1}{2}x^2-8=-6\), so \(x^2=4\) and \(x=\pm2\). 3. On the left side, the region extends from \(x=-4\) to \(x=0\). From \(x=-2\) to \(x=0\), it is a rectangle between \(y=0\) and \(y=-6\), with area \(2\cdot6=12\). 4. From \(x=-4\) to \(x=-2\), the area between the \(x\)-axis and the curve is \(\int_{-4}^{-2}\left(8-\frac{1}{2}x^2\right)\,dx\). 5. Evaluate: \(\left[8x-\frac{x^3}{6}\right]_{-4}^{-2}=\frac{20}{3}\). 6. The total area is \(12+\frac{20}{3}=\frac{56}{3}\).

a) The zeros are \(x=-4\) and \(x=4\). The intersections with \(g\) have \(x\)-coordinates \(-2\) and \(2\). b) The area is \(\frac{56}{3}\approx18.67\) square units.

- First find the total area between the parabola and the \(x\)-axis. - Describe the upper part of the region with an integral involving \(c\). - Find the intersection points of \(y=4-x^2\) and \(y=c\). - Use the symmetry of the parabola.

1. The graph meets the \(x\)-axis at \(x=-2\) and \(x=2\). The total area is \(\int_{-2}^{2}(4-x^2)\,dx=\frac{32}{3}\). 2. Each part must have area \(\frac{16}{3}\). Use the upper part, between the parabola and \(y=c\). 3. The line intersects the parabola where \(4-x^2=c\), so \(x=\pm\sqrt{4-c}\). 4. The upper area is \(\int_{-\sqrt{4-c}}^{\sqrt{4-c}}(4-x^2-c)\,dx=\frac{4}{3}(4-c)^{3/2}\). 5. Set this equal to \(\frac{16}{3}\): \(\frac{4}{3}(4-c)^{3/2}=\frac{16}{3}\). Thus, \((4-c)^{3/2}=4\), so \(4-c=4^{2/3}=\sqrt[3]{16}\). 6. Therefore, \(c=4-\sqrt[3]{16}=4-2\sqrt[3]{2}\).

\(c=4-\sqrt[3]{16}=4-2\sqrt[3]{2}\)

- Find the total area between the parabola and \(y=4\). - Determine what fraction of the total area is the lower part. - Express the lower part as an integral involving \(c\). - Solve the resulting equation for \(c\).

1. The parabola and \(y=4\) intersect at \(x=\pm2\). The total area is \(\int_{-2}^{2}(4-x^2)\,dx=\frac{32}{3}\). 2. Let \(A_{\text{lower}}\) be the lower part. Since the upper part is twice the lower part, \(3A_{\text{lower}}=\frac{32}{3}\), so \(A_{\text{lower}}=\frac{32}{9}\). 3. The line \(y=c\) intersects the parabola at \(x=\pm\sqrt{c}\). Therefore, \(A_{\text{lower}}=\int_{-\sqrt{c}}^{\sqrt{c}}(c-x^2)\,dx=\frac{4}{3}c^{3/2}\). 4. Set \(\frac{4}{3}c^{3/2}=\frac{32}{9}\). Then \(c^{3/2}=\frac{8}{3}\), so \(c=\left(\frac{8}{3}\right)^{2/3}=\sqrt[3]{\frac{64}{9}}\).

\(c=\left(\frac{8}{3}\right)^{2/3}=\sqrt[3]{\frac{64}{9}}\)

- Factor the quadratic to find its zeros. - Use the midpoint of the zeros to find the vertex. - Eliminate \(k\) from the vertex coordinates to find the locus. - Integrate \(h_k\) between its zeros and set the area equal to \(36\).

1. Factor: \(h_k(x)=x\left(1-\frac{x}{k}\right)\). The zeros are \(x=0\) and \(x=k\). The vertex lies halfway between them, at \(x=\frac{k}{2}\). Its y-coordinate is \(h_k\left(\frac{k}{2}\right)=\frac{k}{4}\). Thus, \(S_k=\left(\frac{k}{2}, \frac{k}{4}\right)\). 2. If \((x, y)=S_k\), then \(x=\frac{k}{2}\), so \(k=2x\). Therefore, \(y=\frac{k}{4}=\frac{x}{2}\). The vertex locus is \(y=\frac{1}{2}x\). 3. The enclosed area is \(A(k)=\int_0^k\left(-\frac{1}{k}x^2+x\right)\,dx=\frac{k^2}{6}\). Set this equal to \(36\): \(\frac{k^2}{6}=36\). Since \(k>0\), \(k=\sqrt{216}=6\sqrt{6}\).

