Free math worksheets

- Treat the constant function as having that same constant average. - Apply the coefficients to the known averages.

1. Use linearity of average value. 2. \((3f-2g+4)_{\mathrm{avg}}=3(5)-2(-1)+4\). 3. The result is \(21\).

\(21\)

- Identify the length of the interval. - Find the rectangular area under the constant graph. - Average value is accumulated area divided by interval length.

1. The area under the graph is the area of a rectangle: \(4(6-1)=20\). 2. Divide by the interval length: \(f_{\text{avg}}=\frac{1}{6-1}\int_1^6 4\,\text{d}x=\frac{20}{5}=4\).

\(4\)

- Find the area under the piecewise-linear graph using a familiar geometric shape. - Divide the area by the total time interval. - Interpret the average as the height of an equal-area rectangle.

1. The area under the graph is a triangle with base \(12\,\text{h}\) and height \(900\,\text{W/m}^2\), so the estimated energy per square meter is \(\frac{1}{2}\cdot12\cdot900=5400\,\text{W}\cdot\text{h/m}^2\). 2. Divide by the interval length: \(I_{\mathrm{avg}}=\frac{5400}{12}=450\,\text{W/m}^2\). 3. Geometrically, a rectangle over \([6,18]\) with height \(450\,\text{W/m}^2\) has the same area as the triangular region under the graph.

\(450\,\text{W/m}^2\). A rectangle of this height over \([6,18]\) has the same area as the region under the graph.

- First determine whether any part of the graph lies below the x-axis. - The signed area is the definite integral over the interval. - Average value is signed area divided by interval length.

1. The function is positive on \([-1,2]\), so the signed area is \(\int_{-1}^2(x^3-3x+4)\,\text{d}x=\frac{45}{4}\). 2. Divide the signed area by the interval length \(2-(-1)=3\): \(f_{\mathrm{avg}}=\frac{1}{3}\cdot\frac{45}{4}=\frac{15}{4}\).

a) \(\frac{45}{4}\) square units b) \(\frac{15}{4}\)

- Match the shifted interval to the original interval. - Consider separately what horizontal and vertical shifts do to an average.

1. The substitution \(u=x-4\) maps \([2,7]\) to \([-2,3]\). 2. The horizontal shift does not change the average value, while adding \(7\) raises every function value by \(7\). 3. Therefore, \(g_{\mathrm{avg}}=-1+7=6\).

\(6\)

- Use the symmetry of the interval and the function. - Separate the constant part from the odd part.

1. Because \(f\) is odd, \(\int_{-6}^{6}f(x)\,\text{d}x=0\). 2. Thus \(\int_{-6}^{6}(4+3f(x))\,\text{d}x=4\cdot12=48\). 3. Divide by \(12\) to obtain the average value \(4\).

\(4\)

- Compare the interval length with the period of the oscillating term. - A full cycle contributes equal positive and negative signed area.

1. The interval length \(\frac{\pi}{2}\) is one full period of \(\sin(4x)\). 2. The sine term has integral \(0\) over any full period. 3. The constant term therefore determines the average value, which is \(5\).

\(5\)

- Use how integration responds to sums and constant multiples. - Build one equation from the three stated averages.

1. Average value is linear, so \(h_{\mathrm{avg}}=a(3)+4(-2)\). 2. Set \(3a-8=17\). 3. Solving gives \(a=\frac{25}{3}\).

\(a=\frac{25}{3}\)

- Distinguish between total signed area and average height. - Check which quantity should appear in the denominator.

1. The definite integral gives accumulated signed area, not average height. 2. The interval length is \(9-1=8\). 3. Divide \(\frac{208}{3}\) by \(8\): the correct average value is \(\frac{26}{3}\).

The student failed to divide by the interval length. The correct average value is \(\frac{26}{3}\).

- Treat the force values as heights over the compression interval. - First find the accumulated quantity over the full interval. - Check that the final units are units of force, not work.

1. The accumulated work-related integral is \(\int_0^{2/5}(20+5x^2)\,\text{d}x=\frac{608}{75}\,\text{N}\cdot\text{m}\). 2. The interval length is \(\frac25\,\text{m}\). 3. The average force is \(\frac{608}{75}\div\frac25=\frac{304}{15}\,\text{N}\).

