Chapter 7: Integrals

Ans: The given expression is, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$. 

Assume, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$ 

$\Rightarrow I=\int_{1}^{4}{\left| x-1 \right|dx}+\int_{1}^{4}{\left| x-2 \right|dx}+\int_{1}^{4}{\left| x-3 \right|dx}$

$\therefore I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\,\,\,...\left( 1 \right)$

where, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx},\,{{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx},\,{{I}_{3}}=+\int_{1}^{4}{\left| x-3 \right|dx}$ 

Now, consider, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx}$, where $\left( x-1 \right)\ge 0\,\forall \,1\le x\le 4$  $\therefore {{I}_{1}}=\int_{1}^{4}{\left( x-1 \right)dx}=\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{1}^{4}=\left[ 8-4-\dfrac{1}{2}+1 \right]=\dfrac{9}{2}\,\,\,...\left( 2 \right)\,$

Again, consider, ${{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx}$, where $\left( x-2 \right)\ge 0\,\forall \,2\le x\le 4$ and $\left( x-2 \right)\le 0\,\forall \,1\le x\le 2$.

$\therefore {{I}_{2}}=\int_{1}^{2}{\left( 2-x \right)dx}+\int_{2}^{4}{\left( x-2 \right)dx}=\left[ 2x-\dfrac{{{x}^{2}}}{2} \right]_{1}^{4}+\left[ \dfrac{{{x}^{2}}}{2}-2x \right]_{2}^{4}$

$\Rightarrow {{I}_{2}}=\left[ 4-2-2+\dfrac{1}{2} \right]+\left[ 8-8-2+4 \right]=\dfrac{1}{2}+2=\dfrac{5}{2}\,\,\,...\left( 3 \right)\,$

Also, consider, ${{I}_{3}}=\int_{1}^{4}{\left| x-3 \right|dx}$, where $\left( x-3 \right)\ge 0\,\forall \,3\le x\le 4$ and $\left( x-3 \right)\le 0\,\forall \,1\le x\le 3$.

$\therefore {{I}_{3}}=\int_{1}^{3}{\left( 3-x \right)dx}+\int_{3}^{4}{\left( x-3 \right)dx}=\left[ 3-\dfrac{{{x}^{2}}}{2} \right]_{1}^{3}+\left[ \dfrac{{{x}^{2}}}{2}-3x \right]_{3}^{4}$

$\Rightarrow {{I}_{3}}=\left[ 9-\dfrac{9}{2}-3+\dfrac{1}{2} \right]+\left[ 8-12-\dfrac{9}{2}+9 \right]=2+\dfrac{1}{2}=\dfrac{5}{2}\,\,\,...\left( 4 \right)\,$

Now, from equations $\left( 1 \right)$, $\left( 2 \right)$, $\left( 3 \right)$ and $\left( 4 \right)$ it can be obtained that, 

$I=\dfrac{9}{2}+\dfrac{5}{2}+\dfrac{5}{2}=\dfrac{19}{2}$

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

$\Rightarrow 2I=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}={{\pi }^{2}}-\pi \left[ \tan x-\sec x \right]_{0}^{\pi }$

$\Rightarrow 2I={{\pi }^{2}}-\pi \left[ 0-\left( -1 \right)-0+1 \right]={{\pi }^{2}}-2\pi $

$\Rightarrow I=\dfrac{\pi \left( \pi -2 \right)}{2}$

Prove the following (Exercises 32 to 37)

Ans: The given equation is, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\dfrac{2}{3}+\log \dfrac{2}{3}$. 

Assume, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}$ 

Now consider it as shown below,

$\therefore \dfrac{1}{{{x}^{2}}\left( x+1 \right)}=\dfrac{A}{x}+\dfrac{B}{{{x}^{2}}}+\dfrac{C}{\left( x+1 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 1=Ax\left( x+1 \right)+B\left( x+1 \right)+C\left( {{x}^{2}} \right)$

$\Rightarrow 1=A{{x}^{2}}+Ax+Bx+B+C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=0\,\,\,...\left( 2 \right)$ 

$A+B=0\,\,\,...\left( 3 \right)$

$B=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-1,\,B=1,\,C=1$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$I=\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\int_{1}^{3}{\left\{ \dfrac{-1}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{\left( x+1 \right)} \right\}dx}$ 

$\Rightarrow I=\left[ -\log x-\dfrac{1}{x}+\log \left( x+1 \right) \right]_{1}^{3}=\left[ \log \left( \dfrac{x+1}{x} \right)-\dfrac{1}{x} \right]_{1}^{3}=\log \left( \dfrac{4}{3} \right)-\dfrac{1}{3}-\log \left( 2 \right)+1$

$\Rightarrow I=\log 4-\log 3-\log 2+\dfrac{2}{3}=\log 2-\log 3+\dfrac{2}{3}=\log \left( \dfrac{2}{3} \right)+\dfrac{2}{3}$

Henceforth, it can be proved.

Ans: The given equation is, $\int_{0}^{4}{x{{e}^{x}}dx}=1$. 