1) The zeros are \(0\) and \(k\), and \(S_k=\left(\frac{k}{2}, \frac{k}{4}\right)\). 2) The vertex locus is \(y=\frac{1}{2}x\). 3) \(k=6\sqrt{6}\approx14.7\)

- Find the local minimum using the first and second derivatives. - The tangent line at a local minimum is horizontal. - Factor the function to find the other intersection with the tangent. - Express the enclosed area in terms of \(a\).

1. The first derivative is \(f_a'(x)=a(3x^2-6x)=3ax(x-2)\). The critical points are \(x=0\) and \(x=2\). 2. Since \(f_a''(x)=a(6x-6)\) and \(f_a''(2)=6a>0\), the local minimum occurs at \(x=2\). 3. Because \(f_a(2)=0\), the tangent line at the minimum is \(y=0\). 4. Factor: \(x^3-3x^2+4=(x-2)^2(x+1)\). The graph meets the tangent line at \(x=-1\) and \(x=2\). 5. The area is \(A=\int_{-1}^{2}a(x^3-3x^2+4)\,dx=\frac{27}{4}a\). 6. Set \(\frac{27}{4}a=10.8\). Therefore, \(a=1.6\).

\(a=1.6\)

- Find the zeros of \(f'(x)\) and use \(f''(x)\) to classify them. - The tangent at the local maximum is the horizontal line through its function value. - The tangency point is a repeated solution of the intersection equation. - Integrate the horizontal line minus the cubic over the shaded interval.

1. The first derivative is \(f'(x)=-3x^2+12\). Setting it equal to zero gives \(x=\pm2\). 2. Since \(f''(x)=-6x\), \(f''(2)<0\) and \(f''(-2)>0\). Thus, the local maximum is \((2,16)\), and the local minimum is \((-2,-16)\). 3. The tangent line at the local maximum is \(y=16\). 4. Solve \(-x^3+12x=16\). The point of tangency \(x=2\) is a double root, and the other intersection is \(x=-4\). 5. The area is \(A=\int_{-4}^{2}\left[16-(-x^3+12x)\right] \,dx\). 6. Evaluate: \(\left[\frac{x^4}{4}-6x^2+16x\right]_{-4}^{2}=12-(-96)=108\).

a) Local maximum: \((2,16)\); local minimum: \((-2,-16)\) b) The area is \(108\) square units.

- Find the two critical points and their function values. - Each local extremum produces a horizontal tangent line. - Factor each line-minus-curve equation to find the second intersection. - Compute the two shaded areas separately and compare them.

1. The first derivative is \(g'(x)=3x^2-12x+9=3(x-1)(x-3)\). The critical points are \(x=1\) and \(x=3\). 2. The function values are \(g(1)=4\) and \(g(3)=0\). Thus, the horizontal tangent lines are \(y=4\) and \(y=0\). 3. For \(y=4\), \(g(x)-4=(x-1)^2(x-4)\). The bounded region extends from \(x=1\) to \(x=4\), and \(A_1=\int_1^4[4-g(x)]\,dx=\frac{27}{4}\). 4. For \(y=0\), \(g(x)=x(x-3)^2\). The bounded region extends from \(x=0\) to \(x=3\), and \(A_2=\int_0^3g(x)\,dx=\frac{27}{4}\). 5. Since \(A_1=A_2=\frac{27}{4}\), the two regions have equal area.

The tangent lines are \(y=4\) and \(y=0\). Both regions have area \(\frac{27}{4}=6.75\) square units.

- Find half of the full shaded area. - The horizontal line meets the parabola at \(x=\sqrt{c}\). - Describe the part below the line as an integral plus a rectangle. - Substitute \(u=\sqrt{c}\) after forming the equal-area equation.

