Free math worksheets

- Read the constant function value from the graph. - Find the width and height of the rectangle. - Use a negative signed area for a region below the x-axis.

1. The graph is the horizontal line \(y=-2\), so \(f(x)=-2\). 2. On \([-2, 3]\), the region between the graph and the x-axis is a rectangle below the axis. 3. Its width is \(3-(-2)=5\), and its height is \(2\). 4. Because it lies below the x-axis, its signed contribution is \(-5\cdot 2=-10\).

\(\int_{-2}^{3}f(x)\,\text{d}x=-10\)

- Read the function value from the graph. - Find the width of the integration interval. - Interpret the region as a rectangle.

1. The graph is the horizontal line \(y=1.5\), so the function is constant and positive. 2. The interval width is \(1-(-3)=4\). 3. The region is a rectangle with width \(4\) and height \(1.5\), so the integral is \(4\cdot 1.5=6\).

\(\int_{-3}^{1}f(x)\,\text{d}x=6\)

- Read the left and right boundaries of the shaded region from the x-axis. - Keep the displayed curve name \(f\) as the integrand. - Use one definite integral over the entire shaded interval.

1. The shaded region begins at \(x=0\) and ends at \(x=2\). 2. A definite integral of \(f\) over those bounds represents the net signed area. 3. Therefore the required expression is \(\int_0^2 f(x)\,\text{d}x\).

\(\int_0^2 f(x)\,\text{d}x\)

- Read the left and right endpoints of the marked region from the graph. - Net signed area uses the function itself as the integrand. - One definite integral over the whole marked interval combines regions above and below the x-axis automatically.

1. The graph gives the function \(f(x)=\frac{1}{2}x-1\). 2. The orange marks show that the region extends from \(x=0\) to \(x=4\). 3. Therefore, the net signed area is represented by \(\int_{0}^{4}\left(\frac{1}{2}x-1\right)\,\text{d}x\). The definite integral automatically combines the negative and positive contributions.

\(\int_{0}^{4}\left(\frac{1}{2}x-1\right)\,\text{d}x\)

- Check whether any limit of integration changes when \(x\) changes. - The dummy variable inside the integral disappears after integration. - Constant bounds produce a fixed value, while variable bounds can produce a varying value.

1. a) Both bounds are constants, so the expression is a single number. 2. b) The upper bound depends on \(x\), so the value varies with \(x\); it is a function. 3. c) Both bounds depend on \(x\), so the value also varies with \(x\); it is a function.

a) A single number b) A function of \(x\) c) A function of \(x\)

- Locate the point where the defining rule changes. - Use that point as the shared endpoint of two adjacent intervals. - Preserve the original direction from the lower bound to the upper bound.

1. The formula changes at \(x=1\), so split the interval there. 2. Add the accumulated values over the adjacent intervals. 3. The result is \(\int_{-2}^{1}f(x)\,\text{d}x+\int_{1}^{5}f(x)\,\text{d}x\).

\(\int_{-2}^{1}f(x)\,\text{d}x+\int_{1}^{5}f(x)\,\text{d}x\)

- Use position as the variable of integration because the density depends on position. - Match the section's starting and ending positions to the bounds. - Multiply the integrand's unit by the differential's unit. - The result is an accumulated amount, not a density.

1. a) Accumulate the linear mass density over the coated positions: \(\int_{0.1}^{0.4}m(x)\,\text{d}x\). 2. b) Grams per meter multiplied by meters gives grams.

a) \(\int_{0.1}^{0.4}m(x)\,\text{d}x\) b) Grams

- Treat the differential as carrying the unit of the independent variable. - Cancel only the unit that appears in both the rate denominator and the interval measure. - Distinguish concentration from total amount of substance.

1. The integrand unit is milligrams per liter per minute. 2. Multiplying by the time unit from \(\text{d}t\) cancels minutes. 3. The integral is measured in milligrams per liter and represents a net change in concentration, not a total mass.

