"Integral Calculus" — Practice

Question 1

Evaluate $\displaystyle\int\frac{dx}{x^{2}-6x+13}$.

Accepted Answer

The denominator has no real roots, so complete the square: $x^{2}-6x+13=(x-3)^{2}+4=(x-3)^{2}+2^{2}$.

With $u=x-3$ this is the standard form $\displaystyle\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}	an^{-1}\frac{u}{a}+C$ with $a=2$.

$$\int\frac{dx}{x^{2}-6x+13}=\frac{1}{2}	an^{-1}\!\left(\frac{x-3}{2}ight)+C$$

Check by differentiating: $\frac{1}{2}\cdot\frac{1}{1+\left(\frac{x-3}{2}ight)^{2}}\cdot\frac{1}{2}=\frac{1}{4+(x-3)^{2}}$, which is the integrand.

Question 2

Evaluate $\displaystyle\int\frac{3x+1}{x^{2}+2x+5}\,dx$.

Accepted Answer

The denominator's derivative is $2x+2$. Write $3x+1=\frac{3}{2}(2x+2)-2$, since $\frac{3}{2}(2x+2)=3x+3$.

The integral separates:

$$\frac{3}{2}\int\frac{2x+2}{x^{2}+2x+5}\,dx-2\int\frac{dx}{x^{2}+2x+5}$$

The first is $\int\frac{f'}{f}=\log\lvert fvert$. For the second, complete the square: $x^{2}+2x+5=(x+1)^{2}+4$, giving $\frac{1}{2}	an^{-1}\frac{x+1}{2}$.

$$=\frac{3}{2}\log\left(x^{2}+2x+5ight)-	an^{-1}\!\left(\frac{x+1}{2}ight)+C$$

No modulus is needed on the logarithm: the quadratic is positive definite.

Question 3

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

Accepted Answer

This is the $e^{x}\left[f(x)+f'(x)\right]$ pattern. Take $f(x)=\dfrac{1}{x}=x^{-1}$.

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

$$\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx=\frac{e^{x}}{x}+C$$

Check: $\dfrac{d}{dx}\left(e^{x}x^{-1}\right)=e^{x}x^{-1}-e^{x}x^{-2}$, which is the integrand. Attempting by parts here works but takes far longer.

Question 4

Evaluate $\displaystyle\int x^{2}e^{x}\,dx$.

Accepted Answer

By ILATE, differentiate $x^{2}$. Take $u=x^{2}$, $v'=e^{x}$, so $v=e^{x}$:

$I=x^{2}e^{x}-2\displaystyle\int xe^{x}\,dx$

Apply by parts again to $\int xe^{x}dx$, with $u=x$, $v=e^{x}$:

$\displaystyle\int xe^{x}\,dx=xe^{x}-\int e^{x}dx=xe^{x}-e^{x}$

Substituting back:

$$I=x^{2}e^{x}-2xe^{x}+2e^{x}+C=e^{x}\left(x^{2}-2x+2\right)+C$$

Check: $\dfrac{d}{dx}\left[e^{x}(x^{2}-2x+2)\right]=e^{x}(x^{2}-2x+2)+e^{x}(2x-2)=x^{2}e^{x}$.

Question 5

Evaluate $\displaystyle\int\frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx$.

Accepted Answer

Decompose as $\dfrac{A}{x-1}+\dfrac{B}{(x-1)^{2}}+\dfrac{D}{x+3}$, so

$x^{2}+1=A(x-1)(x+3)+B(x+3)+D(x-1)^{2}$

Put $x=1$: $2=4B$, so $B=\frac{1}{2}$.

Put $x=-3$: $10=16D$, so $D=\frac{5}{8}$.

Comparing coefficients of $x^{2}$: $1=A+D$, so $A=\frac{3}{8}$.

Check with the constant term: $-3A+3B+D=-\frac{9}{8}+\frac{3}{2}+\frac{5}{8}=1$. Correct.

$$\int=\frac{3}{8}\log\lvert x-1\rvert-\frac{1}{2(x-1)}+\frac{5}{8}\log\lvert x+3\rvert+C$$

Question 6

Evaluate $\displaystyle\int\cos\sqrt{x}\,dx$.

Accepted Answer

Put $t=\sqrt{x}$, so $x=t^{2}$ and $dx=2t\,dt$:

$I=\displaystyle\int\cos t\cdot 2t\,dt=2\int t\cos t\,dt$

Now by parts with $u=t$, $v'=\cos t$, so $v=\sin t$:

$\displaystyle\int t\cos t\,dt=t\sin t-\int\sin t\,dt=t\sin t+\cos t$

So $I=2\left(t\sin t+\cos t\right)+C$, and substituting $t=\sqrt{x}$ back:

$$\int\cos\sqrt{x}\,dx=2\left(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x}\right)+C$$

The substitution had to come first: by parts on the original integrand goes nowhere.

