Integrals — Practice

Question 1

Evaluate $\displaystyle\int\frac{dx}{\sqrt{5-4x-x^{2}}}$.

Accepted Answer

$5-4x-x^{2}=9-(x+2)^{2}$.

$\displaystyle\int\frac{dx}{\sqrt{9-(x+2)^{2}}}=\sin^{-1}\frac{x+2}{3}+C$.

Question 2

Evaluate $\displaystyle\int\frac{x^{2}-1}{x^{4}+1}\,dx$.

Accepted Answer

The integrand becomes $\dfrac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}$.

Put $t=x+\dfrac1x$, so $dt=\left(1-\dfrac{1}{x^{2}}ight)dx$ and $x^{2}+\dfrac{1}{x^{2}}=t^{2}-2$.

$\displaystyle\int\frac{dt}{t^{2}-2}=\frac{1}{2\sqrt2}\ln\left|\frac{t-\sqrt2}{t+\sqrt2}ight|+C$, with $t=x+\dfrac1x$.

Question 3

Evaluate $\displaystyle\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx$.

Accepted Answer

Take $f(x)=\dfrac1x$, so $f'(x)=-\dfrac{1}{x^{2}}$ and the bracket is exactly $f+f'$.

Hence the integral is $\dfrac{e^{x}}{x}+C$.

Question 4

Evaluate $\displaystyle\int\frac{3x+5}{x^{2}+4x+8}\,dx$.

Accepted Answer

$3x+5=	frac32(2x+4)-1$.

First part: $	frac32\ln\left(x^{2}+4x+8ight)$.

Second part: $-\displaystyle\int\frac{dx}{(x+2)^{2}+4}=-	frac12	an^{-1}\dfrac{x+2}{2}$.

Sum: $	frac32\ln\left(x^{2}+4x+8ight)-	frac12	an^{-1}\dfrac{x+2}{2}+C$.

Question 5

Evaluate $\displaystyle\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$.

Accepted Answer

Substituting $x\mapsto	frac{\pi}{2}-x$ swaps sine and cosine, giving the same integral with $\sqrt{\cos x}$ on top.

Adding the two, the integrands sum to $1$.

$2I=\dfrac{\pi}{2}$, so $I=\dfrac{\pi}{4}$.

Question 6

Evaluate $\displaystyle\int_{-\pi/4}^{\pi/4}\frac{x^{3}\sin^{2}x}{1+x^{2}}\,dx$.

Accepted Answer

$x^{3}$ is odd, $\sin^{2}x$ is even and $1+x^{2}$ is even, so the whole integrand is odd.

The interval is symmetric about the origin, so the integral is $0$.

No antiderivative is needed, and none exists in elementary terms.

Question 7

Evaluate $\displaystyle\lim_{n	o\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}ight)$.

Accepted Answer

The $r$th term is $\dfrac{1}{n+r}=\dfrac1n\cdot\dfrac{1}{1+\frac rn}$.

So the limit is $\displaystyle\int_0^1\frac{dx}{1+x}=\ln 2$.

Question 8

Evaluate $\displaystyle\int_0^{3\pi}\left|\cos x\right|\,dx$.

Accepted Answer

$\left|\cos xight|$ has period $\pi$, and $\displaystyle\int_0^{\pi}\left|\cos xight|dx=2$.

So the integral is $3	imes2=6$.

Dropping the modulus and integrating $\cos x$ over $[0,3\pi]$ gives $0$.

Question 9

If $F(x)=\displaystyle\int_{0}^{x^{2}}\frac{\sin t}{t}\,dt$ for $x>0$, find $F'(x)$.

Accepted Answer

$F'(x)=\dfrac{\sin\left(x^{2}ight)}{x^{2}}\cdot 2x=\dfrac{2\sin\left(x^{2}ight)}{x}$.

The integrand has no elementary antiderivative, yet the derivative is immediate.

Question 10

Find the area enclosed by $y=x^{2}$ and $y=2-x^{2}$.

Accepted Answer

They meet where $x^{2}=2-x^{2}$, that is $x=\pm1$.

Between them $2-x^{2}$ is above.

$A=\displaystyle\int_{-1}^{1}\left(2-2x^{2}ight)dx=\left[2x-	frac{2x^{3}}{3}ight]_{-1}^{1}=	frac83$.

Question 11

Find the area bounded by $y=x^{3}-3x$ and the $x$-axis between $x=-2$ and $x=2$.

Accepted Answer

Zeros at $x=0,\pm\sqrt3$, all inside $[-2,2]$.

By symmetry the total area is twice that on $[0,2]$.

$\displaystyle\int_0^{\sqrt3}\left(3x-x^{3}ight)dx=	frac94$ and $\displaystyle\int_{\sqrt3}^{2}\left(x^{3}-3xight)dx=	frac14$.

Total $=2\left(	frac94+	frac14ight)=5$.

Question 12

Show that $\dfrac12\le\displaystyle\int_0^{1}\frac{dx}{1+x^{4}}\le1$ without evaluating the integral.

Accepted Answer

On $[0,1]$ we have $0\le x^{4}\le1$, so $1\le1+x^{4}\le2$.

Taking reciprocals reverses the inequalities: $\dfrac12\le\dfrac{1}{1+x^{4}}\le1$.

Integrating over an interval of length $1$ gives $\dfrac12\le I\le1$.

The true value is about $0.867$, comfortably inside the bounds.

Practice — Integrals

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities