Practice questions
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Chapter 6: Applications of integrals
Try each problem first. Then check your work with the solution.
1. Areas between curves. Find the area of each region.
(a) The region enclosed by ${y=x^2}$ and ${y=x+2}$.
(b) The region between ${y=e^x}$ and ${y=x}$, from ${x=0}$ to ${x=1}$.
(c) The region enclosed by ${x=y^2-1}$ and ${x=1-y^2}$.
Solution:
(a)
Where they meet: Set the two equal:
$\begin{align*}x^2&=x+2\\x^2-x-2&=0\\(x+1)(x-2)&=0\end{align*}$
So they meet at ${x=-1}$ and ${x=2}$.
Which is on top: Test ${x=0}$:
- ${x+2=2}$
- ${x^2=0}$
So the line ${y=x+2}$ is on top.
Integrate:
$\begin{align*}&A\\&=\int_{-1}^{2}\left[(x+2)-x^2\right]dx\\&=\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}\\&=\left(2+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right)\\&=\dfrac{10}{3}-\left(-\dfrac{7}{6}\right)\\&=\dfrac{9}{2}\end{align*}$
(b)
Which is on top: On ${[0,1]}$, ${e^x}$ is always bigger than ${x}$. (At ${x=0}$, ${e^0=1}$ and ${x=0}$. At ${x=1}$, ${e\approx 2.72}$ and ${x=1}$.) The curves do not meet on this interval.
Integrate:
$\begin{align*}&A\\&=\int_0^1(e^x-x)\,dx\\&=\left[e^x-\dfrac{x^2}{2}\right]_0^1\\&=\left(e-\dfrac{1}{2}\right)-(1-0)\\&=e-\dfrac{3}{2}\approx 1.22\end{align*}$
(c)
Both curves are written as ${x}$ in terms of ${y}$. So use thin horizontal rectangles, and integrate with respect to ${y}$. Each rectangle goes from the left curve to the right curve.
Where they meet: Set the two equal:
$\begin{align*}y^2-1&=1-y^2\\2y^2&=2\\y&=\pm 1\end{align*}$
Which is on the right: Test ${y=0}$: ${1-y^2=1}$ and ${y^2-1=-1}$. So ${x=1-y^2}$ is on the right.
Integrate: right curve minus left curve:
$\begin{align*}&A\\&=\int_{-1}^{1}\left[(1-y^2)-(y^2-1)\right]dy\\&=\int_{-1}^{1}(2-2y^2)\,dy\\&=\left[2y-\dfrac{2y^3}{3}\right]_{-1}^{1}\\&=\dfrac{4}{3}-\left(-\dfrac{4}{3}\right)\\&=\dfrac{8}{3}\end{align*}$
2. Volumes by slicing. Find the volume of each solid.
(a) The region under ${y=x^2}$, from ${x=0}$ to ${x=2}$, is rotated about the ${x}$-axis.
(b) The region between ${y=2x}$ and ${y=x^2}$ is rotated about the ${x}$-axis.
(c) The region between ${y=x^2}$, the line ${y=4}$, and the ${y}$-axis (for ${x\ge 0}$) is rotated about the ${y}$-axis.
Solution:
(a)
Each slice is a disk. Its radius is the height of the curve, ${x^2}$. So its area is
${A(x)=\pi(x^2)^2=\pi x^4}$.
$\begin{align*}&V\\&=\int_0^2\pi x^4\,dx\\&=\pi\left[\dfrac{x^5}{5}\right]_0^2\\&=\dfrac{32\pi}{5}\end{align*}$
(b)
The curves meet where ${2x=x^2}$, so at ${x=0}$ and ${x=2}$. On ${[0,2]}$, ${2x\ge x^2}$.
Each slice is a washer. The outer radius is ${R=2x}$, and the inner radius is ${r=x^2}$:
$A(x)=\pi\left[(2x)^2-(x^2)^2\right]=\pi(4x^2-x^4)$
$\begin{align*}&V\\&=\int_0^2\pi(4x^2-x^4)\,dx\\&=\pi\left[\dfrac{4x^3}{3}-\dfrac{x^5}{5}\right]_0^2\\&=\pi\left(\dfrac{32}{3}-\dfrac{32}{5}\right)\\&=\dfrac{64\pi}{15}\end{align*}$
(c)
For rotation about the ${y}$-axis, use horizontal slices and integrate with respect to ${y}$, from ${y=0}$ to ${y=4}$.