\(\frac{304}{15}\,\text{N}\approx20.27\,\text{N}\)

- Interpret the integral of density before averaging. - Check that average density times length has units of mass.

1. The total mass is \(\int_0^4\rho(x)\,\text{d}x=\frac{56}{3}\,\text{kg}\). 2. The average linear density is \(\frac14\cdot \frac{56}{3}=\frac{14}{3}\,\text{kg/m}\). 3. Multiplying the average density by \(4\,\text{m}\) recovers \(\frac{56}{3}\,\text{kg}\).

Average density: \(\frac{14}{3}\,\text{kg/m}\) Total mass: \(\frac{56}{3}\,\text{kg}\)

- The area under the depth profile has square units. - Average depth is the constant depth producing the same cross-sectional area.

1. The cross-sectional area is \(A=\int_0^8d(x)\,\text{d}x=\frac{56}{3}\,\text{m}^2\). 2. Divide by the width: \(d_{\mathrm{avg}}=\frac{A}{8}=\frac{7}{3}\,\text{m}\).

Average depth: \(\frac{7}{3}\,\text{m}\) Cross-sectional area: \(\frac{56}{3}\,\text{m}^2\)

- Use the endpoint values of the antiderivative directly. - Do not do extra work that the given antiderivative already replaces.

1. By the Fundamental Theorem of Calculus, \(\int_{-1}^{2}f(x)\,\text{d}x=F(2)-F(-1)\). 2. \(F(2)=14\) and \(F(-1)=-4\), so the integral is \(18\). 3. Divide by the interval length \(3\) to obtain \(6\).

\(6\)

- Connect the stated average to the total signed area over the interval. - Consider what continuity guarantees about a function attaining its average height. - Ask whether endpoint values alone determine an integral for every continuous function.

1. The interval length is \(4\), so \(\int_1^5 f(x)\,\text{d}x=4(-3)=-12\). Statement a) must be true. 2. Since \(f\) is continuous, the Mean Value Theorem for Integrals guarantees some \(c\in[1,5]\) with \(f(c)=-3\). Statement b) must be true. 3. An average value does not determine the sum of the endpoint values for a general continuous function. Statement c) need not be true.

a) Must be true. b) Must be true. c) Need not be true.

- Translate each net change into an average rate over the same interval. - Because the interval lengths match, compare the net changes first.

1. The average of \(u'\) is \(\frac{u(6)-u(1)}{5}=3\). 2. The average of \(v'\) is \(\frac{v(6)-v(1)}{5}=2\). 3. Therefore, \(u'\) has the greater average value by \(1\).

\(u'\) has the greater average value, by \(1\).

- Use the antiderivative associated with reciprocal input. - Keep the exact logarithmic form before approximating.

1. \(\int_1^4\frac1x\,\text{d}x=\ln4\). 2. Divide by \(3\): \(f_{\mathrm{avg}}=\frac{\ln4}{3}\). 3. Numerically, \(\frac{\ln4}{3}\approx 0.462\).

\(\frac{\ln4}{3}\approx 0.462\)

- Set up the average-value expression before using numerical technology. - Check that the result lies between the minimum and maximum displayed heights.

1. Evaluate the definite integral numerically: \(\int_0^2\left(e^{-x^2}+\frac{x}{3}\right)\,\text{d}x\approx 1.548748\). 2. Divide by the interval length \(2\). 3. The average value is approximately \(0.774\).

\(\approx 0.774\)

- Use the average-value formula for a function. - Find an antiderivative of the sinusoidal model. - Interpret the integral unit before dividing by time. - A sinusoid averages to zero over one full period.

1. The average on \([10,22]\) is \(\frac{1}{12}\int_{10}^{22}T(t)\,\text{d}t\). 2. An antiderivative is \(-\frac{120}{\pi}\cos\left(\frac{\pi}{12}(t-10)\right)+22t\). 3. The integral equals \(\frac{240}{\pi}+264\), so the average is \(22+\frac{20}{\pi}\approx28.37\,{}^\circ\text{C}\). 4. The expression in part b is the average temperature over the full 24-hour day. The sine term averages to zero over one full period, so the value is \(22\,{}^\circ\text{C}\).

a) \(22+\frac{20}{\pi}\,{}^\circ\text{C}\approx28.37\,{}^\circ\text{C}\) b) The average temperature over the entire day; its value is \(22\,{}^\circ\text{C}\).

