Random practice problems from Sections 6.1 through 7.2.
Each section contains 10 problems: 2 problems each from difficulty 1/5 through 5/5.
Choose a section, generate a problem, and reveal the answer when ready.
Click “New Problem” to get started.
Section 6.1 · Difficulty 1/5
Find the area between the curves on the stated interval.
\( y=x,\ y=0,\ 0\le x\le 4 \)
Answer: \( \displaystyle \int_0^4 x\,dx = 8 \)
Section 6.1 · Difficulty 1/5
Find the area between the curves on the stated interval.
\( y=5-x,\ y=0,\ 0\le x\le 5 \)
Answer: \( \displaystyle \int_0^5 (5-x)\,dx = \frac{25}{2} \)
Section 6.1 · Difficulty 2/5
Find the area enclosed by the curves.
\( y=x^2,\ y=4 \)
Answer: intersections at \(x=\pm2\), so \( \displaystyle A=\int_{-2}^{2}(4-x^2)\,dx=\frac{32}{3} \)
Section 6.1 · Difficulty 2/5
Find the area enclosed by the curves.
\( y=x^2,\ y=2x \)
Answer: intersections at \(x=0,2\), so \( \displaystyle A=\int_0^2(2x-x^2)\,dx=\frac{4}{3} \)
Section 6.1 · Difficulty 3/5
Find the enclosed area.
\( y=x^2+1,\ y=3x+1 \)
Answer: intersections at \(x=0,3\), so \( \displaystyle A=\int_0^3\big((3x+1)-(x^2+1)\big)\,dx=\frac{9}{2} \)
Section 6.1 · Difficulty 3/5
Find the enclosed area.
\( y=x,\ y=x^3 \)
Answer: intersections at \(x=-1,0,1\), total area \( \displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_0^1(x-x^3)\,dx=\frac{1}{2} \)
Section 6.1 · Difficulty 4/5
Find the enclosed area.
\( y=\sqrt{x},\ y=x \)
Answer: intersections at \(x=0,1\), so \( \displaystyle A=\int_0^1(\sqrt{x}-x)\,dx=\frac{1}{6} \)
Section 6.1 · Difficulty 4/5
Find the enclosed area using integration with respect to \(y\).
\( x=y^2,\ x=2y \)
Answer: intersections at \(y=0,2\), so \( \displaystyle A=\int_0^2(2y-y^2)\,dy=\frac{4}{3} \)
Section 6.1 · Difficulty 5/5
Find the enclosed area.
\( y=x^3,\ y=x^2 \)
Answer: intersections at \(x=0,1\), so \( \displaystyle A=\int_0^1(x^2-x^3)\,dx=\frac{1}{12} \)
Section 6.1 · Difficulty 5/5
Find the area enclosed by the curves.
\( y=x^2-2x,\ y=-x \)
Answer: intersections at \(x=0,1\), so \( \displaystyle A=\int_0^1\big((-x)-(x^2-2x)\big)\,dx=\frac{1}{6} \)
Section 6.2 · Difficulty 1/5
Find the average value of the function on the interval.
\( f(x)=x,\ [0,4] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{4}\int_0^4 x\,dx=2 \)
Section 6.2 · Difficulty 1/5
Find the average value of the function on the interval.
\( f(x)=3,\ [2,7] \)
Answer: \( 3 \)
Section 6.2 · Difficulty 2/5
Find the average value of the function on the interval.
\( f(x)=x^2,\ [0,3] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{3}\int_0^3 x^2\,dx=3 \)
Section 6.2 · Difficulty 2/5
Find the average value of the function on the interval.
\( f(x)=4-x,\ [0,4] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{4}\int_0^4 (4-x)\,dx=2 \)
Section 6.2 · Difficulty 3/5
Find the average value of the function on the interval.
\( f(x)=\sqrt{x},\ [0,4] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{4}\int_0^4 \sqrt{x}\,dx=\frac{4}{3} \)
Section 6.2 · Difficulty 3/5
Find the average value of the function on the interval.