The integral has units of \(\text{mg/L}\). The student incorrectly dropped the “per liter” part of the concentration-rate unit.

- Convert both clock times to elapsed hours from the stated reference time. - Use the earlier elapsed time as the lower limit and the later one as the upper limit. - Combine the power unit with the time unit from the differential.

1. At \(10{:}30\) a.m., \(t=1.5\). At \(2{:}00\) p.m., \(t=5\). 2. Energy is accumulated power, so the expression is \(\int_{1.5}^{5}P(t)\,\text{d}t\). 3. Kilowatts multiplied by hours give kilowatt-hours.

\(\int_{1.5}^{5}P(t)\,\text{d}t\), measured in \(\text{kWh}\).

- Read both vertical boundaries of the shaded region from the x-axis. - Read the line's height at each boundary from the grid. - Use the two vertical heights as the parallel sides of a trapezoid.

1. The graph shows vertical boundaries at \(x=-1\) and \(x=4\), so the integral is \(\int_{-1}^{4}f(x)\,\text{d}x\). 2. The endpoint heights are \(f(-1)=2.5\) and \(f(4)=5\), and the horizontal width is \(5\). 3. The trapezoid area is \(\frac{2.5+5}{2}\cdot5=18.75\). 4. The graph is above the x-axis, so this geometric area equals the definite integral.

The graph gives side lengths \(2.5\) and \(5\) and width \(5\). Thus \(\int_{-1}^{4}f(x)\,\text{d}x=\frac{2.5+5}{2}\cdot5=18.75\).

- The differential identifies the variable of integration. - Treat all other letters as constants. - Use the power rule in the selected variable. - Integrating a constant over an interval multiplies it by the interval length.

1. For a), integrate with respect to \(x\): \(\int(4x^3y-2y^2)\,\text{d}x=x^4y-2xy^2\). Evaluating from \(1\) to \(3\) gives \(80y-4y^2\). 2. For b), integrate with respect to \(y\): \(\int(4x^3y-2y^2)\,\text{d}y=2x^3y^2-\frac{2}{3}y^3\). Evaluating from \(1\) to \(3\) gives \(16x^3-\frac{52}{3}\). 3. For c), \(a^2\) is constant with respect to \(x\): \(\int_0^1a^2\,\text{d}x=a^2\).

a) \(80y-4y^2\) b) \(16x^3-\frac{52}{3}\) c) \(a^2\)

- The differential identifies the variable of integration. - Treat every other letter as a constant. - After applying the bounds, the result should not contain the variable of integration. - Distinguish \(e^z w\) from \(e^{zw}\).

1. For a), integrate with respect to \(t\): \(\left[\frac{t^4}{4}\right]_0^k=\frac{k^4}{4}\). 2. For b), integrate with respect to \(k\): \(\left[\frac{k^4}{4}\right]_0^t=\frac{t^4}{4}\). 3. For c), \(w\) is constant with respect to \(z\): \(w[e^z]_0^1=w(e-1)\). 4. For d), \(e^z\) is constant with respect to \(w\): \(e^z\left[\frac{w^2}{2}\right]_0^1=\frac{e^z}{2}\).

a) \(\frac{k^4}{4}\) b) \(\frac{t^4}{4}\) c) \(w(e-1)\) d) \(\frac{e^z}{2}\)

- Interpret the definite integral as net signed area. - Identify which regions lie above and below the x-axis. - Divide the regions into triangles or other simple shapes. - Find the x-intercepts before combining the signed areas.

