MCQ Questions for Class 12 Commerce Maths Integrals Quiz 4

The value of

$\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}$ is

0%
$\displaystyle \frac{\pi }{1+\alpha ^{2}}$ if $\alpha > 1$
0%
$\displaystyle \frac{\pi }{\alpha ^{2}-1}$ if $\alpha > 1$
0%
$\displaystyle \frac{\pi }{1+\alpha ^{2}}$ if $\alpha < 1$
0%
$\displaystyle \frac{\pi }{\alpha ^{2}-1}$ if $\alpha < 1$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } =\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } }$

Substitute

$\displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt$

$\displaystyle I=\int _{ 0 }^{ \infty } \dfrac { 1 }{ 1-2\alpha \left( \dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } \right) +\alpha ^{ 2 } } .\dfrac { 2t }{ 1+{ t }^{ 2 } } dt$

$\displaystyle =\int _{ 0 }^{ \infty } \dfrac { 2t }{ 1+{ t }^{ 2 }-2\alpha \left( 1-{ t }^{ 2 } \right) +\alpha ^{ 2 }\left( 1+{ t }^{ 2 } \right) } dt$

$\displaystyle =\frac { \pi }{ { \alpha }^{ 2 }-1 }$ if

$\alpha >1$

Given

$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$ the value of

$\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx }$ is?

0%
${ e }^{ 4 }-e$
0%
${ e }^{ 4 }-a$
0%
$2{ e }^{ 4 }-a$
0%
$2{ e }^{ 4 }-e-a$

Explanation

Given

$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$

Let

$I=\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx }$

Put

$\displaystyle \ln { \left( x \right) } ={ t }^{ 2 }\Rightarrow \frac { 1 }{ x } dx=2tdt$

$\therefore I=\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } .{ 2t }^{ 2 }dt={ \left( t{ e }^{ { t }^{ 2 } } \right) }_{ 1 }^{ 2 }-\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } dt={ 2e }^{ 4 }-e-a.$

The value of

$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}+2x}}$ is

0%
$\pi /6$
0%
$\pi /3$
0%
$\pi /2$
0%
$\pi$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }+2x } } dx } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { \left( x+1 \right) }^{ 2 }-1 } } dx }$
Substitute

$u=x+1\Rightarrow du=dx$

$\displaystyle\therefore I=\int _{ 1 }^{ 2 }{ \frac { 1 }{ u\sqrt { { u }^{ 2 }-1 } } du }$
Substitute

$\displaystyle s=\sqrt { { u }^{ 2 }-1 } \Rightarrow ds=\frac { u }{ \sqrt { { u }^{ 2 }-1 } } du$

$\displaystyle\therefore I=\int _{ 0 }^{ \sqrt { 3 } }{ \frac { 1 }{ { s }^{ 2 }+1 } ds } ={ \left[ \tan ^{ -1 }{ s } \right] }_{ 0 }^{ \sqrt { 3 } }=\frac { \pi }{ 3 }$

Value of

$\displaystyle \int_{0}^{\pi /4}\left ( \sqrt{\tan x}-\sqrt{\cot x} \right )\: dx$ is

0%
$\sqrt{2}\log \left ( \sqrt{2}-1 \right )$
0%
$\sqrt{2}\log \left ( \sqrt{2}+1 \right )$
0%
$\log \left ( \sqrt{2}+1 \right )$
0%
$\log \left ( \sqrt{2}-1 \right )$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \sqrt { \tan { x } } -\sqrt { cot{ x } } \right) } dx=-\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { cos{ x }-sin{ x } }{ \sqrt { cos{ x }sin{ x } } } dx }$
Substitute

$\left( sin{ x+cos{ x } } \right) =t\Rightarrow 2sin{ x }cos{ x }={ t }^{ 2 }-1$

$\displaystyle \therefore I=\sqrt { 2 } \int _{ 1 }^{ \sqrt { 2 } }{ \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } \\ =\left[ \sqrt { 2 } \log { \left( t+\sqrt { { t }^{ 2 } } -1 \right) } \right] _{ 0 }^{ \sqrt { 2 } }\\ =\sqrt { 2 } \log { \left( \sqrt { 2 } -1 \right) }$