- Differentiate using the product rule. - Find the zeros of the profile to determine the full base interval. - Use the average-value formula. - Multiply the cross-sectional area by the perpendicular length.

1. By the product rule, \(F'(x)=(4x+20)e^{-0.2x}+(2x^2+20x+50)(-0.2)e^{-0.2x}\). 2. Simplifying gives \(F'(x)=(10-0.4x^2)e^{-0.2x}=f(x)\), so \(F\) is an antiderivative. 3. The zeros satisfy \(10-0.4x^2=0\), so the base extends from \(x=-5\) to \(x=5\). 4. The cross-sectional area is \(A=\int_{-5}^{5}f(x)\,\text{d}x=F(5)-F(-5)=\frac{200}{e}\approx73.58\,\text{m}^2\). 5. The average height is \(\frac{A}{10}=\frac{20}{e}\approx7.36\,\text{m}\). 6. The volume is \(50A=\frac{10{,}000}{e}\approx3678.79\,\text{m}^3\).

a) \(F'(x)=f(x)\) b) \(\frac{20}{e}\,\text{m}\approx7.36\,\text{m}\) c) \(\frac{10{,}000}{e}\,\text{m}^3\approx3678.79\,\text{m}^3\)

- Use the average-value formula. - Set the function equal to the average value. - Solve the resulting equation. - Check that each solution lies in the given interval.

1. \(\int_0^3(x-1)^2\,\text{d}x=\left[\frac{(x-1)^3}{3}\right]_0^3=3\). 2. The average value is \(\frac{1}{3-0}(3)=1\). 3. Set \((x_0-1)^2=1\). Then \(x_0-1=\pm1\), so \(x_0=0\) or \(x_0=2\). 4. Both values lie in \([0,3]\).

\(x_0=0\) and \(x_0=2\)

- Find the average value first. - Set the function equal to that value. - Use the symmetry of sine on \([0,\pi]\). - Use radian mode for numerical approximations.

1. \(\int_0^\pi(\sin x+1)\,\text{d}x=[-\cos x+x]_0^\pi=\pi+2\). 2. The average value is \(\frac{\pi+2}{\pi}=1+\frac{2}{\pi}\). 3. Set \(\sin c+1=1+\frac{2}{\pi}\), so \(\sin c=\frac{2}{\pi}\). 4. On \([0,\pi]\), the solutions are \(c=\arcsin\left(\frac{2}{\pi}\right)\) and \(c=\pi-\arcsin\left(\frac{2}{\pi}\right)\). 5. Numerically, these are approximately \(0.690\) and \(2.451\).

\(c=\arcsin\left(\frac{2}{\pi}\right)\approx0.690\) and \(c=\pi-\arcsin\left(\frac{2}{\pi}\right)\approx2.451\)

- Approximate the area under the data with trapezoids. - Divide the estimated integral by the total time. - Compute the ordinary mean separately. - Compare how the two methods weight the endpoint measurements.

1. With step size \(2\,\text{h}\), the trapezoidal estimate is \(\int_0^{10}c(t)\,\text{d}t\approx\frac{2}{2}(0+2(4.2)+2(6.8)+2(5.1)+2(3.4)+2.1)=41.1\,\text{mg}\cdot\text{h/L}\). 2. The estimated average concentration is \(\frac{41.1}{10}=4.11\,\text{mg/L}\). 3. The arithmetic mean is \(\frac{0+4.2+6.8+5.1+3.4+2.1}{6}=3.6\,\text{mg/L}\). 4. The trapezoidal average is higher because the endpoint values receive half the weight of the interior values in the time-weighted estimate.

Trapezoidal average: \(4.11\,\text{mg/L}\) Arithmetic mean: \(3.6\,\text{mg/L}\)

- Use the product rule and chain rule to differentiate \(f\). - A maximum can occur where the first derivative changes from positive to negative. - To verify an antiderivative, differentiate it. - Use the average-value formula for a function on an interval.

