AP Calculus AB Average Value, Area and Volume Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Average Value, Area and Volume – Maths PDF

Cover of the book AP Calculus AB Math Worksheets with Answers (PDF)

Download as a PDFThe AP Calculus AB worksheets book with answersPrintable PDFSee the book ›

Sommaire

This ap calculus area volume answers for AP Calculus AB covers average value, area and volume (Common Core AP CALC AB 8.1, AP CALC AB 8.2, AP CALC AB 8.3, AP CALC AB 8.4, AP CALC AB 8.5, AP CALC AB 8.6, AP CALC AB 8.7, AP CALC AB 8.8, AP CALC AB 8.9, AP CALC AB 8.10, AP CALC AB 8.11, AP CALC AB 8.12), with a free PDF to print.

This answer key gives a full worked solution for every problem in the area and volume worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 8.1, AP CALC AB 8.2, AP CALC AB 8.3, AP CALC AB 8.4, AP CALC AB 8.5, AP CALC AB 8.6, AP CALC AB 8.7, AP CALC AB 8.8, AP CALC AB 8.9, AP CALC AB 8.10, AP CALC AB 8.11, AP CALC AB 8.12

Ap calculus area volume answers: exercises 1 to 10

Exercise 1 – Average value of x squared

\(\frac{1}{3}\int_{0}^{3}x^{2}dx=\frac{1}{3}\cdot\frac{27}{3}=3\).

Solve \(c^{2}=3\): \(c=\sqrt{3}\approx 1.732\) (in \([0,3]\)).

Final answer: \(f_{\text{avg}}=3,\ c=\sqrt{3}\).

The curve crosses the average height at x = √3.

Exercise 2 – Average value of a root

\(\frac{1}{4}\int_{0}^{4}x^{1/2}dx=\frac{1}{4}\left[\frac{2}{3}x^{3/2}\right]_{0}^{4}=\frac{1}{4}\cdot\frac{16}{3}=\frac{4}{3}\).

Final answer: \(\frac{4}{3}\).

Exercise 3 – Average value of sine

\(\frac{1}{\pi}\int_{0}^{\pi}\sin x\,dx=\frac{1}{\pi}[-\cos x]_{0}^{\pi}=\frac{2}{\pi}\).

Final answer: \(\frac{2}{\pi}\approx 0.637\).

The average value of sin x on [0, π] is 2/π.

Exercise 4 – Average temperature

\(\frac{1}{24}\int_{0}^{24}\left(60+10\sin\frac{\pi t}{12}\right)dt=\frac{1}{24}\left[60t-\frac{120}{\pi}\cos\frac{\pi t}{12}\right]_{0}^{24}\).

The cosine terms cancel because \(\cos 2\pi=\cos 0=1\), leaving \(\frac{1}{24}(1440)=60\).

Final answer: \(60\,^{\circ}\text{F}\).

Exercise 5 – Average value from a table

Integral estimate: \(2\left[\frac{3+5}{2}+\frac{5+9}{2}+\frac{9+11}{2}+\frac{11+10}{2}\right]=2(4+7+10+10.5)=63\).

Average: \(\frac{63}{8}=7.875\).

Final answer: \(\approx 7.9\text{ m/s}\).

Exercise 6 – Displacement and distance

\(v=(t-1)(t-3)\) changes sign at \(t=1\) and \(t=3\).

Displacement: \(\int_{0}^{4}v\,dt=\left[\frac{t^{3}}{3}-2t^{2}+3t\right]_{0}^{4}=\frac{64}{3}-32+12=\frac{4}{3}\).

Distance: \(\int_{0}^{1}=\frac{4}{3}\), \(\left|\int_{1}^{3}\right|=\frac{4}{3}\), \(\int_{3}^{4}=\frac{4}{3}\). Sum \(4\).

Final answer: \(\text{displacement }\frac{4}{3}\text{ ft;}\ \text{distance }4\text{ ft}\).

Exercise 7 – Position from velocity

\(s(3)=s(0)+\int_{0}^{3}(2t+1)dt=5+[t^{2}+t]_{0}^{3}=5+12=17\).

Final answer: \(17\text{ m}\).

Exercise 8 – Velocity from acceleration

\(v(3)=2+\int_{0}^{3}6t\,dt=2+[3t^{2}]_{0}^{3}=2+27=29\).

Final answer: \(29\text{ m/s}\).

Exercise 9 – Net change of a population

Net change: \(\int_{0}^{10}(40-6t)dt=[40t-3t^{2}]_{0}^{10}=400-300=100\). Population: \(300+100=400\).

\(r=0\) at \(t=\frac{20}{3}\), changing from positive to negative, so the maximum is at \(t\approx 6.67\) weeks.

Final answer: \(400\text{ fish;}\ \text{maximum at }t=\frac{20}{3}\).

Exercise 10 – Area between a root and a parabola

They meet where \(\sqrt{x}=x^{2}\): \(x=0\) and \(x=1\). The root curve is on top.

\(\int_{0}^{1}(\sqrt{x}-x^{2})dx=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\).

Final answer: \(\frac{1}{3}\).

Exercises 11 and up

Exercise 11 – Area between two parabolas

\(8-x^{2}=x^{2}\) gives \(x=\pm 2\).

\(\int_{-2}^{2}(8-2x^{2})dx=\left[8x-\frac{2x^{3}}{3}\right]_{-2}^{2}=2\left(16-\frac{16}{3}\right)=\frac{64}{3}\).