Solve ${y=x^2}$ for ${x}$: ${x=\sqrt{y}}$. Each slice is a disk with radius ${\sqrt{y}}$:
${A(y)=\pi(\sqrt{y})^2=\pi y}$
$\begin{align*}&V\\&=\int_0^4\pi y\,dy\\&=\pi\left[\dfrac{y^2}{2}\right]_0^4\\&=8\pi\end{align*}$
3. Volumes by cylindrical shells. Use shells to find the volume of each solid.
(a) The region under ${y=x^2}$, from ${x=0}$ to ${x=2}$, is rotated about the ${y}$-axis.
(b) The region between ${y=2x}$ and ${y=x^2}$ is rotated about the ${y}$-axis.
(c) The region under ${y=x^2}$, from ${x=0}$ to ${x=1}$, is rotated about the line ${x=-1}$.
Solution:
(a)
Radius: ${x}$. Height: ${x^2}$.
$\begin{align*}&V\\&=\int_0^2 2\pi x\cdot x^2\,dx\\&=2\pi\int_0^2 x^3\,dx\\&=2\pi\left[\dfrac{x^4}{4}\right]_0^2\\&=2\pi(4)\\&=8\pi\end{align*}$
(b)
The curves meet at ${x=0}$ and ${x=2}$, and ${2x}$ is on top.
Radius: ${x}$. Height: ${2x-x^2}$.
$\begin{align*}&V\\&=\int_0^2 2\pi x(2x-x^2)\,dx\\&=2\pi\int_0^2(2x^2-x^3)\,dx\\&=2\pi\left[\dfrac{2x^3}{3}-\dfrac{x^4}{4}\right]_0^2\\&=2\pi\left(\dfrac{16}{3}-4\right)\\&=\dfrac{8\pi}{3}\end{align*}$
(c)
The axis is the line ${x=-1}$. A shell at ${x}$ is ${x-(-1)=x+1}$ away from it.
Radius: ${x+1}$. Height: ${x^2}$.
$\begin{align*}&V\\&=\int_0^1 2\pi(x+1)x^2\,dx\\&=2\pi\int_0^1(x^3+x^2)\,dx\\&=2\pi\left[\dfrac{x^4}{4}+\dfrac{x^3}{3}\right]_0^1\\&=2\pi\left(\dfrac{1}{4}+\dfrac{1}{3}\right)\\&=\dfrac{7\pi}{6}\end{align*}$
4. Average value of a function.
(a) Find the average value of ${f(x)=3x^2-2x}$ on ${[0,2]}$. Then find every number ${c}$ in ${[0,2]}$ where ${f(c)=f_{\text{ave}}}$.
(b) Find the average value of ${\cos x}$ on ${\left[0,\dfrac{\pi}{2}\right]}$.
(c) A car speeds up from rest. Its velocity after ${t}$ seconds is ${v(t)=3t^2}$ m/s. Find its average velocity for the first ${4}$ seconds.
Solution:
(a)
The interval has length ${2-0=2}$:
$\begin{align*}&f_{\text{ave}}\\&=\dfrac{1}{2}\int_0^2(3x^2-2x)\,dx\\&=\dfrac{1}{2}\Big[x^3-x^2\Big]_0^2\\&=\dfrac{1}{2}(8-4)\\&=2\end{align*}$
Now solve ${f(c)=2}$:
$\begin{align*}3c^2-2c&=2\\3c^2-2c-2&=0\end{align*}$
Use the quadratic formula. The coefficients are ${3}$, ${-2}$, and ${-2}$:
$\begin{align*}&c\\&=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(3)(-2)}}{2(3)}\\&=\dfrac{2\pm\sqrt{28}}{6}\\&=\dfrac{2\pm 2\sqrt{7}}{6}\\&=\dfrac{1\pm\sqrt{7}}{3}\end{align*}$
The two numbers are about ${1.22}$ and ${-0.55}$. Only ${\dfrac{1+\sqrt{7}}{3}\approx 1.22}$ is in ${[0,2]}$.
(b)
The interval has length ${\dfrac{\pi}{2}}$:
$\begin{align*}&f_{\text{ave}}\\&=\dfrac{1}{\pi/2}\int_0^{\pi/2}\cos x\,dx\\&=\dfrac{2}{\pi}\Big[\sin x\Big]_0^{\pi/2}\\&=\dfrac{2}{\pi}(1-0)\\&=\dfrac{2}{\pi}\approx 0.64\end{align*}$
(c)
The average velocity is the average value of ${v(t)}$ on ${[0,4]}$:
$\begin{align*}&v_{\text{ave}}\\&=\dfrac{1}{4}\int_0^4 3t^2\,dt\\&=\dfrac{1}{4}\Big[t^3\Big]_0^4\\&=\dfrac{1}{4}(64)\\&=16\ \text{m/s}\end{align*}$
The integral, ${64}$ m, is the distance the car travels. Dividing by the time, ${4}$ s, gives the average velocity. A car moving at a steady ${16}$ m/s would go just as far in ${4}$ seconds.