Practice questions

Find an antiderivative of each term. Add ${1}$ to the exponent, and divide by the new exponent:

$\begin{align*}&F(x)\\&=4\cdot\dfrac{x^{3+1}}{3+1}-6\cdot\dfrac{x^{1+1}}{1+1}+2x+C\\&=4\cdot\dfrac{x^4}{4}-6\cdot\dfrac{x^2}{2}+2x+C\\&=x^4-3x^2+2x+C\end{align*}$

Write ${\dfrac{3}{\sqrt{x}}=3x^{-1/2}}$. An antiderivative of ${\sin x}$ is ${-\cos x}$:

$\begin{align*}&G(x)\\&=3\cdot\dfrac{x^{-1/2+1}}{-1/2+1}-\cos x+C\\&=3\cdot\dfrac{x^{1/2}}{1/2}-\cos x+C\\&=6\sqrt{x}-\cos x+C\end{align*}$

The general antiderivative:

$\begin{align*}&f(x)\\&=4\cdot\dfrac{x^{1+1}}{1+1}-3x+C\\&=2x^2-3x+C\end{align*}$

Use ${f(2)=5}$ to find ${C}$:

$\begin{align*}2(2)^2-3(2)+C&=5\\2+C&=5\\C&=3\end{align*}$

So ${f(x)=2x^2-3x+3}$.

Take the direction the car moves as positive. Start the clock (${t=0}$) when the driver brakes. Let ${s(t)}$ be the distance the car has gone since then, so ${s(0)=0}$.

The car is slowing down, so its acceleration is negative: ${a(t)=-5}$ m/s${^2}$. At the start, its velocity is ${v(0)=20}$ m/s.

The velocity is an antiderivative of the acceleration:

${v(t)=-5t+C_1}$

At ${t=0}$, the velocity is ${20}$:

$\begin{align*}v(0)&=20\\-5(0)+C_1&=20\\C_1&=20\end{align*}$

So ${v(t)=-5t+20}$. The car stops when its velocity is ${0}$:

$\begin{align*}-5t+20&=0\\t&=4\ \text{s}\end{align*}$

The position is an antiderivative of the velocity:

$\begin{align*}&s(t)\\&=-5\cdot\dfrac{t^{1+1}}{1+1}+20t+C_2\\&=-2.5t^2+20t+C_2\end{align*}$

Since ${s(0)=0}$, we get ${C_2=0}$. At ${t=4}$:

$\begin{align*}&s(4)\\&=-2.5(4)^2+20(4)\\&=-40+80\\&=40\ \text{m}\end{align*}$

The car travels ${40}$ m before it stops.

Cut ${[0,2]}$ into ${4}$ equal pieces. Each has width ${\Delta x=\dfrac{2}{4}=0.5}$. The cut points are ${0}$, ${0.5}$, ${1}$, ${1.5}$, ${2}$.

The left endpoints are ${0}$, ${0.5}$, ${1}$, ${1.5}$. The heights are

${f(0)=1}$,   ${f(0.5)=1.25}$,

${f(1)=2}$,   ${f(1.5)=3.25}$.

Add the areas of the rectangles. Each one is height times width:

$\begin{align*}&L_4\\&=(1+1.25+2+3.25)(0.5)\\&=(7.5)(0.5)\\&=3.75\end{align*}$

The right endpoints are ${0.5}$, ${1}$, ${1.5}$, ${2}$. The heights are

${f(0.5)=1.25}$,   ${f(1)=2}$,

${f(1.5)=3.25}$,   ${f(2)=5}$.

$\begin{align*}&R_4\\&=(1.25+2+3.25+5)(0.5)\\&=(11.5)(0.5)\\&=5.75\end{align*}$

The curve rises from left to right on ${[0,2]}$.

• With right endpoints, each rectangle is as tall as the highest point of the curve on its piece. So each sticks out above the curve, and ${R_4=5.75}$ is too big.
• With left endpoints, each rectangle is as tall as the lowest point. So each stays under the curve, and ${L_4=3.75}$ is too small.

