Key takeaway: This AP calculus applications worksheet answer key shows justified solutions to all 24 problems. Each entry gives the integral, the antiderivative, the substitution and the final value in bold, with units where needed. Compare the setup first. Then compare the arithmetic, because most lost AP points come from a wrong integrand or wrong limits, not from the antiderivative.
Every worked solution for the applications of integration worksheet is below. Each shows the setup, the antiderivative and the evaluation, with the final answer in bold. Compare your setup first, because most lost points come from the integrand or the limits.
AP Calculus Applications Worksheet Answer Key: problems 1–5
1. Average value of a polynomial
\(\int_{1}^{4}(3x^{2}-2x)\,dx=[x^{3}-x^{2}]_{1}^{4}=48-0=48\).
\(g_{\text{avg}}=\frac{1}{4-1}\cdot 48=16\).
16
2. Average value of sine
\(\int_{0}^{\pi}\sin x\,dx=[-\cos x]_{0}^{\pi}=1-(-1)=2\).
\(f_{\text{avg}}=\frac{2}{\pi}\approx 0.6366\).
\(\frac{2}{\pi}\approx 0.6366\)
3. Average value of a quadratic
An antiderivative is \(x^{3}-x^{2}+x\). At 3: \(27-9+3=21\). At 1: \(1-1+1=1\). The integral is \(20\).
\(f_{\text{avg}}=\frac{20}{2}=10\).
10
4. Where the average value occurs
The average value is 3 (from the first problem). Solve \(c^{2}=3\): \(c=\pm\sqrt{3}\). Only \(\sqrt{3}\) is in \((0,3)\).
\(c=\sqrt{3}\approx 1.732\)
5. Displacement and distance
\(v(t)=(t-1)(t-3)\), so the sign changes at \(t=1\) and \(t=3\). An antiderivative is \(\frac{t^{3}}{3}-2t^{2}+3t\).
\(\int_{0}^{1}v\,dt=\frac{4}{3}\), \(\int_{1}^{3}v\,dt=-\frac{4}{3}\), \(\int_{3}^{4}v\,dt=\frac{4}{3}\).
Displacement: \(\frac{4}{3}-\frac{4}{3}+\frac{4}{3}=\frac{4}{3}\). Distance: \(\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4\).
Displacement \(\frac{4}{3}\); total distance 4
AP Calculus Applications Worksheet Answer Key: problems 6–10
6. Final position
\(\int_{0}^{2}(3t^{2}-6t)\,dt=[t^{3}-3t^{2}]_{0}^{2}=8-12=-4\).
\(s(2)=2+(-4)=-2\).
\(s(2)=-2\)
7. Cosine motion
Displacement: \(\int_{0}^{\pi}\cos t\,dt=\sin\pi-\sin 0=0\).
\(\cos t\ge 0\) on \([0,\frac{\pi}{2}]\) and \(\le 0\) after. Distance: \(\int_{0}^{\pi/2}\cos t\,dt-\int_{\pi/2}^{\pi}\cos t\,dt=1-(-1)=2\).
Displacement 0; distance 2
8. Area between a parabola and a line
Intersections: \(x^{2}=3x+4\) gives \((x-4)(x+1)=0\), so \(x=-1\) and \(x=4\). On \([-1,4]\), the line is on top.
\(A=\int_{-1}^{4}(3x+4-x^{2})\,dx=[\frac{3x^{2}}{2}+4x-\frac{x^{3}}{3}]_{-1}^{4}=\frac{56}{3}-(-\frac{13}{6})=\frac{125}{6}\).
\(\frac{125}{6}\)
9. Area with two intersection points
Set \(x+2=x^{2}\): \(x^{2}-x-2=0\), so \(x=-1\) or \(x=2\). The line is on top.
\(A=\int_{-1}^{2}(x+2-x^{2})\,dx=[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}]_{-1}^{2}=\frac{10}{3}-(-\frac{7}{6})=\frac{9}{2}\).
\(\frac{9}{2}\)
10. Area between exponentials
For \(x\ge 0\), \(e^{x}\ge e^{-x}\). \(A=\int_{0}^{\ln 3}(e^{x}-e^{-x})\,dx=[e^{x}+e^{-x}]_{0}^{\ln 3}\).