1. The total area is \(\int_0^2x^2\,dx=\frac{8}{3}\), so each part must have area \(\frac{4}{3}\). 2. The line \(y=c\) meets the parabola at \(x=\sqrt{c}\). 3. The area below \(y=c\) consists of the area under the parabola from \(0\) to \(\sqrt{c}\) and a rectangle from \(\sqrt{c}\) to \(2\): \(A_{\text{lower}}=\int_0^{\sqrt{c}}x^2\,dx+(2-\sqrt{c})c\). 4. Simplify: \(A_{\text{lower}}=\frac{1}{3}c^{3/2}+2c-c^{3/2}=2c-\frac{2}{3}c^{3/2}\). 5. Set this equal to \(\frac{4}{3}\): \(3c-c^{3/2}=2\). Let \(u=\sqrt{c}\). Then \(u^3-3u^2+2=0\). 6. Factor: \((u-1)(u^2-2u-2)=0\). The only solution with \(0<u<2\) is \(u=1\). Therefore, \(c=1\).

\(c=1\)

- Find the full shaded area and divide it by \(2\). - Represent the left portion with an integral whose upper limit is \(k\). - After evaluating the integral, substitute \(u=e^k\). - Select the root that places \(k\) inside the original interval.

1. The total area is \(\int_0^{\ln(3)}(e^x-e^{-x})\,dx=\left[e^x+e^{-x}\right]_0^{\ln(3)}=\frac{4}{3}\). 2. Each part has area \(\frac{2}{3}\). Thus, \(\int_0^k(e^x-e^{-x})\,dx=\frac{2}{3}\). 3. Evaluating gives \(e^k+e^{-k}-2=\frac{2}{3}\), so \(e^k+e^{-k}=\frac{8}{3}\). 4. Let \(u=e^k\). Then \(u+\frac{1}{u}=\frac{8}{3}\), so \(3u^2-8u+3=0\). 5. The solutions are \(u=\frac{4\pm\sqrt{7}}{3}\). Because \(0<k<\ln(3)\), \(1<u<3\), so \(u=\frac{4+\sqrt{7}}{3}\). 6. Therefore, \(k=\ln\left(\frac{4+\sqrt{7}}{3}\right)\).

\(k=\ln\left(\frac{4+\sqrt{7}}{3}\right)\)

- Use symmetry about the \(y\)-axis throughout the problem. - The tangent slopes at \(-k\) and \(k\) are opposites. - The chord \(AB\) is horizontal, while the tangent-bounded region changes lower boundary at \(x=0\). - Factor out the common power of \(k\) before forming the ratio.

1. Since \(f'(x)=2x\), the tangent slopes at \(x=-k\) and \(x=k\) are \(-2k\) and \(2k\). The tangent lines are \(y=-2kx-k^2\) and \(y=2kx-k^2\). They intersect at \(C(0,-k^2)\). 2. The chord \(AB\) is \(y=k^2\). Thus, \(A_1=\int_{-k}^{k}(k^2-x^2)\,dx=\frac{4}{3}k^3\). 3. By symmetry, \(A_2=2\int_0^k\left[x^2-(2kx-k^2)\right] \,dx=2\int_0^k(x-k)^2\,dx=\frac{2}{3}k^3\). 4. Therefore, \(A_1{:}A_2=\frac{4}{3}k^3{:}\frac{2}{3}k^3=2{:}1\).

1. Tangent lines: \(y=-2kx-k^2\) and \(y=2kx-k^2\); \(C(0,-k^2)\) 2. \(A_1=\frac{4}{3}k^3\) 3. \(A_2=\frac{2}{3}k^3\) 4. \(A_1{:}A_2=2{:}1\)

- An even fourth-degree polynomial uses only even powers. - Use the conditions at \(x=0\) and \(x=2\). - Use the second derivative to determine the sign of the leading coefficient. - Use symmetry to express the total area.