1. The graph suggests a maximum near \(t=5\). By the product and chain rules, \(f'(t)=(10-2t)e^{-0.2t}\). Since the exponential factor is positive, \(f'(t)=0\) when \(10-2t=0\), so \(t_{\max}=5\) hours. The derivative changes from positive to negative there. 2. \(f(5)=50e^{-1}\approx 18.39\,\text{mg/L}\). 3. Differentiate \(F\): \(F'(t)=-50\left[e^{-0.2t}-0.2(t+5)e^{-0.2t}\right]=10te^{-0.2t}=f(t)\). Therefore, \(F\) is an antiderivative of \(f\). 4. The average value on \([0,20]\) is \(\frac{1}{20}\int_0^{20}f(t)\,dt=\frac{F(20)-F(0)}{20}\). Since \(F(20)=-1250e^{-4}\approx-22.89\) and \(F(0)=-250\), the average is \(\frac{-22.89+250}{20}\approx 11.36\,\text{mg/L}\).

a) \(t_{\max}=5\) hours b) About \(18.39\,\text{mg/L}\) c) \(F'(t)=f(t)\), so \(F\) is an antiderivative of \(f\) d) About \(11.36\,\text{mg/L}\)

- Use the maximum value of sine. - Energy is the integral of power over time. - Divide the total energy by the interval length to find average power.

1. The sine function reaches its maximum when \(\frac{\pi}{12}t=\frac{\pi}{2}\), so \(t=6\). The maximum is \(P(6)=800\,\text{W}\), at noon. 2. The energy is \(E=\int_0^{12}800\sin\left(\frac{\pi}{12}t\right)\,\text{d}t\). 3. \(E=\left[-\frac{9600}{\pi}\cos\left(\frac{\pi}{12}t\right)\right]_0^{12}=\frac{19{,}200}{\pi}\approx6111.55\,\text{Wh}\). 4. The average power is \(\frac{E}{12}=\frac{1600}{\pi}\approx509.30\,\text{W}\).

a) \(800\,\text{W}\) at noon b) \(\frac{19{,}200}{\pi}\,\text{Wh}\approx6111.55\,\text{Wh}\) c) \(\frac{1600}{\pi}\,\text{W}\approx509.30\,\text{W}\)

- Write the average-value formula on \([0,3]\). - Factor the constant \(k\) outside the integral. - Set the resulting average equal to \(12\) and solve for \(k\).

1. \(\int_0^3(x^2+2x)\,\text{d}x=18\). 2. The average value of \(f_k\) is \(\frac{k}{3}\cdot 18\). 3. Solve \(\frac{k}{3}\cdot 18=12\), giving \(k=2\).

\(k=2\)

- For a linear function, relate the average height to the two endpoint heights. - Substitute the endpoint values \(x=1\) and \(x=b\). - Set the endpoint mean equal to \(10\) and solve for \(b\).

1. For a linear function, the average value over an interval equals the mean of the endpoint values. 2. Set \(\frac{f(1)+f(b)}{2}=10\), so \(\frac{4+(3b+1)}{2}=10\). 3. Solving gives \(3b+5=20\), hence \(b=5\).

\(b=5\)

- Identify where the expression inside the absolute value changes sign. - Treat the two parts of the interval separately before dividing by the total length.

1. Split at \(x=2\): \(|x-2|=2-x\) on \([0,2]\) and \(|x-2|=x-2\) on \([2,5]\). 2. The total integral is \(\int_0^2(2-x)\,\text{d}x+\int_2^5(x-2)\,\text{d}x=\frac{13}{2}\). 3. Divide by \(5\): \(f_{\mathrm{avg}}=\frac{13}{10}\).

\(\frac{13}{10}\)

- Convert each known average into an accumulated value over its interval. - Use the way adjacent definite integrals combine. - The requested interval has a different length from either given interval.

1. The integral over \([0,4]\) is \(4\cdot5=20\). 2. The integral over \([0,10]\) is \(10\cdot\frac{16}{5}=32\). 3. Therefore, the integral over \([4,10]\) is \(32-20=12\). 4. Divide by the interval length \(6\) to obtain an average value of \(2\).