Assume, $I=\int_{0}^{4}{x{{e}^{x}}dx}$ 

Using integration by parts, it can be obtained that,

$I=x\int_{0}^{4}{{{e}^{x}}dx}-\int_{0}^{4}{\left\{ \left( \dfrac{d\left( x \right)}{dx} \right)\int{{{e}^{x}}} \right\}}=\left[ x{{e}^{x}} \right]_{0}^{1}-\left[ {{e}^{x}} \right]_{0}^{1}=e-e+1=1$ 

Henceforth, it can be proved.

Ans: The given equation is, $\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$. 

Assume, $I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}$ 

Now, consider $f(x)={{x}^{17}}{{\cos }^{4}}x$ 

$\therefore f\left( -x \right)={{\left( -x \right)}^{17}}{{\cos }^{4}}\left( -x \right)=-{{x}^{17}}{{\cos }^{4}}x=-f\left( x \right)$

$\Rightarrow f\left( x \right)$ is an odd function and henceforth it is known to us that when $f(x)$ is an odd function, then $\int_{-a}^{a}{f\left( x \right)dx}=0$. 

$\therefore I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$

Henceforth, it can be proved.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}=\dfrac{2}{3}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\left( 1-{{\cos }^{2}}x \right)\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\sin xdx}-\int_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}x\sin xdx}$

$\Rightarrow I=\left[ -\cos x \right]_{0}^{\dfrac{\pi }{2}}+\left[ \dfrac{co{{s}^{3}}x}{3} \right]_{0}^{\dfrac{\pi }{2}}=1-\dfrac{1}{3}=\dfrac{2}{3}$

Henceforth, it can be proved.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}=1-\log 2$. 

Assume, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{2}}x\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\left( 1-{{\sec }^{2}}x \right)\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\tan xdx}-\int_{0}^{\dfrac{\pi }{4}}{{{\sec }^{2}}x\tan xdx}$

$\Rightarrow I=2\left[ \dfrac{{{\tan }^{2}}x}{2} \right]_{0}^{\dfrac{\pi }{4}}+2\left[ \log \cos x \right]_{0}^{\dfrac{\pi }{4}}=1+2\left[ \log \cos \dfrac{\pi }{4}-\log \cos 0 \right]_{0}^{\dfrac{\pi }{2}}=1-\log 2-\log 1$

$\Rightarrow I=1-\log 2$

Henceforth, it can be proved.

Ans: The given equation is, $\int_{0}^{1}{{{\sin }^{-1}}xdx}=\dfrac{\pi }{2}-1$. 

Assume, $I=\int_{0}^{1}{{{\sin }^{-1}}xdx}$ 

$\Rightarrow I=\int_{0}^{1}{{{\sin }^{-1}}x.1dx}$

Using integration by parts, it can be obtained that,

$I=\left[ {{\sin }^{-1}}x.x \right]_{0}^{1}-\int_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}xdx}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{\left( -2x \right)}{\sqrt{1-{{x}^{2}}}}dx}$ 

Now, substitute $1-{{x}^{2}}=t$ 

$\therefore \left( -2x \right)dx=dt$

And also when $x=0,\,t=1$ and when $x=1,\,t=0$.

$\therefore I=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{\sqrt{t}}}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\left[ 2\sqrt{t} \right]_{1}^{0}={{\sin }^{-1}}1-\sqrt{1}=\dfrac{\pi }{2}-1$ Henceforth, it can be clearly proved.

Choose the correct answers in Exercises 38 to 40

Ans: The given expression is, $\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$. 

Assume, $I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$ 

Now, consider ${{e}^{x}}=t$ 

$\therefore {{e}^{x}}dx=dt$

$\therefore I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}=\int{\dfrac{1}{1+{{t}^{2}}}dt}=\int{{{\tan }^{-1}}\operatorname{t}dt}+C$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (A). 

Ans: The given expression is, $\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$. 

Assume, $I=\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$ 

$\Rightarrow I=\int{\dfrac{(\sin x+\cos x)(\cos x-\sin x)}{{{\left( \sin x+\cos x \right)}^{2}}}dx}=\int{\dfrac{(\cos x-\sin x)}{\left( \sin x+\cos x \right)}dx}$

Now, substitute $\left( \sin x+\cos x \right)=t$ 

$\therefore (\cos x-\sin x)dx=dt$

$\therefore I=I=\int{\dfrac{1}{t}dt=\log \left| t \right|+C=\log \left| \cos x+\sin x \right|+C}$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (B). 

Ans: Assume, $I=\int_{a}^{b}{xf\left( x \right)dx}\,\,\,...\left( 1 \right)$ 

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( a+b-x \right)dx}\,\,\,\left[ \because \int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx} \right]$

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( x \right)dx}=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}-I$   using $\left( 1 \right)$

$\Rightarrow 2I=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}$

$\Rightarrow I=\dfrac{\left( a+b \right)}{2}\int_{a}^{b}{f\left( x \right)dx}$

Hence, the correct answer is option (D).