1. Write an even fourth-degree polynomial as \(f(x)=ax^4+bx^2+c\). Since \(f(0)=0\), \(c=0\). 2. The condition \(f(2)=0\) gives \(16a+4b=0\), so \(b=-4a\). Thus, \(f(x)=ax^4-4ax^2\). 3. The second derivative is \(f''(x)=12ax^2-8a\). For a local maximum at \(x=0\), \(f''(0)=-8a<0\), so \(a>0\). 4. The zeros are \(-2, 0,\) and \(2\), and the two bounded regions are symmetric. Therefore, \(12.8=2\left|\int_0^2(ax^4-4ax^2)\,dx\right|\). 5. The integral is \(-\frac{64}{15}a\), so \(\frac{128}{15}a=12.8\). Thus, \(a=1.5\), and \(b=-6\). 6. Therefore, \(f(x)=1.5x^4-6x^2\).

\(f(x)=1.5x^4-6x^2\)

- Identify the rectangle formed by the endpoint \((\sqrt[4]{b},b)\) and its reflection. - Write the rectangle area as \(b\sqrt[4]{b}\). - Subtract the area under \(x^4\) from the rectangle. - Simplify \((\sqrt[4]{b})^5\) before combining terms.

1. The function \(g(x)=\sqrt[4]{x}\) is the inverse of \(f(x)=x^4\) on \([0,\infty)\). Their graphs and corresponding regions are reflections across \(y=x\). 2. The rectangle with side lengths \(\sqrt[4]{b}\) and \(b\) has area \(b\sqrt[4]{b}\). 3. The two reflected regions partition the rectangle, so \(\int_0^b\sqrt[4]{x}\,dx=b\sqrt[4]{b}-\int_0^{\sqrt[4]{b}}x^4\,dx\). 4. Evaluate the second integral: \(\int_0^{\sqrt[4]{b}}x^4\,dx=\frac{1}{5}(\sqrt[4]{b})^5=\frac{1}{5}b\sqrt[4]{b}\). 5. Therefore, \(\int_0^b\sqrt[4]{x}\,dx=\frac{4}{5}b\sqrt[4]{b}\).

The rectangle has area \(b\sqrt[4]{b}\), and the reflected \(x^4\) region has area \(\frac{1}{5}b\sqrt[4]{b}\). Their difference is \(\frac{4}{5}b\sqrt[4]{b}\).

- Clear the denominator, then substitute \(u=x^2\) in the intersection equation. - Use symmetry about the \(y\)-axis to calculate only the right-hand region. - On the positive interval, the downward-opening parabola is the upper graph. - Integrate the \(x^{-2}\) term carefully.

1. Set the functions equal: \(\frac{8}{x^2}=9-x^2\). After multiplying by \(x^2\), obtain \(x^4-9x^2+8=0\). Let \(u=x^2\). Then \(u^2-9u+8=0\), so \(u=1\) or \(u=8\). Thus, the intersections occur at \(x=\pm1\) and \(x=\pm2\sqrt{2}\). 2. By symmetry, calculate the right-hand region on \([1,2\sqrt{2}]\), where \(g\) is above \(f\). 3. Its area is \(\int_1^{2\sqrt{2}}\left(9-x^2-\frac{8}{x^2}\right)\,dx\). An antiderivative is \(9x-\frac{x^3}{3}+\frac{8}{x}\). 4. The right-hand area is \(\frac{44\sqrt{2}-50}{3}\). Doubling gives \(A=\frac{88\sqrt{2}-100}{3}\).

The combined area is \(\frac{88\sqrt{2}-100}{3}\approx8.15\) square units.

- Begin by finding half of the full shaded area. - For the vertical split, use an accumulation integral with upper limit \(k\). - For the horizontal split, locate the intersection of \(y=c\) with the curve. - The area below the horizontal line is a rectangle plus the tail area under the curve.

1. The total area is \(A=\int_1^4\frac{4}{x^2}\,dx=\left[-\frac{4}{x}\right]_1^4=3\). 2. For the vertical division, \(\int_1^k\frac{4}{x^2}\,dx=\frac{3}{2}\). Thus, \(4-\frac{4}{k}=\frac{3}{2}\), so \(k=\frac{8}{5}=1.6\). 3. For the horizontal division, let \(x_c\) be the intersection of \(y=c\) and the curve. Then \(c=\frac{4}{x_c^2}\), so \(x_c=\frac{2}{\sqrt{c}}\). 4. The area below \(y=c\) within the original region is \(c(x_c-1)+\int_{x_c}^4\frac{4}{x^2}\,dx=4\sqrt{c}-c-1\). Set this equal to \(\frac{3}{2}\). With \(u=\sqrt{c}\), \(u^2-4u+\frac{5}{2}=0\). The valid root is \(u=2-\frac{\sqrt{6}}{2}\), so \(c=u^2=\frac{11}{2}-2\sqrt{6}\approx0.601\).