At \(\ln 3\): \(3+\frac{1}{3}=\frac{10}{3}\). At 0: \(2\). \(A=\frac{10}{3}-2=\frac{4}{3}\).
\(\frac{4}{3}\)
AP Calculus Applications Worksheet Answer Key: problems 11–15
11. Area using y
\(y^{2}=2y\) gives \(y=0\) and \(y=2\). For \(0< y< 2\), \(2y>y^{2}\), so \(x=2y\) is the right curve.
\(A=\int_{0}^{2}(2y-y^{2})\,dy=[y^{2}-\frac{y^{3}}{3}]_{0}^{2}=4-\frac{8}{3}=\frac{4}{3}\).
\(\frac{4}{3}\)
12. Three intersection points
\(x^{3}=x\) gives \(x=-1,0,1\). On \([-1,0]\), \(x^{3}\ge x\); on \([0,1]\), \(x\ge x^{3}\).
\(\int_{-1}^{0}(x^{3}-x)\,dx=[\frac{x^{4}}{4}-\frac{x^{2}}{2}]_{-1}^{0}=0-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}\).
\(\int_{0}^{1}(x-x^{3})\,dx=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\).
Total area \(\frac{1}{2}\)
13. Sine and cosine area
On \((\frac{\pi}{4},\frac{5\pi}{4})\), \(\sin x\ge\cos x\). \(A=\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx=[-\cos x-\sin x]_{\pi/4}^{5\pi/4}\).
At \(\frac{5\pi}{4}\): \(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}\). At \(\frac{\pi}{4}\): \(-\sqrt{2}\). \(A=\sqrt{2}-(-\sqrt{2})=2\sqrt{2}\).
\(2\sqrt{2}\approx 2.828\)
14. Square cross sections
The curves meet where \(\sqrt{x}=\frac{x}{2}\), that is \(x=0\) and \(x=4\). The side of the square at \(x\) is \(\sqrt{x}-\frac{x}{2}\).
\(V=\int_{0}^{4}(\sqrt{x}-\frac{x}{2})^{2}dx=\int_{0}^{4}(x-x^{3/2}+\frac{x^{2}}{4})dx=8-\frac{64}{5}+\frac{16}{3}=\frac{8}{15}\).
\(\frac{8}{15}\)
15. Semicircle cross sections
Diameter \(=\sqrt{x}\), radius \(=\frac{\sqrt{x}}{2}\). Area \(=\frac{\pi}{2}\cdot\frac{x}{4}=\frac{\pi x}{8}\).
\(V=\int_{0}^{4}\frac{\pi x}{8}\,dx=\frac{\pi}{8}\cdot 8=\pi\).
\(\pi\)
AP Calculus Applications Worksheet Answer Key: problems 16–20
16. Triangle cross sections
Side \(s=6x-x^{2}\). Area \(=\frac{\sqrt{3}}{4}s^{2}\). Limits \(0\) to \(6\).
\(V=\frac{\sqrt{3}}{4}\int_{0}^{6}(36x^{2}-12x^{3}+x^{4})\,dx=\frac{\sqrt{3}}{4}(2592-3888+\frac{7776}{5})=\frac{\sqrt{3}}{4}\cdot\frac{1296}{5}=\frac{324\sqrt{3}}{5}\).
\(\frac{324\sqrt{3}}{5}\approx 112.2\)
17. Disc method about the x-axis
Radius \(\sqrt{x}\). \(V=\pi\int_{0}^{4}x\,dx=\pi\cdot\frac{16}{2}=8\pi\).
\(8\pi\)
18. Disc method about the y-axis
Write \(x=\sqrt{y}\), so the radius is \(\sqrt{y}\). \(V=\pi\int_{0}^{4}y\,dy=\pi\cdot 8=8\pi\).
\(8\pi\)
19. Washer about the x-axis
They intersect at \(x=0\) and \(x=1\); \(y=x\) is outer. \(V=\pi\int_{0}^{1}(x^{2}-x^{6})dx=\pi(\frac{1}{3}-\frac{1}{7})=\frac{4\pi}{21}\).