\( f(x)=\dfrac{1}{x},\ [1,4] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{3}\int_1^4 \frac{1}{x}\,dx=\frac{\ln 4}{3} \)
Section 6.2 · Difficulty 4/5
Find the average value of the function on the interval.
\( f(x)=x+\dfrac{1}{x},\ [1,2] \)
Answer: \( \displaystyle f_{\text{avg}}=\int_1^2 \left(x+\frac{1}{x}\right)dx=\frac{3}{2}+\ln 2 \)
Section 6.2 · Difficulty 4/5
Find the average value of the function on the interval.
\( f(x)=x^3-2x,\ [-1,2] \)
Answer: \( \displaystyle f_{\text{avg}}=\frac{1}{3}\int_{-1}^{2}(x^3-2x)\,dx=\frac{1}{4} \)
Section 6.2 · Difficulty 5/5
A cross-sectional area perpendicular to the \(x\)-axis is \(A(x)=x^2+1\) for \(0\le x\le 2\). Find the volume.
\( A(x)=x^2+1,\ 0\le x\le 2 \)
Answer: \( \displaystyle V=\int_0^2 (x^2+1)\,dx=\frac{14}{3} \)
Section 6.2 · Difficulty 5/5
A solid has cross-sectional area \(A(x)=9-x^2\) on \([-3,3]\). Find the volume.
\( A(x)=9-x^2,\ -3\le x\le 3 \)
Answer: \( \displaystyle V=\int_{-3}^{3}(9-x^2)\,dx=36 \)
Section 6.3 · Difficulty 1/5
Set up and evaluate the volume using disks or washers.
Rotate the region under \( y=x \) on \( [0,2] \) about the \(x\)-axis.
Answer: \( \displaystyle V=\pi\int_0^2 x^2\,dx=\frac{8\pi}{3} \)
Section 6.3 · Difficulty 1/5
Set up and evaluate the volume using disks or washers.
Rotate the region under \( y=3 \) on \( [0,4] \) about the \(x\)-axis.
Answer: \( \displaystyle V=\pi\int_0^4 3^2\,dx=36\pi \)
Section 6.3 · Difficulty 2/5
Find the volume of the solid of revolution.
Rotate the region bounded by \( y=\sqrt{x},\ x=4,\ y=0 \) about the \(x\)-axis.
Answer: \( \displaystyle V=\pi\int_0^4 (\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=8\pi \)
Section 6.3 · Difficulty 2/5
Find the volume of the solid of revolution.
Rotate the region between \( y=4 \) and \( y=x \) on \( [0,4] \) about the \(x\)-axis.
Answer: \( \displaystyle V=\pi\int_0^4(4^2-x^2)\,dx=\frac{128\pi}{3} \)
Section 6.3 · Difficulty 3/5
Find the volume of the solid of revolution.
Rotate the region bounded by \( y=x^2,\ y=0,\ 0\le x\le 2 \) about the \(y\)-axis using washers with respect to \(y\).
Answer: \( x=\sqrt{y},\ 0\le y\le 4,\ \displaystyle V=\pi\int_0^4 (\sqrt{y})^2\,dy=8\pi \)
Section 6.3 · Difficulty 3/5
Find the volume of the solid of revolution.
Rotate the region between \( y=x^2 \) and \( y=4 \) about the \(x\)-axis.
Answer: intersections at \(x=\pm2\), \( \displaystyle V=\pi\int_{-2}^{2}(4^2-(x^2)^2)\,dx=\frac{256\pi}{5} \)
Section 6.3 · Difficulty 4/5
Find the volume of the solid of revolution.
Rotate the region between \( y=x \) and \( y=x^2 \) about the \(x\)-axis.
Answer: on \(0\le x\le1\), \( \displaystyle V=\pi\int_0^1(x^2-x^4)\,dx=\frac{2\pi}{15} \)
Section 6.3 · Difficulty 4/5
Find the volume of the solid of revolution.
Rotate the region between \( y=2x \) and \( y=x^2 \) about the \(x\)-axis.
Answer: on \(0\le x\le2\), \( \displaystyle V=\pi\int_0^2\big((2x)^2-(x^2)^2\big)\,dx=\frac{64\pi}{15} \)
Section 6.3 · Difficulty 5/5
Find the volume of the solid of revolution.