Value of

$\displaystyle \int_{0}^{2a}\dfrac{x^{3/2}}{\sqrt{2a-x}} dx$ is

0%
$\displaystyle \frac{3\pi a^{2}}{2}$
0%
$\pi a^{3}$
0%
$\sqrt{2}\pi a^{3}$
0%
$2\pi a^{3}$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 2a }{ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \sqrt { 2a-x } } dx }$

Substitute

$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$

$\displaystyle I=2\int _{ 0 }^{ \sqrt { 2a } }{ \frac { { u }^{ 4 } }{ \sqrt { 2a-{ u }^{ 2 } } } du }$

Substitute

$u=\sqrt { 2a } \sin { t } \Rightarrow du=\sqrt { 2a } \cos { t } dt$

$\displaystyle I=2\sqrt { 2a } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sqrt { 2 } { a }^{ \frac { 3 }{ 2 } }\sin ^{ 4 }{ t } \right) dt } =8{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sin ^{ 4 }{ t } dt }$

Using reduction formulae

$\displaystyle \int { \sin ^{ m }{ x } dx } =\frac { -\cos { x } \sin ^{ m-1 }{ x } }{ m } +\frac { m-1 }{ m } \int { \sin ^{ m-2 }{ x } dx }$

$\displaystyle I={ \left[ -2{ a }^{ 2 }\sin ^{ 3 }{ t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ t } dt }$

$\displaystyle =0+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \frac { 1 }{ 2 } -\frac { 1 }{ 2 } \cos { 2t } \right) dt }$

$\displaystyle ={ \left[ 3{ a }^{ 2 }t-3{ a }^{ 2 }\sin { t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { 3\pi { a }^{ 2 } }{ 2 }$

The value of

$\displaystyle \int_{a}^{b}\displaystyle \frac{\log x}{x}\: dx$ is

0%
$\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$
0%
$\displaystyle \frac{1}{2}\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$
0%
$\log \left ( a^{2}-b^{2} \right )$
0%
$\left ( a+b \right )\log \left ( a+b \right )$

Explanation

$\displaystyle I=\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx={ \left[ \log { x } .\log { x } \right] }_{ a }^{ b }-\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx\\ \Rightarrow 2I={ \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b }={ \left( \log { b } \right) }^{ 2 }-{ \left( \log { a } \right) }^{ 2 }$

$\displaystyle \Rightarrow I=\frac { 1 }{ 2 } \left( \log { b } +\log { a } \right) \left( \log { b } -\log { a } \right) =\frac { 1 }{ 2 } \log { ab } \log { \frac { b }{ a } }$

Value of

$\displaystyle \int_{0}^{25}\displaystyle \frac{1}{\sqrt{4+\sqrt{x}}}\: dx$ is

0%
$2\left ( \sqrt{29}-1 \right )$
0%
$2\left ( \sqrt{29}-5 \right )$
0%
$3 \sqrt{29}-1$
0%
none of these

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 25 }{ \frac { 1 }{ \sqrt { \sqrt { x } +4 } } dx }$

Substitute

$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$

$\displaystyle I=2\int _{ 0 }^{ 5 }{ \frac { u }{ \sqrt { u+4 } } du }$

Substitute

$s=u+4\Rightarrow ds=du$

$\displaystyle I=2\int _{ 4 }^{ 9 }{ \frac { s-4 }{ \sqrt { s } } ds } =2\int _{ 4 }^{ 9 }{ \left( \sqrt { s } -\frac { 4 }{ \sqrt { s } } \right) ds }$

$\displaystyle={ \left[ \frac { 4{ s }^{ 3/2 } }{ 3 } \right] }_{ 4 }^{ 9 }-{ \left[ 16\sqrt { s } \right] }_{ 4 }^{ 9 }=36-\frac { 32 }{ 3 } -48-32=-\frac { 164 }{ 3 }$

$\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$

0%
is equal to zero
0%
is equal to one
0%
is equal to $\displaystyle \frac{1}{2}$
0%
can not be evaluated

Explanation

$I = \displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$

Substitute

$\displaystyle x = \frac{1}{t}\Rightarrow dx =-\frac{1}{t^2}dt$

$I = \displaystyle \int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln (1/t)}{1/t} .\frac{-dt}{t^2}$

$\quad \displaystyle =\int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln t}{t} dt$