The true area is between them. (It is ${\dfrac{14}{3}\approx 4.67}$. You can check this with the Fundamental Theorem.)

The graph of ${y=\sqrt{9-x^2}}$ is the top half of the circle ${x^2+y^2=9}$, with radius ${3}$. The integral is the area of this half circle:

$\begin{align*}&\int_{-3}^{3}\sqrt{9-x^2}\,dx\\&=\dfrac{1}{2}\pi(3)^2\\&=\dfrac{9\pi}{2}\approx 14.14\end{align*}$

The graph of ${y=2-x}$ is a line. It crosses the ${x}$-axis at ${x=2}$.

• From ${0}$ to ${2}$, the line is above the axis. It makes a triangle with base ${2}$ and height ${2}$. Its area is ${\dfrac{1}{2}(2)(2)=2}$.
• From ${2}$ to ${4}$, the line is below the axis. It makes a triangle with the same area, ${2}$. Area below the axis counts as negative.

So

$\displaystyle\int_0^4(2-x)\,dx=2-2=0$.

The two pieces add up to the whole:

$\begin{align*}&\int_0^2 f(x)\,dx+\int_2^5 f(x)\,dx\\&=\int_0^5 f(x)\,dx\end{align*}$

So

$\displaystyle\int_2^5 f(x)\,dx=8-3=5$.

For the second integral, split it, and take the constant ${3}$ out:

$\begin{align*}&\int_2^5\big(3f(x)+1\big)\,dx\\&=3\int_2^5 f(x)\,dx+\int_2^5 1\,dx\\&=3(5)+(5-2)\\&=18\end{align*}$

Here $\displaystyle\int_2^5 1\,dx$ is the area of a rectangle with height ${1}$ and width ${3}$.

The integrand ${\cos(t^2)}$ is continuous. By Part 1, replace ${t}$ by ${x}$:

$\dfrac{d}{dx}\displaystyle\int_0^x\cos(t^2)\,dt=\cos(x^2)$

The upper limit is ${x^3}$, not ${x}$. So use Part 1 with the chain rule. Let ${u=x^3}$, so ${u'=3x^2}$:

$\begin{align*}&\dfrac{d}{dx}\int_1^{x^3}\ln t\,dt\\&=\ln(x^3)\cdot 3x^2\\&=3\ln x\cdot 3x^2\\&=9x^2\ln x\end{align*}$

Find an antiderivative:

$\dfrac{x^{3+1}}{3+1}-2\cdot\dfrac{x^{1+1}}{1+1}=\dfrac{x^4}{4}-x^2$

Now use Part 2:

$\begin{align*}&\int_1^2(x^3-2x)\,dx\\&=\Big[\dfrac{x^4}{4}-x^2\Big]_1^2\\&=(4-4)-\left(\dfrac{1}{4}-1\right)\\&=0-\left(-\dfrac{3}{4}\right)\\&=\dfrac{3}{4}\end{align*}$

On ${\left[0,\dfrac{\pi}{2}\right]}$, ${\cos x\ge 0}$. So the area is the integral. An antiderivative of ${\cos x}$ is ${\sin x}$:

$\begin{align*}&\int_0^{\pi/2}\cos x\,dx\\&=\Big[\sin x\Big]_0^{\pi/2}\\&=\sin\dfrac{\pi}{2}-\sin 0\\&=1-0\\&=1\end{align*}$

Write ${\dfrac{1}{x^2}=x^{-2}}$. Integrate one term at a time:

$\begin{align*}&\int\left(x^2-x^{-2}+e^x\right)dx\\&=\dfrac{x^{2+1}}{2+1}-\dfrac{x^{-2+1}}{-2+1}+e^x+C\\&=\dfrac{x^3}{3}-\dfrac{x^{-1}}{-1}+e^x+C\\&=\dfrac{x^3}{3}+\dfrac{1}{x}+e^x+C\end{align*}$

Expand first: ${(x+2)^2=x^2+4x+4}$. Then integrate one term at a time:

$\begin{align*}&\int(x^2+4x+4)\,dx\\&=\dfrac{x^{2+1}}{2+1}+4\cdot\dfrac{x^{1+1}}{1+1}+4x+C\\&=\dfrac{x^3}{3}+2x^2+4x+C\end{align*}$

The displacement is the change in position: where the particle ends, compared with where it started. It is the integral of the velocity. The total distance counts every meter moved, forward or backward.