1. Split the interval at the zeros \(x=-2\) and \(x=3\). 2. On \([-4, -2]\), the graph forms a triangle below the x-axis with base \(2\) and height \(2\). Its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 2=-2\). 3. On \([-2, 3]\), the graph forms a triangle above the x-axis with base \(5\) and height \(2\). Its contribution is \(\frac{1}{2}\cdot 5\cdot 2=5\). 4. On \([3, 4]\), the graph forms a triangle below the x-axis with base \(1\) and height \(2\). Its signed contribution is \(-\frac{1}{2}\cdot 1\cdot 2=-1\). 5. Therefore, \(\int_{-4}^{4}f(x)\,\text{d}x=-2+5-1=2\).

\(2\)

- Interpret the definite integral as net signed area. - Separate regions above and below the x-axis. - Divide the regions into triangles and rectangles. - Combine the areas with the correct signs.

1. Divide the regions between the graph and the x-axis into triangles and a rectangle. 2. Above the x-axis, the triangle on \([-3, 0]\) has area \(\frac{1}{2}\cdot 3\cdot 2=3\), and the triangle on \([4, 5]\) has area \(\frac{1}{2}\cdot 1\cdot 2=1\). The total positive contribution is \(4\). 3. Below the x-axis, the triangle on \([0, 1]\) has area \(1\), the rectangle on \([1, 3]\) has area \(4\), and the triangle on \([3, 4]\) has area \(1\). The total magnitude of the negative contribution is \(6\). 4. Therefore, \(\int_{-3}^{5}f(x)\,\text{d}x=4-6=-2\).

\(-2\)

- Compare the horizontal boundaries in each figure with the limits of integration. - Identify whether each graph is linear, quadratic, or reciprocal. - Check key features such as an intercept, a vertex, or points on the graph. - Distinguish a shifted parabola from \(y=x^2\).

1. Figure a) shows a line with \(y\)-intercept \(1\), slope \(0.5\), and boundaries at \(x=0\) and \(x=4\). It matches integral (1). 2. Figure b) shows an upward-opening parabola with vertex \((2, 0)\), so its equation is \(y=(x-2)^2\). The boundaries are \(x=0\) and \(x=2\), so it matches integral (2). 3. Figure c) shows a reciprocal graph through \((1, 4)\) and \((4, 1)\). For \(y=\frac{k}{x}\), the point \((1, 4)\) gives \(k=4\). The boundaries are \(x=1\) and \(x=4\), so it matches integral (3).

a) Integral (1) b) Integral (2) c) Integral (3)

- Identify the regions above and below the x-axis. - Find the dimensions of the two triangles. - Add their signed areas.

1. The graph crosses the x-axis at \(x=2\). 2. On \([-2, 2]\), the graph forms a triangle above the axis with base \(4\) and height \(2\). Its contribution is \(\frac{1}{2}\cdot 4\cdot 2=4\). 3. On \([2, 4]\), the graph forms a triangle below the axis with base \(2\) and height \(1\). Its signed contribution is \(-\frac{1}{2}\cdot 2\cdot 1=-1\). 4. The integral is \(4-1=3\).

\(\int_{-2}^{4}f(x)\,\text{d}x=3\)

- Divide the region into triangles. - Find where the graph changes sign. - Add the signed areas above and below the x-axis.

1. The graph has zeros at \(x=-3\) and \(x=1\), with a maximum at \((-1, 2)\). 2. On \([-3, 1]\), the graph forms a triangle above the x-axis with base \(4\) and height \(2\). Its area is \(\frac{1}{2}\cdot 4\cdot 2=4\). 3. On \([1, 2]\), the graph forms a triangle below the axis with base \(1\) and height \(1\). Its signed contribution is \(-\frac{1}{2}\cdot 1\cdot 1=-0.5\). 4. Therefore, \(\int_{-3}^{2}f(x)\,\text{d}x=4-0.5=3.5\).

\(\int_{-3}^{2}f(x)\,\text{d}x=3.5\)

- Find the x-intercept that separates the positive and negative regions. - Use signs when evaluating the definite integral. - Use positive magnitudes when finding total geometric area. - Compare subtraction with addition of the same two triangular areas.