$\quad \displaystyle =\int_{\infty }^0 f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x} dx=-I$

$\Rightarrow 2I = 0\Rightarrow I = 0$

$\displaystyle \int_{0}^{\pi /3}\frac{\cos }{3+4\sin x}dx=K\log \frac{\left ( 3+2\sqrt{3} \right )}{3}$ then K is

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{8}$

Explanation

$\displaystyle I=\int _{ 0 }^{ \pi /3 }{ \cfrac { \cos { x } }{ 3+4\sin { x } } dx }$

Substituting

$t=3+4\sin { x } \Rightarrow dt=4\cos { x }$

$\displaystyle I=\frac { 1 }{ 4 } \int _{ 3 }^{ 3+2\sqrt { 3 } }{ \frac { 1 }{ t } dt } ={ \left[ \frac { \log { t } }{ 4 } \right] }_{ 3 }^{ 3+2\sqrt { 3 } }$

$\displaystyle=\frac { \log { \left( 3+2\sqrt { 3 } \right) } -\log { 3 } }{ 4 } =\frac { 1 }{ 4 } \log { \left( \frac { 3+2\sqrt { 3 } }{ 3 } \right) } \Rightarrow K=\frac { 1 }{ 4 }$

Suppose that F(x) is an antiderivative of f(x)

$\displaystyle =\frac{\sin x}{x},x> 0$ then

$\displaystyle \int_{1}^{3}\frac{\sin 2x}{x}$ can be expressed as

0%
$F(6) - F(2)$
0%
$\displaystyle \frac{1}{2}\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$
0%
$\displaystyle \frac{1}{2}\left ( F\left ( 3 \right )-F\left ( 1 \right ) \right )$
0%
$\displaystyle 2\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$

Explanation

$\displaystyle F\left ( x \right )=\int \frac{\sin x}{x}dx$ Now

$\displaystyle I = \int_{1}^{3}\frac{\sin 2x}{x}dx\left [ put2x=t \right ]=\int_{2}^{6}\frac{2}{2}\frac{\sin }{t}dt=\left [ F\left ( x \right ) \right ]_{2}^{6}=F(6)-F(2)$

The value of

$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{e^{x}+e^{-x}}$ is

0%
$\tan ^{-1}e$
0%
$\tan ^{-1}\left ( e \right )-\pi /4$
0%
$\tan ^{-1}\left ( e \right )-\tan ^{-1}\left ( 1/e \right )$
0%
$\tan ^{-1}\left ( 1/e \right )+\pi /4$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dx }{ e^{ x }+e^{ -x } } =\int _{ 0 }^{ 1 } \frac { { e }^{ x }dx }{ e^{ 2x }+1 }$

Substitute

${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$

$\displaystyle I=\int _{ 1 }^{ e }{ \frac { t }{ { t }^{ 2 }+1 } dt } ={ \left[ \tan ^{ -1 }{ t } \right] }_{ 1 }^{ e }=\tan ^{ -1 }{ e } -\frac { \pi }{ 4 }$

$\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx$ has the value equal to

0%
$0$
0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{5}{4}$
0%
$2$

Explanation

$\displaystyle I=\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ x } } \sin { \left( x-\frac { 1 }{ x } \right) dx }$

Put

$\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=\frac { 1 }{ { t }^{ 2 } } dt$

$\displaystyle I=\int _{ 2 }^{ \frac { 1 }{ 2 } }{ t\sin { \left( \frac { 1 }{ t } -t \right) \left( \frac { -1 }{ { t }^{ 2 } } \right) dt } } =\int _{ 2 }^{ \frac { 1 }{ 2 } }{ \frac { 1 }{ t } \sin { \left( t-\frac { 1 }{ t } \right) dt } }$

$\displaystyle =-\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ t } } \sin { \left( t-\frac { 1 }{ t } \right) dt } =-I.$

$\Rightarrow 2I=0\quad \Rightarrow I=0$

$\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt$ where

$x > 0$ , then the value(s) of

$x$ satisfying the equation,

$f(x) +f(1/x)=2$ is

0%
$2$
0%
$e$
0%
$\displaystyle e^{-2}$
0%
$\displaystyle e^{2}$

Explanation

$\displaystyle f\left( x \right)=\int _{ 1 }^{ x }{ \frac { \ln { t } }{ 1+t } dt }$