Final answer: \(\frac{64}{3}\).

Exercise 12 – Area under a line above an exponential

They meet where \(e^{x}=2\): \(x=\ln 2\). The line is on top for \(0\le x\le\ln 2\).

\(\int_{0}^{\ln 2}(2-e^{x})dx=[2x-e^{x}]_{0}^{\ln 2}=(2\ln 2-2)-(0-1)=2\ln 2-1\).

Final answer: \(2\ln 2-1\approx 0.386\).

Exercise 13 – Sine and cosine

On this interval \(\cos x\ge\sin x\).

\([\sin x+\cos x]_{0}^{\pi/4}=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-1=\sqrt{2}-1\).

Final answer: \(\sqrt{2}-1\approx 0.414\).

Exercise 14 – Integrating with respect to y

Intersections: \(y^{2}=y+2\) gives \(y=-1\) and \(y=2\). The line is to the right.

\(\int_{-1}^{2}(y+2-y^{2})dy=\left[\frac{y^{2}}{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2}=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}\).

Final answer: \(\frac{9}{2}\).

Exercise 15 – Two enclosed regions

\(x^{3}-3x=x\) gives \(x(x^{2}-4)=0\), so \(x=-2,0,2\). For \(0 < x < 2\) the line is on top.

By symmetry: \(2\int_{0}^{2}(x-(x^{3}-3x))dx=2\int_{0}^{2}(4x-x^{3})dx=2\left[2x^{2}-\frac{x^{4}}{4}\right]_{0}^{2}=2(8-4)=8\).

Final answer: \(8\).

Two regions of area 4 each.

Exercise 16 – Square cross sections

A square has side \(\sqrt{x}\), so \(A(x)=x\).

\(V=\int_{0}^{4}x\,dx=8\).

Final answer: \(8\).

The base and one slice.

Exercise 17 – Rectangles on an exponential base

Each rectangle has width \(e^{x}\) and height \(3\), so \(A(x)=3e^{x}\).

\(V=\int_{0}^{1}3e^{x}dx=3[e^{x}]_{0}^{1}=3(e-1)\).

Final answer: \(3(e-1)\approx 5.15\).

Exercise 18 – Disc method about the x-axis

Each disc has radius \(\sqrt{x}\): \(V=\pi\int_{0}^{4}(\sqrt{x})^{2}dx=\pi\int_{0}^{4}x\,dx=\pi\cdot 8\).

Final answer: \(8\pi\approx 25.13\).

Exercise 19 – Washer with a horizontal cap

Outer radius \(3\), inner radius \(x^{2}+1\).

\(V=\pi\int_{0}^{\sqrt{2}}\left(9-(x^{2}+1)^{2}\right)dx=\pi\int_{0}^{\sqrt{2}}(8-2x^{2}-x^{4})dx=\pi\left[8x-\frac{2x^{3}}{3}-\frac{x^{5}}{5}\right]_{0}^{\sqrt{2}}\).

\(=\pi\sqrt{2}\left(8-\frac{4}{3}-\frac{4}{5}\right)=\frac{88\sqrt{2}\,\pi}{15}\approx 26.06\).

Final answer: \(\frac{88\sqrt{2}\,\pi}{15}\approx 26.06\).

Exercise 20 – Revolving about the y-axis

The region to the left of the curve is \(x=\sqrt{y}\) for \(0\le y\le 4\). Discs of radius \(\sqrt{y}\):

\(V=\pi\int_{0}^{4}y\,dy=\pi\cdot 8=8\pi\).

Final answer: \(8\pi\approx 25.13\).

Exercise 21 – Volume of a cone

\(V=\pi\int_{0}^{2}x^{2}dx=\pi\cdot\frac{8}{3}=\frac{8\pi}{3}\).

Cone with radius \(r=2\) and height \(h=2\): \(\frac{1}{3}\pi(4)(2)=\frac{8\pi}{3}\). They agree.

Final answer: \(\frac{8\pi}{3}\).

Exercise 22 – A vase

\(V=\pi\int_{0}^{3}\left(\frac{x^{2}}{3}\right)^{2}dx=\frac{\pi}{9}\int_{0}^{3}x^{4}dx=\frac{\pi}{9}\cdot\frac{243}{5}=\frac{27\pi}{5}\).

Final answer: \(\frac{27\pi}{5}\approx 16.96\text{ in}^{3}\).

The vase: the curve and its mirror image.

Exercise 23 – Average rate of production

\(\int_{0}^{6}(50+6t-t^{2})dt=[50t+3t^{2}-\frac{t^{3}}{3}]_{0}^{6}=300+108-72=336\).

Average rate: \(\frac{336}{6}=56\) items per hour.

Final answer: \(336\text{ items;}\ 56\text{ items/hour}\).

Key facts behind this ap calculus area volume answers

Key facts

Average value = (1/(b – a)) times the integral of f over [a, b].
Displacement is the integral of v; total distance is the integral of |v|.
Area between curves = integral of (upper – lower); split the interval at every crossing.
Volume by cross sections = integral of A(x); disc method = pi times the integral of (radius squared); washers subtract the inner radius squared.

How to use this ap calculus area volume answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the ap calculus area volume answers as a PDF with the button below and keep it with your notes.

All AP Calculus AB chapters: AP Calculus AB math.

Voter.. post

Free printable PDF