Rotate the region bounded by \( y=\sqrt{x},\ y=0,\ x=4 \) about \( y=-1 \).
Answer: \( \displaystyle V=\pi\int_0^4\left((\sqrt{x}+1)^2-1^2\right)\,dx=\pi\int_0^4(x+2\sqrt{x})\,dx=\frac{56\pi}{3} \)
Section 6.3 · Difficulty 5/5
Find the volume of the solid of revolution.
Rotate the region between \( y=4-x^2 \) and \( y=0 \) about the \(x\)-axis.
Answer: \( \displaystyle V=\pi\int_{-2}^{2}(4-x^2)^2\,dx=\frac{512\pi}{15} \)
Section 6.4 · Difficulty 1/5
Use the shell method to find the volume.
Rotate the region under \( y=x \) on \( [0,2] \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^2 x(x)\,dx=2\pi\int_0^2 x^2\,dx=\frac{16\pi}{3} \)
Section 6.4 · Difficulty 1/5
Use the shell method to find the volume.
Rotate the region under \( y=3 \) on \( [0,4] \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^4 x(3)\,dx=48\pi \)
Section 6.4 · Difficulty 2/5
Use the shell method to find the volume.
Rotate the region bounded by \( y=x^2,\ y=0,\ x=3 \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^3 x(x^2)\,dx=\frac{81\pi}{2} \)
Section 6.4 · Difficulty 2/5
Use the shell method to find the volume.
Rotate the region under \( y=\sqrt{x} \) on \( [0,4] \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^4 x\sqrt{x}\,dx=\frac{128\pi}{5} \)
Section 6.4 · Difficulty 3/5
Use the shell method to find the volume.
Rotate the region between \( y=x \) and \( y=x^2 \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^1 x(x-x^2)\,dx=\frac{\pi}{6} \)
Section 6.4 · Difficulty 3/5
Use the shell method to find the volume.
Rotate the region between \( y=2x \) and \( y=x^2 \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^2 x(2x-x^2)\,dx=\frac{8\pi}{3} \)
Section 6.4 · Difficulty 4/5
Use the shell method to find the volume.
Rotate the region bounded by \( y=4-x^2,\ y=0 \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^2 x(4-x^2)\,dx=8\pi \)
Section 6.4 · Difficulty 4/5
Use the shell method to find the volume.
Rotate the region bounded by \( y=x^2,\ y=4 \) about the \(y\)-axis.
Answer: \( \displaystyle V=2\pi\int_0^2 x(4-x^2)\,dx=8\pi \)
Section 6.4 · Difficulty 5/5
Use the shell method to find the volume.
Rotate the region bounded by \( y=\sqrt{x},\ y=0,\ x=4 \) about \( x=-1 \).
Answer: radius \(=x+1\), height \(=\sqrt{x}\), so \( \displaystyle V=2\pi\int_0^4 (x+1)\sqrt{x}\,dx=\frac{368\pi}{15} \)
Section 6.4 · Difficulty 5/5
Use the shell method to find the volume.
Rotate the region between \( y=x \) and \( y=x^2 \) about \( x=2 \).
Answer: radius \(=2-x\), height \(=x-x^2\), so \( \displaystyle V=2\pi\int_0^1 (2-x)(x-x^2)\,dx=\frac{\pi}{2} \)
Section 6.5 · Difficulty 1/5
Find the work done by a constant force.
A force of \(10\) N moves an object \(6\) m.
Answer: \( W=Fd=10\cdot 6=60\text{ J} \)
Section 6.5 · Difficulty 1/5
Find the work done by a variable force.
\( F(x)=3x,\ 0\le x\le 4 \)
Answer: \( \displaystyle W=\int_0^4 3x\,dx=24\text{ J} \)
Section 6.5 · Difficulty 2/5
Find the work required to stretch the spring.
A spring has force law \(F(x)=8x\). Find the work to stretch it from \(x=0\) to \(x=3\).