$\displaystyle \Rightarrow f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \ln { t } }{ 1+t } dt }$

Substituting

$t=\dfrac { 1 }{ u } \Rightarrow dt=\left( -\dfrac { 1 }{ { u }^{ 2 } } \right) du$

Therefore

$\displaystyle f\left( \dfrac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \dfrac { \ln { \left( \dfrac { 1 }{ u } \right) \left( -1 \right) } }{ \left( 1+\dfrac { 1 }{ u } \right) { u }^{ 2 } } du }$

$\displaystyle=\int _{ 1 }^{ x }{ \frac { \ln { u } }{ u\left( u+1 \right) } du } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }$

Now,

$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \cfrac { \ln { t } }{ \left( 1+t \right) } dt } +\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }$

$\displaystyle=\int _{ 1 }^{ x }{ \frac { \left( 1+t \right) \ln { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t } dt } =\frac { 1 }{ 2 } { \left( \ln { x } \right) }^{ 2 }$

Hence

$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =2\Rightarrow x={ e }^{ \pm 2 }$

Solve

$\displaystyle \int_{2}^{-13}\frac{dx}{\sqrt[5]{\left ( 3-x \right )^{4}}}$

0%
$\displaystyle -5\left ( \sqrt[5]{16}-1 \right )$
0%
$\displaystyle 5\left ( \sqrt[5]{16}-1 \right )$
0%
$\displaystyle -5\left ( \sqrt[5]{16}+1 \right )$
0%
None of these

Explanation

Let

$\displaystyle I=\int _{ 2 }^{ -13 }{ \frac { dx }{ 5\sqrt { { \left( 3-x \right) }^{ 4 } } } }$

Put

$3-x=t\Rightarrow -dx=dt$

$\displaystyle I=\int _{ 1 }^{ 16 }{ \frac { dt }{ { t }^{ \frac { 4 }{ 5 } } } } ={ \left[ \frac { { t }^{ \frac { 1 }{ 5 } } }{ \frac { 1 }{ 5 } } \right] }_{ 1 }^{ 16 }$

$=-5\left( \sqrt [ 5 ]{ 16 } -1 \right)$

Choose a function

$f(x)$ such that it is integrable over every interval on the real line

0%
$f(x) = [x]$
0%
$f(x)=x|x|$
0%
$f(x)=[sinx]$
0%
$f(x)=\dfrac{|x-1|}{x-1}$

$\displaystyle \int_{0}^{\infty }\frac{x}{\left ( 1+x \right )\left ( 1+x^{2} \right )}dx$

0%
$\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \frac{\pi }{2}$
0%
is sme as $\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( 1+x \right )\left ( 1+x^{2} \right )}$
0%
cannot be evaluated

Explanation

Let

$I=\int _{ 0 }^{ \infty }{ \cfrac { xdx }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } }$
Using partial fraction

$=\int _{ 0 }^{ \infty }{ \left( \cfrac { x+1 }{ 2\left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2\left( 1+x \right) } \right) dx } \\ ={ \left( \lim _{ b\rightarrow \infty }{ \cfrac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+x \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }x } \right) }_{ 0 }^{ b }\\ =\lim _{ b\rightarrow \infty }{ \left( \cfrac { 1 }{ 2 } \log { \left( 1+b \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+b \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }b \right) } \\ =\cfrac { \pi }{ 4 }$

State true or false:

The average value of the function

$f(x) = sin^2xcos^3x$ on the interval

$[ -\pi ,\pi ]$ is 0.

0%
True
0%
False

Explanation

$\int _{ -\pi }^{ \pi }{ f\left( x \right) dx } =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos ^{ 3 }{ x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos { x } dx } -\int _{ -\pi }^{ \pi }{ \sin ^{ 4 }{ x } \cos { x } dx }$

$\displaystyle ={ \left[ \frac { \sin ^{ 3 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }-{ \left[ \frac { \sin ^{ 5 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }=0$

$I_{1}$ is equal to

0%
$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$
0%
$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$
0%
$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$
0%
$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$

Explanation

$\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx$

$x=sin\theta, dx=cos\theta d\theta$

$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta$

$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta$

Ans:

$A$

Evaluate

$\displaystyle \int_{0}^{\pi /2} \frac{dx}{2+\sin 2x}$

0%
$\displaystyle \frac{2\pi }{{3}}$
0%
$\displaystyle \frac{\pi }{{3}}$
0%
$\displaystyle \frac{2\pi }{{5}}$
0%
None of these

Explanation

$\displaystyle \int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\sin { 2x } } } =\int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\frac { 2\tan { x } }{ 1+\tan ^{ 2 }{ x } } } } dx$

$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 2x }{ \frac { \sec ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } +\tan { x+1 } } dx }$

Put

$\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=dt$

$\displaystyle I=\frac { 1 }{ 2 } \int { \frac { 1 }{ { { t }^{ 2 }+t+1 } } } dt.=\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( t+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 3 }{ 2 } } } =\frac { 2\pi }{ 3 }$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

Reason is true
Assertion
Put

$\displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^{2}}dt$

$\displaystyle l=-\int_{3}^{1/3}tcosec ^{99}\left ( \frac{1}{t}-t \right )\frac{1}{t^{2}}dt$

$\displaystyle l=-\int_{1/3}^{3}\frac{1}{t}cosec^{99}\left ( t-\frac{1}{t} \right )dt$

$\displaystyle l=-1$

$\Rightarrow 2l=0$

$\Rightarrow l=0$

$\displaystyle \int_{\tfrac{1}{\sqrt{3}}}^{0}\dfrac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$

0%
$\displaystyle - \tan^{-1} \dfrac{1}{2}$
0%
$\displaystyle \tan^{-1} 1$
0%
$\displaystyle - \tan^{-1} \dfrac{1}{3}$
0%
$\displaystyle \tan^{-1} \dfrac{1}{\sqrt 2}$

Explanation

$I=\displaystyle \int_{\frac{1}{\sqrt{3}}}^{0}\frac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$

Substitute

$x=\displaystyle \frac{1}{z}$

$\Rightarrow\displaystyle dx=-\frac{1}{z^2}dz$

$I=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -\dfrac { 1 }{ { z }^{ 2 } } dz }{ \left( \dfrac { 2 }{ { z }^{ 2 } } +1 \right) \sqrt { \dfrac { 1 }{ { z }^{ 2 } } +1 } }$

$=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -zdz }{ \left( 2+{ z }^{ 2 } \right) \sqrt { 1+{ z }^{ 2 } } }$

Substitute

$\sqrt { 1+{ z }^{ 2 } } =t$

$\Rightarrow z^2+1=t^2$

$\Rightarrow zdz=tdt$

$I=\displaystyle \int _{ 2 }^{ \infty } \frac { -tdt }{ \left( 1+{ t }^{ 2 } \right) t }$

$I=[\cot^{-1}t]_{2}^{\infty}$

$\Rightarrow I=-\cot^{-1}2$

$\Rightarrow I=-\tan^{-1}{\dfrac{1}{2}}$

The value of definite integral

$\displaystyle \int_{\infty }^{0}\frac{Ze^{-z}}{\sqrt{1-e^{-2z}}}dz$

0%
$\displaystyle -\frac{\pi }{2}ln2$
0%
$\displaystyle \frac{\pi }{2}ln2$
0%
$\displaystyle -\pi ln2$
0%
$\displaystyle \pi ln\frac{1}{\sqrt{2}}$

Explanation

$\displaystyle l=\int_{\infty }^{0}\frac{ze^{-z}}{\sqrt{1-e^{-2z}}}dz$ put

$\displaystyle e^{-z}=\sin \theta$

$\displaystyle l=-\int_{0}^{\pi /2}\frac{ln\left ( \sin \theta \right )\left ( -\cos \theta \right )d\theta }{\sqrt{1-\sin ^{2}\theta }}=\int_{0}^{\pi /2}ln\sin \theta d\theta$

$\displaystyle \frac{-\pi }{2}ln2$

$\displaystyle \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0$ , then

0%
$1<\alpha <2$
0%
$\alpha <2$
0%
$0<\alpha<1$
0%
$\alpha=0$

Explanation

$\displaystyle\because \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0,$

$\therefore { e }^{ { x }^{ 2 } }\left( x-\alpha \right)$ must be

$+ve$ and

$-ve$ both for

$x\in (0,1)$

i.e.