1. a) The graph crosses the x-axis at \(x=1\). On \([-2, 1]\), the triangle above the axis has area \(\frac12\cdot3\cdot3=4.5\). On \([1, 3]\), the triangle below the axis has area \(\frac12\cdot2\cdot2=2\), so its signed contribution is \(-2\). Therefore, the integral is \(4.5-2=2.5\). 2. b) Total geometric area counts both regions positively: \(4.5+2=6.5\). 3. c) A definite integral uses signed area, so the region below the x-axis is subtracted. Total geometric area uses the magnitude of every region.

a) \(2.5\) b) \(6.5\) square units c) The integral subtracts area below the x-axis, while total geometric area counts it positively.

- Split the first integral where the graph crosses the x-axis. - Use a negative contribution for the region below the axis. - Recall the property that interchanging the lower and upper limits changes the sign. - The geometric regions do not change when the limits are reversed; only the orientation changes.

1. a) The graph crosses the x-axis at \(x=1\). The triangle below the axis on \([-1, 1]\) has signed area \(-\frac12\cdot2\cdot2=-2\). The triangle above the axis on \([1, 2]\) has area \(\frac12\cdot1\cdot1=0.5\). Thus, \(\int_{-1}^{2}f(x)\,\text{d}x=-2+0.5=-1.5\). 2. b) Reversing the limits changes the sign of a definite integral. Therefore, \(\int_{2}^{-1}f(x)\,\text{d}x=1.5\).

a) \(-1.5\) b) \(1.5\); reversing the limits negates the integral.

- Use the graph's symmetry to reduce the work. - Identify the positive and negative triangular regions. - Check whether their signed contributions cancel.

1. The graph is symmetric about the y-axis and crosses the x-axis at \(x=-1\) and \(x=1\). 2. By symmetry, calculate the integral on \([0, 2]\) and double it. 3. On \([0, 1]\), the graph forms a triangle below the axis with base \(1\) and height \(1\), contributing \(-0.5\). 4. On \([1, 2]\), the graph forms a congruent triangle above the axis, contributing \(0.5\). 5. Thus, \(\int_{0}^{2}f(x)\,\text{d}x=-0.5+0.5=0\), and \(\int_{-2}^{2}f(x)\,\text{d}x=2\cdot 0=0\).

The integral is \(0\).

- Every complete integral needs a differential that matches its integration variable. - Keep the free variable outside the integrand’s dummy role when it also appears as a bound. - Translate the starting and ending x-values directly into lower and upper limits.

1. a) Include the differential: \(F(x)=\int_0^x(t^2+1)\,\text{d}t\). 2. b) Use a dummy variable distinct from the upper-limit variable: \(G(x)=\int_1^xg(t)\,\text{d}t\). 3. c) The signed area is represented by \(\int_{-2}^{3}h(x)\,\text{d}x\).

a) \(F(x)=\int_0^x(t^2+1)\,\text{d}t\) b) \(G(x)=\int_1^xg(t)\,\text{d}t\) c) \(\int_{-2}^{3}h(x)\,\text{d}x\)

- Distinguish a net change from a final amount that includes the value at the starting time. - Look for the expression that normalizes by the length of the time interval. - Use the bounds to identify the time span represented.

1. A accumulates the signed rate over \([2, 7]\), so A matches 2. 2. B adds the accumulated change from \(0\) to \(7\) to the volume at \(t=0\), so B matches 1. 3. C divides the accumulated change by the interval length \(5\), so C matches 3.

A–2 B–1 C–3

- Decide whether the requested quantity keeps or removes the sign of regions below the x-axis. - Split the interval at the known sign-change point. - Use the sign information to convert the negative signed contribution into positive geometric area.