a) \(A=3\) square units b) \(k=\frac{8}{5}=1.6\) c) \(c=\frac{11}{2}-2\sqrt{6}\approx0.601\)

- Translate the local-minimum information into value and derivative conditions. - Use the second x-intercept to relate the coefficients. - Express the Quadrant I area as a definite integral. - Verify the second-derivative condition at the end.

1. Write \(f(x)=Ax^3+Bx^2+Cx+D\). Since the origin is a stationary point on the graph, \(f(0)=0\) and \(f'(0)=0\), so \(C=D=0\). 2. The intercept condition \(f(4)=0\) gives \(64A+16B=0\), so \(B=-4A\). Thus, \(f(x)=A(x^3-4x^2)\). 3. The enclosed area is \(\int_0^4f(x)\,dx=-\frac{64A}{3}=32\), so \(A=-\frac{3}{2}\). 4. Then \(B=6\), giving \(f(x)=-\frac{3}{2}x^3+6x^2\). Also, \(f''(0)=12>0\), confirming a local minimum.

\(f(x)=-\frac{3}{2}x^3+6x^2\)

- Use only even powers because the function is even. - Use the given points to determine two coefficient relationships. - Use symmetry to halve the area calculation. - Verify the local-maximum condition.

1. Because the function is even, write \(f(x)=Ax^4+Cx^2+E\). From \(f(0)=4.5\), \(E=4.5\). 2. The intercept \(f(3)=0\) gives \(81A+9C+4.5=0\), so \(C=-9A-0.5\). 3. By symmetry, the area condition is \(2\int_0^3f(x)\,dx=16.2\). Substitution and integration give \(2(-32.4A+9)=16.2\), so \(A=\frac{1}{36}\). 4. Then \(C=-\frac{3}{4}\). Therefore, \(f(x)=\frac{x^4}{36}-\frac{3x^2}{4}+\frac{9}{2}\). Since \(f''(0)=-\frac{3}{2}<0\), the point at \(x=0\) is a local maximum.

\(f(x)=\frac{x^4}{36}-\frac{3x^2}{4}+\frac{9}{2}\)

- Write the tangent line at a general x-coordinate \(u\). - Substitute the coordinates of \(P\) into that line to determine \(u\). - Use concavity or the graph to decide which curve is above. - Integrate upper minus lower from the y-axis to the tangency point.

1. Let the point of tangency have x-coordinate \(u\). Since \(f'(x)=e^{0.5x}\), the tangent line at \(u\) is \(y=e^{0.5u}(x-u)+2e^{0.5u}\). 2. Because the tangent passes through \(P(2,0)\), \(0=e^{0.5u}(2-u)+2e^{0.5u}=e^{0.5u}(4-u)\). The exponential factor is positive, so \(u=4\). 3. The point of tangency is \(B(4,2e^2)\), and the tangent line is \(t: y=e^2x-2e^2\). 4. Since \(f\) is concave up, its graph lies above its tangent line. The enclosed area is \(A=\int_0^4\left[2e^{0.5x}-(e^2x-2e^2)\right]\,dx\). 5. Evaluate: \(A=\left[4e^{0.5x}-\frac{1}{2}e^2x^2+2e^2x\right]_0^4=4e^2-4\approx25.56\) square units.

a) \(B(4,2e^2)\); \(t: y=e^2x-2e^2\) b) \(A=4e^2-4\approx25.56\) square units

- Set the rational function equal to \(3\) to find both intersection values. - Use the graph to confirm that the horizontal line is the upper boundary. - Integrate the reciprocal terms using logarithms. - Use symmetry in the intersection values and logarithm properties when simplifying.