${ e }^{ x }\left( x-\alpha \right) =0$ for one

$x\in (0,1)$

$\therefore \alpha \in(0,1)$

$\displaystyle \int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } }$ is equal to

0%
$\displaystyle \frac { \pi }{ 2 } +1$
0%
$\displaystyle \frac { \pi }{ 2 } -1$
0%
$\displaystyle 1-\frac { \pi }{ 2 }$
0%
none of these

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\int _{ 0 }^{ \pi /2 }{ \frac { a\cos { \theta } d\theta }{ a+a\cos { \theta } } }$

By putting

$x=a\sin { \theta } \Rightarrow dx=a\cos { \theta } d\theta$

$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \cos { \theta } }{ 1+\cos { \theta } } d\theta } =\int _{ 0 }^{ \pi /2 }{ \left( 1-\frac { 1 }{ 1+\cos { \theta } } \right) d\theta }$

$\displaystyle =\int _{ 0 }^{ \pi /2 }{ d\theta } -\frac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \sec ^{ 2 }{ \frac { \theta }{ 2 } } d\theta }$

$\displaystyle ={ \left[ \theta -\tan { \frac { \theta }{ 2 } } \right] }_{ 0 }^{ \pi /2 }=\frac { \pi }{ 2 } -\tan { \frac { \pi }{ 4 } } =\left( \frac { \pi }{ 2 } -1 \right)$

What is

$\displaystyle\int _{ 1 }^{ 2 }{ \ln { x } dx }$ equal to?

0%
$\ln { 2 }$
0%
$1$
0%
$\ln { \left( \dfrac { 4 }{ e } \right) }$
0%
$\ln { \left( \dfrac { e }{ 4 } \right) }$

Explanation

Let,

$y=\ln x$ then,

$x={ e^{ y } }$

Now at,

$x=1,y=0$

$x=2,y=\ln 2$

$\Rightarrow dx={ e }^{ y }.dy$

Putting the values, we get

$\displaystyle \int_{1}^{2} \ln x dx=\int _{ 0 }^{ \ln2 }{ { e }^{ y }.y.dy }$

$=\displaystyle { \left[ y.{ e }^{ y } \right] }_{ 0 }^{\ln2 }-\int _{ 0 }^{\ln2 }{ { e }^{ y }dy }$

$=2\ln 2-1$

$=2\ln 2-\ln(e)$

$=\ln\left(\dfrac { 4 }{ e }\right)$

Hence, C is correct.

$\displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\ dx$ is equal to

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Let

$I = \displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\, dx\,$

$=\, \displaystyle \int_0^\pi \frac{1}{1\, +\,\displaystyle \frac{2 \tan \frac {x}{2}}{1\, +\, \tan^2 \frac {x}{2}}}\, dx$

$=\, \displaystyle\int_0^\pi \frac{\sec^2 \dfrac {x}{2}}{\left(1\, +\,\tan \dfrac {x}{2}\right)^2}\, dx$

Put

$1+\tan$

$\dfrac{x}{2}\,= t\,\Rightarrow\,\dfrac{1}{2}\,\sec^2\,\dfrac{x}{2}\,=\,dt$

$\therefore\,I\,=\,\displaystyle\int_1^\infty\dfrac{2dt}{1\,+\,t^2}\,=-\dfrac2t\bigg]_1^\infty = 2$

The value of

$\displaystyle\int _{ 0 }^{ { x }/{ 4 } }{ \dfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } dx }$ is

0%
$1+\sqrt{2}$
0%
$-11+\sqrt{2}$
0%
$-\sqrt{2}$
0%
None of these

Explanation

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } } dx$

$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { 1 }/{ \cos { x } } }{ { \left[ \cfrac { 1 }{ \cos { x } } +\cfrac { \sin { x } }{ \cos { x } } \right] }^{ 2 } } } dx$

$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { \cos ^{ 2 }{ x } }/{ \cos { x } } }{ { \left( 1+\sin { x } \right) }^{ 2 } } } dx$

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( 1+\sin { x } \right) }^{ -2 } } \cos { x } dx$