Answer: \( \displaystyle W=\int_0^3 8x\,dx=36\text{ J} \)
Section 6.5 · Difficulty 2/5
Find the work required to lift the object.
Lift a \(20\)-kg mass upward \(5\) m. Use \(g=9.8\text{ m/s}^2\).
Answer: \( W=mgh=20(9.8)(5)=980\text{ J} \)
Section 6.5 · Difficulty 3/5
A cable weighs \(4\) N/m and is \(10\) m long. Find the work required to lift the entire cable vertically.
Cable weight density \(=4\text{ N/m}\), length \(=10\text{ m}\)
Answer: \( \displaystyle W=\int_0^{10} 4x\,dx=200\text{ J} \)
Section 6.5 · Difficulty 3/5
A \(15\)-m rope weighs \(3\) N/m. How much work is required to pull the entire rope up to the top of a building?
Rope weight density \(=3\text{ N/m}\), length \(=15\text{ m}\)
Answer: \( \displaystyle W=\int_0^{15} 3x\,dx=\frac{675}{2}\text{ J} \)
Section 6.5 · Difficulty 4/5
A spring requires \(12\) N of force to hold it stretched \(0.5\) m beyond equilibrium. Find the work required to stretch it from equilibrium to \(1\) m.
Use Hooke’s Law.
Answer: \(k=\frac{12}{0.5}=24\), so \( \displaystyle W=\int_0^1 24x\,dx=12\text{ J} \)
Section 6.5 · Difficulty 4/5
A tank is full of water. Water weighs \(9800\text{ N/m}^3\). The tank is \(2\) m deep and has constant cross-sectional area \(3\text{ m}^2\). Find the work to pump all water to the top.
Rectangular tank, depth \(2\) m, area \(3\text{ m}^2\)
Answer: \( \displaystyle W=\int_0^2 9800(3)(2-y)\,dy=58800\text{ J} \)
Section 6.5 · Difficulty 5/5
A bucket weighing \(30\) N is lifted \(20\) m by a rope weighing \(2\) N/m. Find the total work.
Include both bucket and rope.
Answer: bucket work \(=30(20)=600\), rope work \( \displaystyle =\int_0^{20}2x\,dx=400\), total \(=1000\text{ J}\)
Section 6.5 · Difficulty 5/5
A cylindrical tank of radius \(1\) m and depth \(3\) m is full of water. Water weighs \(9800\text{ N/m}^3\). Find the work required to pump all water to the top rim.
Radius \(1\), depth \(3\)
Answer: area \(=\pi\), so \( \displaystyle W=\int_0^3 9800\pi(3-y)\,dy=44100\pi\text{ J} \)
Section 7.1 · Difficulty 1/5
Differentiate.
\( \dfrac{d}{dx}(e^x) \)
Answer: \( e^x \)
Section 7.1 · Difficulty 1/5
Differentiate.
\( \dfrac{d}{dx}(2^x) \)
Answer: \( 2^x\ln 2 \)
Section 7.1 · Difficulty 2/5
Differentiate.
\( \dfrac{d}{dx}(e^{3x}) \)
Answer: \( 3e^{3x} \)
Section 7.1 · Difficulty 2/5
Differentiate.
\( \dfrac{d}{dx}(5^{2x}) \)
Answer: \( 2\cdot 5^{2x}\ln 5 \)
Section 7.1 · Difficulty 3/5
Differentiate.
\( \dfrac{d}{dx}\left(xe^x\right) \)
Answer: \( e^x+xe^x=e^x(1+x) \)
Section 7.1 · Difficulty 3/5
Find the antiderivative.
\( \displaystyle \int e^{4x}\,dx \)
Answer: \( \displaystyle \frac{1}{4}e^{4x}+C \)
Section 7.1 · Difficulty 4/5
Differentiate.
\( y=e^{x^2+1} \)
Answer: \( y'=2x\,e^{x^2+1} \)
Section 7.1 · Difficulty 4/5
Find the antiderivative.
\( \displaystyle \int 3^x\,dx \)
Answer: \( \displaystyle \frac{3^x}{\ln 3}+C \)
Section 7.1 · Difficulty 5/5
Solve for the population.