1. The net signed area is \(\int_0^5f(x)\,\text{d}x\). 2. The region on \([0, 2]\) contributes positively, while the integral on \([2, 5]\) is nonpositive. 3. The total geometric area is \(\int_0^2f(x)\,\text{d}x-\int_2^5f(x)\,\text{d}x\).

a) \(\int_0^5f(x)\,\text{d}x\) b) \(\int_0^2f(x)\,\text{d}x-\int_2^5f(x)\,\text{d}x\)

- Use the stated starting and ending x-values as the direct bounds. - Split an integral by using the same interior point as the upper bound of one part and the lower bound of the next. - Reversing the order of integration changes the sign. - Check that all three equations describe the same oriented accumulation.

1. a) The direct equation is \(\int_3^9f(x)\,\text{d}x=42\). 2. b) Additivity over adjacent intervals gives \(\int_3^6f(x)\,\text{d}x+\int_6^9f(x)\,\text{d}x=42\). 3. c) Reversing the limits changes the sign, so \(\int_9^3f(x)\,\text{d}x=-42\).

a) \(\int_3^9f(x)\,\text{d}x=42\) b) \(\int_3^6f(x)\,\text{d}x+\int_6^9f(x)\,\text{d}x=42\) c) \(\int_9^3f(x)\,\text{d}x=-42\)

- Use the time interval's endpoints as integration bounds. - Account for the stated monitoring endpoint when interpreting “after.” - A set of two separate time intervals requires two integrals. - Because the rate is nonnegative, each integral represents an amount transferred.

1. a) The amount transferred between the two times is \(\int_{1.2}^{2}r(t)\,\text{d}t\). 2. b) The monitoring period ends at \(t=5\), so the amount after \(t=2\) is \(\int_2^5r(t)\,\text{d}t\). 3. c) The requested times are \([0, 1.2]\) and \([2, 5]\). Add the two accumulations: \(\int_0^{1.2}r(t)\,\text{d}t+\int_2^5r(t)\,\text{d}t\).

a) \(\int_{1.2}^{2}r(t)\,\text{d}t\) b) \(\int_2^5r(t)\,\text{d}t\) c) \(\int_0^{1.2}r(t)\,\text{d}t+\int_2^5r(t)\,\text{d}t\)

- Use the graph to identify which rate is higher throughout the displayed interval. - Read the interval endpoints from the horizontal axis. - Accumulating the higher rate minus the lower rate gives the difference in total production.

1. The graph shows that \(f(x)>g(x)\) throughout the displayed interval, which runs from \(x=0\) to \(x=4\). 2. At time \(x\), the difference in production rates is \(f(x)-g(x)\). 3. Accumulating that rate difference over the displayed interval gives \(\int_0^4[f(x)-g(x)]\,\text{d}x\).

\(\int_0^4[f(x)-g(x)]\,\text{d}x\)

- Identify the upper and lower boundaries of a vertical slice in the shaded region. - The signed integral of a nonpositive function is not itself a positive area. - Write the area either as top minus bottom or by negating the signed integral.

1. Since \(f(x)\le0\) on \([1, 4]\), the signed integral \(\int_1^4f(x)\,\text{d}x\) is nonpositive. 2. The positive geometric area is its negative: \(-\int_1^4f(x)\,\text{d}x\). 3. Equivalently, use top minus bottom: \(\int_1^4[0-f(x)]\,\text{d}x\).

\(-\int_1^4f(x)\,\text{d}x\) and \(\int_1^4[0-f(x)]\,\text{d}x\)

- Replace the free variable in every bound before considering their order. - Keep the dummy variable unchanged because its name does not depend on the parameter. - Do not automatically swap numerical bounds when the defining formula produces them in decreasing order.

1. Substitute \(s=3\) into both bounds: \(F(3)=\int_3^6p(u)\,\text{d}u\). 2. Substitute \(s=-1\): the lower bound is \(-1\) and the upper bound is \(-2\). 3. Thus \(F(-1)=\int_{-1}^{-2}p(u)\,\text{d}u\); the decreasing order is part of the definition and need not be reordered.