Take

$1+\sin x=t \implies \cos x dx=dt$

When

$x=0, t=1$ and

$x=\dfrac{\pi}{4}, t=1+\dfrac{1}{\sqrt{2}}$

$\therefore \quad I=\int_{1}^{1+\frac{1}{\sqrt{2}}} t^{-2}\ dt$

$\therefore \quad I=\left|\dfrac{t^{-1}}{-1}\right|_{1}^{1+\frac{1}{\sqrt{2}}}$

$=-\left[ \cfrac { 1 }{ 1+\frac { 1 }{ \sqrt { 2 } } } -1 \right]$

$=-\cfrac { \sqrt { 2 } }{ \sqrt { 2 } +1 } +1$

$=\cfrac { -\sqrt { 2 } +\sqrt { 2 } +1 }{ \sqrt { 2 } +1 } =\cfrac { 1 }{ \sqrt { 2 } +1 }$

$=\cfrac { 1 }{ \sqrt { 2 } +1 } \ast \cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } -1 } =\cfrac { \sqrt { 2 } -1 }{ 2-1 }$

$=\sqrt { 2 } -1$

$\therefore$ None of these is correct option

$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$ , then

$\int _{ 0 }^{ 2 }{ f(x)dx }$ is

0%
$10$
0%
$50/3$
0%
$1/3$
0%
$47/2$

Explanation

$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$

from 0 to 1 we have

$2x^2+1$ and from 1 to 2 we have

$4x^2-1$

$=\int _{ 0 }^{ 1 }{ \left( 2{ x }^{ 2 }+1 \right) dx+\int _{ 1 }^{ 2 }{ \left( 4{ x }^{ 2 }-1 \right) dx } }$

$={ \left[ \frac { 2{ x }^{ 3 } }{ 3 } +x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { 4{ x }^{ 3 } }{ 3 } -x \right] }_{ 1 }^{ 2 }$

$=\frac { 2 }{ 3 } +1-0-0+\frac { 4{ (2) }^{ 3 } }{ 3 } -2-\left( \frac { 4 }{ 3 } -1 \right)$

$=\frac { 2 }{ 3 } +1+\frac { 32 }{ 3 } -2-\frac { 4 }{ 3 } +1$

$=\frac { 2+32-4 }{ 3 }$

$=\dfrac{30}{3}$

=10.

What is

$\displaystyle \int_0^1 {\frac{\tan^{-1}x}{1 + x^2} dx}$ equal to ?

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{8}$
0%
$\dfrac{\pi^2}{8}$
0%
$\dfrac{\pi^2}{32}$

Explanation

Let

$I=\displaystyle \int _{ 0 }^{ 1 } \dfrac { {\tan^{ -1 } }x }{ { 1+x^{ 2 } } } dx$

Let,

$y={ {\tan^{ -1 } }x }$

$\Rightarrow dy=\dfrac { 1 }{ { 1+x^{ 2 } } } dx$

Now for

$x=1, y=\dfrac { \pi }{ 4 }$ and for

$x=0, y=0$

Putting the values in the integral we have,

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ y.dy }$

$= { \left[\dfrac { y^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }$

$=\dfrac { \pi ^{ 2 } }{ 32 }$

Hence, D is correct.

$\dfrac {dy}{dt} = ky$ and

$k\neq 0$ , which of the following could be the equation of

$y$ ?

0%
$y = kx - 7$
0%
$y = 95e^{kt}$
0%
$y = 5 + ln\ k$
0%
$y = (x - k)^{2}$
0%
$y = \sqrt [k]{x}$

Explanation

Given,

$\dfrac{dy}{dt} = ky$ and

$k \ne 0$
We have

$\dfrac{dy}{y} = kdt$
Apply integral on both sides, so we get

$\displaystyle \int \dfrac{dy}{y} = \int kdt$

$\Rightarrow \ln(y) = kt+c$ (where

$c$ is constant)

$\Rightarrow y = {e}^{c}{e}^{kt} = a{e}^{kt}$ (

$a = {e}^{c}$ )
The possible solution is

$y = 95{e}^{kt}$

Solve

$\int_{0}^{\dfrac {\pi}{2}}\sqrt {\sin \phi}\cos^{5}\phi d\phi$ .