A population is modeled by \( P(t)=200e^{2t} \). Find the initial population and the time when the population is \(1000\).
Answer: \(P(0)=200\). For \(1000\): \(200e^{2t}=1000\Rightarrow e^{2t}=5\Rightarrow 2t=\ln 5\Rightarrow t=\dfrac{\ln 5}{2}\)
Section 7.1 · Difficulty 5/5
Find the tangent line to the curve at the given point.
\( y=e^{2x},\ x=0 \)
Answer: point \(=(0,1)\), slope \(=2e^{0}=2\), so tangent line is \( y-1=2(x-0) \), i.e. \( y=2x+1 \)
Section 7.2 · Difficulty 1/5
Differentiate.
\( \dfrac{d}{dx}(\ln x) \)
Answer: \( \dfrac{1}{x} \)
Section 7.2 · Difficulty 1/5
Differentiate.
\( \dfrac{d}{dx}(\ln(4x-2)) \)
Answer: \( \dfrac{4}{4x-2} \)
Section 7.2 · Difficulty 2/5
Differentiate.
\( \dfrac{d}{dx}(\log_2 x) \)
Answer: \( \dfrac{1}{x\ln 2} \)
Section 7.2 · Difficulty 2/5
Differentiate.
\( y=\ln(x^2+1) \)
Answer: \( y'=\dfrac{2x}{x^2+1} \)
Section 7.2 · Difficulty 3/5
Use the inverse-function derivative formula.
If \( f(x)=x^3+1 \), find \( (f^{-1})'(9) \).
Answer: \( f^{-1}(9)=2 \) since \(2^3+1=9\). Then \( f'(x)=3x^2 \), so \( (f^{-1})'(9)=\dfrac{1}{f'(2)}=\dfrac{1}{12} \)
Section 7.2 · Difficulty 3/5
Use the inverse-function derivative formula.
If \( f(x)=2x+5 \), find \( (f^{-1})'(11) \).
Answer: \( f^{-1}(11)=3 \), \(f'(x)=2\), so \( (f^{-1})'(11)=\dfrac{1}{2} \)
Section 7.2 · Difficulty 4/5
Find the derivative.
\( y=x^{3x}+\ln(4x-2) \)
Answer: \( \dfrac{d}{dx}(x^{3x})=x^{3x}(3\ln x+3) \), so \( y'=x^{3x}(3\ln x+3)+\dfrac{4}{4x-2} \)
Section 7.2 · Difficulty 4/5
Differentiate logarithmically.
\( y=(x^2+1)^x \)
Answer: \( \ln y=x\ln(x^2+1) \Rightarrow \dfrac{y'}{y}=\ln(x^2+1)+\dfrac{2x^2}{x^2+1} \), so \( y'=(x^2+1)^x\left[\ln(x^2+1)+\dfrac{2x^2}{x^2+1}\right] \)
Section 7.2 · Difficulty 5/5
Find the tangent line to the inverse function.
\( f(x)=x^2+13x,\ x\ge -\dfrac{13}{2} \). Find the slope of the tangent line to \(f^{-1}\) at \(x=30\).
Answer: solve \(x^2+13x=30\Rightarrow x=2\) or \(-15\). Domain gives \(x\ge -13/2\), so use \(x=2\). Then \(f'(x)=2x+13\), so \( (f^{-1})'(30)=\dfrac{1}{f'(2)}=\dfrac{1}{17} \)
Section 7.2 · Difficulty 5/5
Use inverse-function differentiation.
If \( f(x)=\sqrt{x^2+13x} \) with domain \( x\ge -\dfrac{13}{2} \), find \( (f^{-1})'(\sqrt{30}) \).
Answer: \( f^{-1}(\sqrt{30})=2 \) since \( \sqrt{2^2+13\cdot2}=\sqrt{30} \). Also \( f'(x)=\dfrac{2x+13}{2\sqrt{x^2+13x}} \), so \( f'(2)=\dfrac{17}{2\sqrt{30}} \). Therefore \( (f^{-1})'(\sqrt{30})=\dfrac{2\sqrt{30}}{17} \)