Question 7

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

Accepted Answer

The limits are symmetric about the origin, so test parity first.

$x^{3}$ is odd and $\cos x$ is even, and odd times even is odd. Formally, if $f(x)=x^{3}\cos x$ then

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

So $f$ is odd, and by property P2 the integral over symmetric limits vanishes.

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

No antiderivative was needed, and in fact finding one would require by parts three times.

Question 8

Evaluate $\displaystyle\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{	an x}}$. *(JEE Main 2013)*

Accepted Answer

Here $a+b=\dfrac{\pi}{6}+\dfrac{\pi}{3}=\dfrac{\pi}{2}$, so King's property reflects $x\mapsto\dfrac{\pi}{2}-x$, which sends $	an x$ to $\cot x$.

$I=\displaystyle\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\cot x}}=\int_{\pi/6}^{\pi/3}\frac{\sqrt{	an x}\;dx}{\sqrt{	an x}+1}$

using $\sqrt{\cot x}=1/\sqrt{	an x}$ and multiplying through.

Adding the two expressions, the numerators combine to the denominator:

$2I=\displaystyle\int_{\pi/6}^{\pi/3}\frac{1+\sqrt{	an x}}{1+\sqrt{	an x}}\,dx=\int_{\pi/6}^{\pi/3}1\,dx=\frac{\pi}{6}$

$$I=\frac{\pi}{12}$$

Question 9

Evaluate $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^{2}x}{1+e^{x}}\,dx$.

Accepted Answer

The limits are symmetric, so $a+b=0$ and King's property gives $x\mapsto-x$. Since $\cos^{2}$ is even:

$I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^{2}x}{1+e^{-x}}\,dx=\int_{-\pi/2}^{\pi/2}\frac{e^{x}\cos^{2}x}{1+e^{x}}\,dx$

Adding the two expressions for $I$, the denominators combine:

$2I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\left(1+e^{x}\right)\cos^{2}x}{1+e^{x}}\,dx=\int_{-\pi/2}^{\pi/2}\cos^{2}x\,dx$

Now $\cos^{2}x$ is even, so this is $2\displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx=2\cdot\frac{\pi}{4}=\frac{\pi}{2}$.

$$I=\frac{\pi}{4}$$

The general result: for even $g$, the integral of $g(x)/(1+c^{x})$ over $[-a,a]$ equals the integral of $g$ over $[0,a]$.

Question 10

Evaluate $\displaystyle\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx$.

Accepted Answer

Here $a+b=\pi$. Under $x\mapsto\pi-x$ we have $\sin(\pi-x)=\sin x$ and $\cos^{2}(\pi-x)=\cos^{2}x$, so only the $x$ changes:

$I=\displaystyle\int_{0}^{\pi}\frac{(\pi-x)\sin x}{1+\cos^{2}x}\,dx$

Adding the two expressions, the $x$ terms cancel:

$2I=\pi\displaystyle\int_{0}^{\pi}\frac{\sin x}{1+\cos^{2}x}\,dx$

Now substitute $u=\cos x$, so $du=-\sin x\,dx$, with limits $1$ down to $-1$:

$2I=\pi\displaystyle\int_{-1}^{1}\frac{du}{1+u^{2}}=\pi\left[	an^{-1}uight]_{-1}^{1}=\pi\left(\frac{\pi}{4}+\frac{\pi}{4}ight)=\frac{\pi^{2}}{2}$

$$I=\frac{\pi^{2}}{4}$$

Question 11

Evaluate $\displaystyle\int_{0}^{3/2}x\left\lfloor x^{2}\right\rfloor dx$. *(AIEEE 2011)*

Accepted Answer

On $\left[0,	frac{3}{2}ight]$, the value $x^{2}$ runs from $0$ to $2.25$, crossing $1$ at $x=1$ and $2$ at $x=\sqrt{2}$.

So $\left\lfloor x^{2}ightfloor=0$ on $[0,1)$, $=1$ on $\left[1,\sqrt{2}ight)$, and $=2$ on $\left[\sqrt{2},	frac{3}{2}ight]$.

The first piece contributes nothing. For the second:

$\displaystyle\int_{1}^{\sqrt{2}}x\,dx=\left[\frac{x^{2}}{2}ight]_{1}^{\sqrt{2}}=\frac{2-1}{2}=\frac{1}{2}$

For the third:

$\displaystyle\int_{\sqrt{2}}^{3/2}2x\,dx=\left[x^{2}ight]_{\sqrt{2}}^{3/2}=\frac{9}{4}-2=\frac{1}{4}$

$$I=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$$

Question 12

Evaluate $\displaystyle\int_{0}^{2\pi}\lvert\sin x\rvert\,dx$.