1. Solve \(f(x)=3\): \(\frac{-18}{x(x-6)}=3\), so \(x^2-6x+6=0\). The intersection values are \(x=3-\sqrt3\) and \(x=3+\sqrt3\). 2. The line lies above the curve between these values. Therefore, \(A=\int_{3-\sqrt3}^{3+\sqrt3}\left(3-\frac{3}{x}+\frac{3}{x-6}\right)\,dx\). 3. An antiderivative is \(3x-3\ln|x|+3\ln|x-6|\). 4. Evaluating and simplifying gives \(A=6\sqrt3-6\ln(2+\sqrt3)\approx2.491\) square units.

The area is \(6\sqrt3-6\ln(2+\sqrt3)\approx2.491\) square units.

- Differentiate \(f\) to find the tangent slope at \(x=1\). - The y-axis and the tangency point give the horizontal bounds. - Use the graph to identify the curve as the upper boundary. - Integrate the polynomial and reciprocal terms separately.

1. The derivative is \(f'(x)=\frac{1}{2}x-\frac{4}{(x+1)^2}\), so \(f'(1)=-\frac{1}{2}\). 2. The tangent line at \(P\) is \(t(x)=-\frac{1}{2}(x-1)+\frac{9}{4}=-\frac{1}{2}x+\frac{11}{4}\). 3. On \([0,1]\), the curve lies above the tangent. Hence, \(A=\int_0^1\left(\frac{1}{4}x^2+\frac{4}{x+1}+\frac{1}{2}x-\frac{11}{4}\right)\,dx\). 4. Evaluating, \(A=\left[\frac{x^3}{12}+\frac{x^2}{4}-\frac{11x}{4}+4\ln(x+1)\right]_0^1=4\ln2-\frac{29}{12}\) square units.

The exact area is \(4\ln2-\frac{29}{12}\) square units, approximately \(0.356\).

- Find the x-intercept of the rational graph and its intersection with \(y=x\). - Follow the shaded boundary to see where the lower function changes. - Split the area at \(x=-\frac{2}{3}\). - Rewrite the rational function as separate powers of \(x\) before integrating.

1. The curve and line intersect when \(\frac{x^3-3x-2}{x^2}=x\), which gives \(x=-\frac{2}{3}\). 2. The relevant x-intercept of \(f\) is \(x=-1\), and \(y=x\) meets the x-axis at \(x=0\). 3. Split the area where the lower boundary changes: \(A=\int_{-1}^{-2/3}\left[-f(x)\right]\,dx+\int_{-2/3}^{0}(-x)\,dx\). 4. Since \(f(x)=x-\frac{3}{x}-\frac{2}{x^2}\), \(A=\left[-\frac{x^2}{2}+3\ln|x|-\frac{2}{x}\right]_{-1}^{-2/3}+\left[-\frac{x^2}{2}\right]_{-2/3}^{0}=\frac{3}{2}+3\ln\left(\frac{2}{3}\right)\approx0.284\) square units.

The area is \(\frac{3}{2}+3\ln\left(\frac{2}{3}\right)\approx0.284\) square units.

- Differentiate the quotient and factor the numerator of the derivative. - Use the positive critical point as the upper integration bound. - Compare the numerator of the integrand with the derivative of \(x^3+k^3\). - Use a logarithm property to cancel the parameter.

1. Differentiate: \(g_k'(x)=\frac{x(2k^3-x^3)}{(x^3+k^3)^2}\). For \(x>0\), the derivative is zero when \(x^3=2k^3\), so \(x_M=k\sqrt[3]{2}\). 2. The area is \(A=\int_0^{k\sqrt[3]{2}}\frac{x^2}{x^3+k^3}\,dx\). 3. An antiderivative is \(\frac{1}{3}\ln(x^3+k^3)\). Therefore, \(A=\frac{1}{3}\left[\ln(3k^3)-\ln(k^3)\right]=\frac{1}{3}\ln3\). 4. This value does not depend on \(k\).

\(x_M=k\sqrt[3]{2}\), and the area is \(\frac{1}{3}\ln3\approx0.366\) square units, independent of \(k\).

- Set the numerator equal to zero. - Factor the derivative to locate the critical points. - Substitute \(u=\ln x\) when integrating. - Remember that \(\ln c<0\) for \(0<c<1\).