\(\frac{4\pi}{21}\)
20. Washer about a horizontal line
Distances from \(y=-1\): outer \(R=x+1\), inner \(r=x^{2}+1\).
\(V=\pi\int_{0}^{1}[(x+1)^{2}-(x^{2}+1)^{2}]dx=\pi\int_{0}^{1}(2x-x^{2}-x^{4})\,dx=\pi(1-\frac{1}{3}-\frac{1}{5})=\frac{7\pi}{15}\).
\(\frac{7\pi}{15}\)
AP Calculus Applications Worksheet Answer Key: problems 21–24
21. Rotation about a vertical line
Use \(y\): the curve is \(x=y^{2}\) for \(0\le y\le 2\). The radius from \(x=4\) to the curve is \(4-y^{2}\).
\(V=\pi\int_{0}^{2}(4-y^{2})^{2}\,dy=\pi\int_{0}^{2}(16-8y^{2}+y^{4})\,dy=\pi(32-\frac{64}{3}+\frac{32}{5})=\frac{256\pi}{15}\).
\(\frac{256\pi}{15}\)
22. Water in a tank
\(W^{\prime}(t)=(5+2t)-9=2t-4\). \(W(6)=100+\int_{0}^{6}(2t-4)\,dt=100+(36-24)=112\).
\(W^{\prime}=0\) at \(t=2\), and \(W^{\prime}\) changes from negative to positive. \(W(2)=100+(4-8)=96\). Endpoints: \(W(0)=100\), \(W(6)=112\).
\(W(6)=112\) gallons; least at \(t=2\) minutes (96 gallons)
23. Oil leak
\(\int_{0}^{5}12e^{-0.2t}\,dt=[-60e^{-0.2t}]_{0}^{5}=60(1-e^{-1})\approx 37.93\).
About 37.93 gallons
24. Average temperature
\(\int_{0}^{12}10\sin\frac{\pi t}{12}\,dt=10\cdot\frac{12}{\pi}[-\cos\frac{\pi t}{12}]_{0}^{12}=\frac{120}{\pi}\cdot 2=\frac{240}{\pi}\).
Average \(=60+\frac{1}{12}\cdot\frac{240}{\pi}=60+\frac{20}{\pi}\approx 66.4\).
About \(66.4^{\circ}\text{F}\)
Which results confirm the average value answers?
The averages are 16 for problem 1 and \(\frac{2}{\pi}\approx0.6366\) for problem 2. Next, problem 3 gives 10, and problem 4 gives \(c=\sqrt{3}\). Each average equals the integral divided by the interval length.
What are the motion and area values?
For problem 6, the antiderivative \(t^3-3t^2\) gives \(s(2)=2-4=-2\). Problem 7 has displacement 0 and distance 2. In problem 8, the area is \(\frac{125}{6}\). Likewise, the answers to problems 10, 11 and 12 are \(\frac{4}{3}\), \(\frac{4}{3}\) and \(\frac{1}{2}\), and problem 13 gives \(2\sqrt{2}\).
Which volume answers should you double-check?
Problem 17 gives \(8\pi\), and problem 18 also gives \(8\pi\), since \(\pi\int_0^4y\,dy=8\pi\). Problem 19 gives \(\frac{4\pi}{21}\) after you subtract the inner radius squared. Check that each washer integrand is nonnegative on the interval.
How do you verify the accumulation answers?
Problem 23 evaluates to \(60(1-e^{-1})\approx37.93\) gallons. Because the rate is always positive, the leak total must exceed zero and stay below \(12\cdot5=60\). That quick bound catches most setup mistakes.
How can you learn from errors in a setup?
Mark whether the mistake came from limits, the integrand or the formula. Then rewrite the setup with the correct pieces and color the changed part. After that, solve a nearby problem on a new region, so the repaired habit gets tested right away.
Printable PDF · Worksheet Answer Key
AP Calculus AB Applications of Integration worksheet answer key
Download the PDF
See every chapter on the AP Calculus AB math page.