An antiderivative of ${v}$ is

${F(t)=\dfrac{t^3}{3}-2t^2+3t}$.

Its values: ${F(0)=0}$, ${F(1)=\dfrac{4}{3}}$, ${F(3)=0}$, and ${F(4)=\dfrac{4}{3}}$.

The displacement:

$\begin{align*}&\int_0^4 v(t)\,dt\\&=F(4)-F(0)\\&=\dfrac{4}{3}\ \text{m}\end{align*}$

For the distance, factor the velocity to see when it changes sign:

${v(t)=(t-1)(t-3)}$

• On ${(0,1)}$: ${v(t)>0}$. The particle moves forward.
• On ${(1,3)}$: ${v(t)<0}$. It moves backward.
• On ${(3,4)}$: ${v(t)>0}$. It moves forward again.

Integrate each piece:

$\displaystyle\int_0^1 v(t)\,dt=F(1)-F(0)=\dfrac{4}{3}$

$\displaystyle\int_1^3 v(t)\,dt=F(3)-F(1)=-\dfrac{4}{3}$

$\displaystyle\int_3^4 v(t)\,dt=F(4)-F(3)=\dfrac{4}{3}$

Add the sizes:

$\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3}=4\ \text{m}$

So the particle travels ${4}$ m in all, but it ends only ${\dfrac{4}{3}}$ m from where it started.

Let ${u=x^2+3}$. Then ${du=2x\,dx}$.

So ${x\,dx=\dfrac{1}{2}\,du}$.

$\begin{align*}&\int x(x^2+3)^4\,dx\\&=\dfrac{1}{2}\int u^4\,du\\&=\dfrac{1}{2}\cdot\dfrac{u^{4+1}}{4+1}+C\\&=\dfrac{u^5}{10}+C\\&=\dfrac{(x^2+3)^5}{10}+C\end{align*}$

Let ${u=\sin x}$. Then ${du=\cos x\,dx}$.

$\begin{align*}&\int\sin^2 x\cos x\,dx\\&=\int u^2\,du\\&=\dfrac{u^3}{3}+C\\&=\dfrac{\sin^3 x}{3}+C\end{align*}$

Let ${u=1+e^x}$. Then ${du=e^x\,dx}$.

$\begin{align*}&\int\dfrac{e^x}{1+e^x}\,dx\\&=\int\dfrac{1}{u}\,du\\&=\ln|u|+C\\&=\ln(1+e^x)+C\end{align*}$

We can drop the absolute value, because ${1+e^x}$ is always positive.

Let ${u=x^2}$. Then ${du=2x\,dx}$, so ${x\,dx=\dfrac{1}{2}\,du}$. Change the limits:

• When ${x=0}$: ${u=0}$.
• When ${x=1}$: ${u=1}$.

$\begin{align*}&\int_0^1 xe^{x^2}\,dx\\&=\dfrac{1}{2}\int_0^1 e^u\,du\\&=\dfrac{1}{2}\Big[e^u\Big]_0^1\\&=\dfrac{1}{2}(e-1)\\&\approx 0.859\end{align*}$

Let ${f(x)=x^3\cos x}$. Then

${f(-x)=(-x)^3\cos(-x)=-x^3\cos x=-f(x)}$.

So ${f}$ is odd. The interval ${[-1,1]}$ is symmetric about ${0}$. The area above the axis on one side cancels the area below on the other side. So

$\displaystyle\int_{-1}^{1}x^3\cos x\,dx=0$.