1. The zero is \(x=1\). 2. Since \(f'(x)=\frac{\ln x(2-\ln x)}{x^2}\), the local maximum occurs at \(x=e^2\), with value \(\frac{4}{e^2}\). 3. An antiderivative is \(F(x)=\frac{1}{3}(\ln x)^3\). Thus, \(A=\int_1^{e^2}f(x)\,dx=\frac{8}{3}\). 4. Require \(\int_c^1f(x)\,dx=-\frac{1}{3}(\ln c)^3=\frac{8}{3}\). Then \(\ln c=-2\), so \(c=e^{-2}\).

a) Zero: \(x=1\); local maximum: \(\left(e^2, \frac{4}{e^2}\right)\) b) \(A=\frac{8}{3}\) square units c) \(c=e^{-2}\approx0.135\)

- Use the known maximum of \(\sin x\) on \([0,\pi]\). - The three points determine the coefficients of the quadratic. - Use the graph to identify the quadratic as the upper boundary. - Factor out \(a\) before evaluating the area integral.

1. The maximum occurs at \(x=\frac{\pi}{2}\), so it is \(\left(\frac{\pi}{2},a+1\right)\). 2. Solving for the quadratic through the three points gives \(q_a(x)=-\frac{4a}{\pi^2}x^2+\frac{4a}{\pi}x+1\). 3. On \([0,\pi]\), \(q_a(x)\ge f_a(x)\). Therefore, \(A(a)=\int_0^\pi\left(q_a(x)-f_a(x)\right)\,dx=a\left(\frac{2\pi}{3}-2\right)\). 4. Set \(a\left(\frac{2\pi}{3}-2\right)=1\). Thus, \(a=\frac{3}{2\pi-6}\approx10.594\).

1) \(\left(\frac{\pi}{2},a+1\right)\) 2) \(q_a(x)=-\frac{4a}{\pi^2}x^2+\frac{4a}{\pi}x+1\) 3) \(A(a)=a\left(\frac{2\pi}{3}-2\right)\) 4) \(a=\frac{3}{2\pi-6}\approx10.594\)

- Set the functions equal and factor the difference. - Equal positive region areas make the signed integral of \(f-g_k\) over the full interval equal to zero. - Use integration by parts for \(\int x\sin x\,dx\). - On the first subinterval, identify which coefficient of \(\sin x\) is larger.

1. Setting the functions equal gives \((x-k)\sin x=0\), so the intersections are \(x=0,k,\pi\). 2. Equal region areas imply \(\int_0^\pi\left(f(x)-g_k(x)\right)\,dx=0\). 3. Integration by parts gives \(\int_0^\pi x\sin x\,dx=\pi\), while \(\int_0^\pi\sin x\,dx=2\). Thus, \(\pi-2k=0\), so \(k=\frac{\pi}{2}\). 4. One region has area \(A=\int_0^{\pi/2}\left(\frac{\pi}{2}-x\right)\sin x\,dx=\frac{\pi}{2}-1\) square units.

a) The intersections are \(x=0,k,\pi\). b) \(k=\frac{\pi}{2}\) c) \(A=\frac{\pi}{2}-1\approx0.571\) square units

- Find each tangent equation from the point and derivative. - Use the shaded diagram to identify every boundary and intercept. - In part a), integrate curve minus tangent. - In part b), subtract the triangular tangent contribution from the area under the curve.

1. For a), \(f(2)=5\) and \(f'(2)=4\), so the tangent is \(t(x)=4x-3\). Since \(f(x)-t(x)=(x-2)^2\), \(A=\int_0^2(x-2)^2\,dx=\frac{8}{3}\). 2. For b), \(f(1)=3\) and \(f'(1)=-8\), so the tangent is \(t(x)=-8x+11\). The curve meets the x-axis at \(x=2\), and the tangent meets it at \(x=\frac{11}{8}\). 3. The area is \(A=\int_1^2\left(\frac{4}{x^2}-1\right)\,dx-\int_1^{11/8}(-8x+11)\,dx=1-\frac{9}{16}=\frac{7}{16}\).

a) \(A=\frac{8}{3}\) square units b) \(A=\frac{7}{16}\) square unit