\(F(3)=\int_3^6p(u)\,\text{d}u\) \(F(-1)=\int_{-1}^{-2}p(u)\,\text{d}u\)

- Treat disconnected time windows as separate intervals. - Add the accumulated quantities from the requested intervals. - Check what extra times would be included by one continuous pair of bounds.

1. Accumulate power separately over the two requested windows and add: \(\int_1^3P(t)\,\text{d}t+\int_6^8P(t)\,\text{d}t\). 2. The integral from \(1\) to \(8\) also includes energy used during the unrequested gap \([3, 6]\). 3. Therefore, it represents a different time set unless the power in the gap is known to contribute zero.

a) \(\int_1^3P(t)\,\text{d}t+\int_6^8P(t)\,\text{d}t\) b) \(\int_1^8P(t)\,\text{d}t\) incorrectly includes the interval \([3, 6]\).

- Use only whether the graph is above or below the x-axis for part a. - For part b, compare the sizes of the two displayed regions rather than trying to find an antiderivative. - Net signed area subtracts the region below the x-axis from the region above it.

1. On \([0,2]\), the graph lies below the x-axis except at \(x=2\), so \(\int_0^2 f(x)\,\text{d}x<0\). 2. On \([0,4]\), the negative region is from \(0\) to \(2\) and the positive region is from \(2\) to \(4\). The displayed positive region is larger in area than the negative region, so the net signed area is positive. 3. Therefore \(\int_0^4 f(x)\,\text{d}x>0\).

a) Negative. b) Positive; the positive signed-area region on \([2,4]\) is larger than the negative signed-area region on \([0,2]\).

- Break the displayed region where the graph changes direction or crosses the x-axis. - Identify the rectangle and the two triangles directly from the grid. - Treat the triangle below the x-axis as a negative contribution.

1. From \(x=-2\) to \(x=1\), the graph forms a rectangle of width \(3\) and height \(2\), contributing \(6\). 2. From \(x=1\) to \(x=3\), the graph forms a triangle above the x-axis with base \(2\) and height \(2\), contributing \(2\). 3. From \(x=3\) to \(x=5\), the graph forms a congruent triangle below the x-axis, contributing \(-2\). 4. The net signed area is \(6+2-2=6\).

Rectangle: \(6\); triangle above the x-axis: \(2\); triangle below the x-axis: \(-2\). Therefore \(\int_{-2}^{5}f(x)\,\text{d}x=6+2-2=6\).

- Identify the horizontal and slanted line segments in the graph. - Find where the graph crosses the x-axis. - Divide the signed region into rectangles and triangles. - Treat areas below the x-axis as negative contributions.

1. The segment from \(B(2, 2)\) to \(C(5, -1)\) has slope \(-1\), so its equation is \(y=-x+4\). It crosses the x-axis at \(x=4\). 2. Above the x-axis, the rectangle from \(x=0\) to \(x=2\) contributes \(2\cdot 2=4\), and the triangle from \(x=2\) to \(x=4\) contributes \(\frac{1}{2}\cdot 2\cdot 2=2\). 3. Below the x-axis, the triangle from \(x=4\) to \(x=5\) has area \(\frac{1}{2}\cdot 1\cdot 1=0.5\), and the rectangle from \(x=5\) to \(x=7\) has area \(2\cdot 1=2\). These contribute negatively. 4. The net signed area is \(4+2-0.5-2=3.5\).

\(\int_{0}^{7}f(x)\,\text{d}x=3.5\)

- Relate each definite integral to signed area between the graph and the x-axis. - Identify where the graph is above and below the axis. - Bound curved regions with triangles or rectangles. - Interpret the integral of the constant function \(1\) geometrically.