Accepted Answer

On $[0,\pi]$ we have $\sin x\ge 0$, so $\lvert\sin x\rvert=\sin x$. On $[\pi,2\pi]$ we have $\sin x\le 0$, so $\lvert\sin x\rvert=-\sin x$. Split at $\pi$:

$\displaystyle\int_{0}^{\pi}\sin x\,dx=\left[-\cos x\right]_{0}^{\pi}=1+1=2$

$\displaystyle\int_{\pi}^{2\pi}(-\sin x)\,dx=\left[\cos x\right]_{\pi}^{2\pi}=1+1=2$

$$I=2+2=4$$

Without the modulus the second piece would have been $-2$ and the total $0$. That cancellation is exactly what the modulus prevents, and it is the same reason area requires $\int\lvert y\rvert$.

Question 13

If $g(x)=\displaystyle\int_{0}^{x}\cos 4t\,dt$, prove that $g(x+\pi)=g(x)+g(\pi)$. *(AIEEE 2012)*

Accepted Answer

Split the defining integral at $\pi$:

$g(x+\pi)=\displaystyle\int_{0}^{x+\pi}\cos 4t\,dt=\int_{0}^{\pi}\cos 4t\,dt+\int_{\pi}^{x+\pi}\cos 4t\,dt$

The first piece is $g(\pi)$ by definition.

In the second, substitute $t=u+\pi$, so $dt=du$ and the limits become $0$ to $x$:

$\displaystyle\int_{\pi}^{x+\pi}\cos 4t\,dt=\int_{0}^{x}\cos\left(4u+4\pi\right)du$

Since $\cos$ has period $2\pi$, adding $4\pi$ changes nothing, so $\cos(4u+4\pi)=\cos 4u$ and the integral is $g(x)$.

$$g(x+\pi)=g(\pi)+g(x)$$

The key step is that $4\pi$ is a whole number of periods of $\cos 4u$.

Question 14

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

Accepted Answer

Intersections: $x^{2}=x$ gives $x(x-1)=0$, so $x=0$ and $x=1$.

Between them, test $x=	frac{1}{2}$: the line gives $0.5$ and the parabola gives $0.25$, so the **line is upper**.

$A=\displaystyle\int_{0}^{1}\left(x-x^{2}ight)dx=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}ight]_{0}^{1}=\frac{1}{2}-\frac{1}{3}$

$$A=\frac{1}{6}	ext{ square units}$$

Testing an interior point is what settles which curve is upper. Guessing from the equations is where sign errors come from.

Question 15

The area bounded by the parabola $y^{2}=4\lambda x$ and the line $y=\lambda x$ is $\dfrac{1}{9}$. Find $\lambda$, given $\lambda>0$.

Accepted Answer

Substitute $y=\lambda x$ into $y^{2}=4\lambda x$:

$\lambda^{2}x^{2}=4\lambda x\;\Rightarrow\;\lambda x(\lambda x-4)=0$, so $x=0$ or $x=\dfrac{4}{\lambda}$.

On that range the parabola $y=2\sqrt{\lambda x}$ lies above the line, so

$A=\displaystyle\int_{0}^{4/\lambda}\left(2\sqrt{\lambda}\,x^{1/2}-\lambda x\right)dx=\left[\frac{4\sqrt{\lambda}}{3}x^{3/2}-\frac{\lambda x^{2}}{2}\right]_{0}^{4/\lambda}$

With $\left(\dfrac{4}{\lambda}\right)^{3/2}=\dfrac{8}{\lambda^{3/2}}$:

$A=\dfrac{4\sqrt{\lambda}}{3}\cdot\dfrac{8}{\lambda^{3/2}}-\dfrac{\lambda}{2}\cdot\dfrac{16}{\lambda^{2}}=\dfrac{32}{3\lambda}-\dfrac{8}{\lambda}=\dfrac{8}{3\lambda}$

Setting $\dfrac{8}{3\lambda}=\dfrac{1}{9}$ gives $3\lambda=72$.

$$\lambda=24$$

Question 16

Find the area bounded by $y=x^{2}-1$ and the $x$-axis between $x=-1$ and $x=1$.

Accepted Answer

On $(-1,1)$ we have $x^{2}<1$, so $y=x^{2}-1$ is **negative** throughout: the region lies below the axis.

The plain integral would therefore come out negative, and area must use the modulus:

$A=\displaystyle\int_{-1}^{1}\left\lvert x^{2}-1ightvert dx=\int_{-1}^{1}\left(1-x^{2}ight)dx$

The integrand is even, so use symmetry:

$A=2\displaystyle\int_{0}^{1}\left(1-x^{2}ight)dx=2\left[x-\frac{x^{3}}{3}ight]_{0}^{1}=2\left(1-\frac{1}{3}ight)$

$$A=\frac{4}{3}	ext{ square units}$$

Reporting $-\frac{4}{3}$ loses the mark: area is never negative.

Practice — "Integral Calculus"

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