\(2\)

- Distinguish an average over every point of the interval from an average of only two values. - Use the interval length when converting the integral to an average. - Compare the exact expressions before using decimals.

1. The average value is \(\frac{2}{\pi}\int_0^{\pi/2}\cos x\,\text{d}x=\frac{2}{\pi}\). 2. The arithmetic mean of the endpoint values is \(\frac{f(0)+f(\pi/2)}{2}=\frac12\). 3. Since \(4>\pi\), \(\frac{2}{\pi}>\frac12\). 4. The difference is \(\frac{2}{\pi}-\frac12=\frac{4-\pi}{2\pi}\).

The average value is \(\frac{2}{\pi}\), which exceeds the endpoint mean \(\frac12\) by \(\frac{4-\pi}{2\pi}\).

- Separate the known part of the function from the unknown constant. - A vertical translation changes every height by the same amount.

1. The average of \(2x^2-x\) on \([-1,2]\) is \(\frac13\int_{-1}^2(2x^2-x)\,\text{d}x=\frac{3}{2}\). 2. Adding \(c\) adds \(c\) to the average. 3. Solve \(\frac{3}{2}+c=5\), yielding \(c=\frac{7}{2}\).

\(c=\frac{7}{2}\)

- Compare the functions point by point on the interval. - Look at the sign of the added or subtracted term.

1. On \([0,2]\), \(x(2-x)\ge0\), and it is positive except at the endpoints. 2. Thus \(h(x)\le f(x)\le g(x)\) throughout the interval, with strict inequality on the interior. 3. Integration and division by the same positive interval length preserve the order, so \(h_{\mathrm{avg}}<f_{\mathrm{avg}}<g_{\mathrm{avg}}\).

\(h_{\mathrm{avg}}<f_{\mathrm{avg}}<g_{\mathrm{avg}}\)

- Use the rule that applies on each subinterval. - Combine the two accumulated contributions before averaging.

1. Split the integral at \(x=2\). 2. \(\int_0^2(2x+1)\,\text{d}x=6\) and \(\int_2^5(7-x)\,\text{d}x=\frac{21}{2}\). 3. The total integral is \(\frac{33}{2}\). 4. Divide by \(5\) to get \(\frac{33}{10}\).

\(\frac{33}{10}\)

- Compute signed area separately on each interval. - Both intervals have the same length, so their signed areas can also be compared directly.

1. On \([0,4]\), the graph forms a trapezoid with area \(\frac12(1+5)(4)=12\), so the average is \(3\). 2. On \([4,8]\), the signed area is \(\frac12(5+1)(2)+\frac12(1-3)(2)=6-2=4\), so the average is \(1\). 3. Therefore, \([0,4]\) has the greater average value.

Average on \([0,4]\): \(3\) Average on \([4,8]\): \(1\) The interval \([0,4]\) has the greater average.

- Factor the unknown constant out of the integral. - The interval length happens to simplify the average-value denominator.

1. The average is \(\int_0^1ae^x\,\text{d}x=a(e-1)\), since the interval length is \(1\). 2. Set \(a(e-1)=10\). 3. Therefore, \(a=\frac{10}{e-1}\).

\(a=\frac{10}{e-1}\)

- Use the symmetry of the absolute-value graph. - The negative half-cycle is reflected above the x-axis.

1. \(|\sin x|\) has two identical positive arches on \([0,2\pi]\). 2. The total integral is \(2\int_0^\pi\sin x\,\text{d}x=4\). 3. Divide by \(2\pi\): \(f_{\mathrm{avg}}=\frac{2}{\pi}\).

\(\frac{2}{\pi}\)

- Assign a sign to each geometric region. - Use the full width from the first endpoint to the last.

1. The semicircle contributes signed area \(\frac12\pi(2)^2=2\pi\). 2. The triangle contributes signed area \(-\frac12(4)(3)=-6\). 3. The total signed area is \(2\pi-6\), and the interval length is \(8\). 4. The average value is \(\frac{2\pi-6}{8}=\frac{\pi-3}{4}\).

\(\frac{\pi-3}{4}\)

- Convert each local average into an accumulated amount. - The two intervals have different lengths, so do not simply average \(7\) and \(3\).

1. The integral over \([0,2]\) is \(7\cdot2=14\). 2. The integral over \([2,8]\) is \(3\cdot6=18\). 3. The integral over \([0,8]\) is \(32\). 4. Divide by \(8\) to obtain an average value of \(4\).

\(4\)

- Recognize the given arc length as the integral of the stated function. - Apply the average-value idea over the horizontal interval.

1. By definition, \(\int_2^8g(x)\,\text{d}x=15\). 2. The interval length is \(8-2=6\). 3. The average value is \(\frac{15}{6}=\frac52\). It is the average amount of curve length per unit of horizontal change.

\(\frac52\), meaning an average of \(2.5\) units of curve length per horizontal unit.

- Split the absolute-value expression where its sign changes. - After forming the average, look for symmetry or minimize the resulting quadratic.

1. Split at \(x=m\): \(A(m)=\frac16\left[\int_0^m(m-x)\,\text{d}x+\int_m^6(x-m)\,\text{d}x\right]\). 2. This simplifies to \(A(m)=\frac{m^2+(6-m)^2}{12}\). 3. The quadratic is minimized at \(m=3\). 4. The minimum value is \(A(3)=\frac32\).

The minimizing value is \(m=3\), and the minimum average value is \(\frac32\).

- Write the average with a variable interval length. - Simplify the difference of cubes before solving. - Use the stated restriction to select the valid root.

1. \(\frac{1}{3-a}\int_a^3x^2\,\text{d}x=\frac{27-a^3}{3(3-a)}=\frac{a^2+3a+9}{3}\). 2. Set \(\frac{a^2+3a+9}{3}=\frac{13}{3}\), so \(a^2+3a-4=0\). 3. The roots are \(a=1\) and \(a=-4\); only \(a=1\) satisfies \(0\le a<3\).

\(a=1\)

- Write a separate average-value expression for each subinterval. - The symmetry of the function can help you anticipate the split point.

1. The average on \([0,c]\) is \(\frac{1}{c}\int_0^c(x-2)^2\,\text{d}x=\frac{c^2}{3}-2c+4\). 2. The average on \([c,4]\) is \(\frac{1}{4-c}\int_c^4(x-2)^2\,\text{d}x=\frac{c^2}{3}-\frac{2c}{3}+\frac43\). 3. Equating the averages gives \(-\frac43(c-2)=0\), so \(c=2\).

\(c=2\)

- Convert the known average into a definite integral. - Look for a way to transfer the derivative from one factor to the other. - After finding the integral of the new function, account for the interval length.

1. The given average implies \(\int_0^2f(x)\,\text{d}x=2\cdot3=6\). 2. Integration by parts gives \(\int_0^2xf'(x)\,\text{d}x=[xf(x)]_0^2-\int_0^2f(x)\,\text{d}x\). 3. Thus \(\int_0^2h(x)\,\text{d}x=2f(2)-6=10-6=4\). 4. Divide by \(2\) to obtain the average value \(2\).

\(2\)

- Pair inputs that are equally far from the midpoint. - The transformed integral covers the same interval in reverse.

1. Integrate the identity over \([0,10]\): \(\int_0^{10}f(x)\,\text{d}x+\int_0^{10}f(10-x)\,\text{d}x=140\). 2. By the substitution \(u=10-x\), the two integrals are equal. 3. Hence \(2\int_0^{10}f(x)\,\text{d}x=140\), so the integral is \(70\). 4. Divide by \(10\) to obtain the average value \(7\).

\(7\)

- Rewrite the average statement as an equation for accumulated area. - Think about how an accumulation changes when its upper endpoint changes. - Use the given formula only after expressing the accumulation itself.

1. The average-value statement gives \(\int_0^c f(x)\,\text{d}x=cA(c)=c^3+c\). 2. Differentiate both sides with respect to \(c\): \(f(c)=3c^2+1\). 3. Therefore, \(f(2)=3(2)^2+1=13\).

\(13\)