1. For a), the positive region on \([-2, 0]\) is smaller than the negative region on \([0, 3]\). The net signed area is negative, so the statement is true. Numerically, the integral is about \(-5.21\). 2. For b), the region below the x-axis contains the triangle with vertices \((0, 0)\), \((2, -4)\), and \((3, 0)\), whose area is \(\frac{1}{2}\cdot 3\cdot 4=6\). The curve lies below the two slanted sides of this triangle, so the magnitude of the negative area is greater than \(6\). Thus, \(\int_{0}^{3}f(x)\,\text{d}x<-6\), and the statement is false. 3. For c), \(\int_{-2}^{0}1\,\text{d}x=2\). The region under \(f\) contains the triangle with vertices \((-2, 0)\), \((-1, 2)\), and \((0, 0)\), which has area \(2\). Since the graph lies above the triangle's slanted sides, \(\int_{-2}^{0}f(x)\,\text{d}x>2\). The statement is true.

a) True. The negative signed area on \([0, 3]\) has greater magnitude than the positive signed area on \([-2, 0]\). b) False. In fact, \(\int_{0}^{3}f(x)\,\text{d}x<-6\). c) True. \(\int_{-2}^{0}f(x)\,\text{d}x>2=\int_{-2}^{0}1\,\text{d}x\).

- Read the rectangle widths and heights from the grid. - Add the three rectangle areas before comparing them with the region under the curve. - A strict inequality requires the graph to be above the rectangle tops on more than isolated points.

1. The central rectangle has width \(4\) and height \(1.2\), so its area is \(4.8\). 2. Each outer rectangle has width \(1\) and height \(0.6\), so together they contribute \(1.2\). 3. The three rectangles therefore have total area \(6\). 4. Each outlined rectangle lies below the graph on its interval, with the curve strictly above the rectangle top except at isolated points. Therefore the area under the graph is strictly greater than \(6\), so \(\int_{-3}^{3}f(x)\,\text{d}x>6\).

The rectangles have total area \(4(1.2)+2(1)(0.6)=6\). Because all three rectangles lie below the graph and the curve is strictly above their tops on nonzero-length portions, \(\int_{-3}^{3}f(x)\,\text{d}x>6\).

- Use the differential to identify the symbol that is bound by the integral. - A symbol not matched by the differential remains available outside the integration process. - When renaming a dummy variable, change every bound occurrence of that symbol and the differential together.

1. The symbol \(t\) is the variable of integration. Both occurrences of \(x\) are free: one appears in the upper bound and one acts as a parameter in the integrand. 2. Renaming only the dummy variable gives \(H(x)=\int_0^x(x+u)\,\text{d}u\).

a) \(t\) is the variable of integration; \(x\) is free in both the bound and the integrand. b) \(H(x)=\int_0^x(x+u)\,\text{d}u\)

- Decide which symbol should vary during the accumulation and which should stay fixed. - The bounds must be expressions in the fixed parameter, not the dummy variable. - Check that the symbol in the differential matches the symbol that varies inside the integrand.

1. The draft uses \(k\) as the variable of integration even though \(k\) is intended to remain fixed. 2. The draft also uses the changing symbol \(x\) in the bounds instead of the fixed endpoints determined by \(k\). 3. Use \(x\) as the dummy variable and \(k\) in the bounds: \(\int_k^{k+2}(x-k)^2\,\text{d}x\).

The differential should be \(\text{d}x\), and the bounds should be \(k\) and \(k+2\). The intended integral is \(\int_k^{k+2}(x-k)^2\,\text{d}x\).

- Check that the symbol inside the function matches the differential when it is intended as a dummy variable. - Remember what happens to an integral when its limits are reversed. - Treat unmatched symbols as fixed parameters rather than silently renaming them.

1. A is equal because the variable of integration is only renamed. 2. B is equal because reversing the limits changes the sign, and the outside negative sign reverses it again. 3. In C, \(t\) is not the variable of integration; \(f(t)\) is constant with respect to \(x\), so it is not generally equal. 4. In D, \(x\) is not bound by \(\text{d}t\), so it is not generally equal.

A and B only.