0%
$\dfrac{64}{231}$
0%
$\dfrac{24}{231}$
0%
$\dfrac{54}{231}$
0%
None of these

Explanation

Let

$\begin{matrix} I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \cos }^{ 5 } }\phi d\phi } \\ =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \cos }^{ 4 } }\phi \cos \phi d\phi } \\ =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \left( { 1-{ { \sin }^{ 2 } }\phi } \right) }^{ 2 } }\cos \phi d\phi } \\ put\, \, \, t=\sin \phi \\ \dfrac { { dt } }{ { d\phi } } =\cos \phi \\ dt=\cos \phi d\phi \\ When\, \, \phi =0,\, \, t=0\, \, when\, \, \phi =\dfrac { \pi }{ 2 } ,\, t=1 \\ \therefore I=\int _{ 0 }^{ 1 }{ \sqrt { t } { { \left( { 1-{ t^{ 2 } } } \right) }^{ 2 } }dt } \\ =\int _{ 0 }^{ 1 }{ \sqrt { t } \left( { 1-2{ t^{ 2 } }+{ t^{ 4 } } } \right) dt } \\ =\int _{ 0 }^{ 1 }{ { t^{ \dfrac { 1 }{ 2 } } }+{ t^{ \dfrac { 9 }{ 2 } } }-2{ t^{ \dfrac { 5 }{ 2 } } }dt } \\ =\left[ { \dfrac { 2 }{ 3 } { t^{ \dfrac { 3 }{ 2 } } }+\dfrac { 2 }{ { 11 } } { t^{ \dfrac { { 11 } }{ 2 } } }-\dfrac { 4 }{ 7 } { t^{ \dfrac { 7 }{ 2 } } } } \right] _{ 0 }^{ 1 } \\ =\dfrac { 2 }{ 3 } +\dfrac { 2 }{ { 11 } } -\dfrac { 4 }{ 7 } \\ =\dfrac { { 154+42-132 } }{ { 3\times 11\times 7 } } =\dfrac { { 64 } }{ { 231 } } \\ \end{matrix}$

$\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 3 }\sin { \left( { x }^{ 2 } \right) } dx }$ is equal to

0%
$\dfrac { 1 }{ \sqrt { 2 } } \left( 1+\dfrac { \pi }{ 4 } \right)$
0%
$\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 4 } \right)$
0%
$\dfrac { 1 }{ \sqrt { 2 } } \left( \dfrac { \pi }{ 2 } -1 \right)$
0%
$\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 2 } \right)$
0%
$\dfrac { 1 }{ \sqrt { 2 } } \left( \dfrac { \pi }{ 4 } -1 \right)$

Explanation

$I=\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 3 }\sin { \left( { x }^{ 2 } \right) } dx }$

$=\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 2 }\cdot x\sin { \left( { x }^{ 2 } \right) } dx }$
Put

${ x }^{ 2 }=t$

$\Rightarrow 2xdx=dt$
Also, when

$x=0$ , then

$t=0$
and when

$x=\dfrac { \sqrt { \pi } }{ 2 }$ , then

$t=\dfrac { \pi }{ 4 }$

$\Rightarrow I=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \underset { I }{ t } \underset { II }{ \sin { t } } dt }$

$={ \left[ t\left( -\cos { t } \right) \right] }_{ 0 }^{ { \pi }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ -\cos { t } \left( 1 \right) dt }$

$={ \left[ -t\cos { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }+\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \cos { t } dt }$

$={ \left[ -t\cos { t } +\sin { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }$

$=\left[ -\dfrac { \pi }{ 4 } \cdot \dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } } \right]$

$=\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 4 } \right)$

$\displaystyle\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx }$ is equal to

0%
$\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
0%
$-\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
0%
$-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
0%
$-7\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
0%
$\cos { \left( x+70 \right) } +C$

Explanation

Let

$I=\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx }$

$=\cfrac { 1 }{ 7 } \int { \sin { \left( \cfrac { x }{ 7 } +10 \right) } } =\cfrac { 1 }{ 7 } \cfrac { -\cos { \left( \cfrac { x }{ 7 } +10 \right) } }{ \cfrac { 1 }{ 7 } }$

$=-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$

$\int _0^1 xdx = \dfrac {\pi}{4} - \dfrac {1}{2} ln 2$ then the value of definite integral

$\int _0^1 \tan^{-1} (1-x+x^2) dx$ equals :

0%
$ln2$
0%
$\dfrac {\pi}{4} + ln 2$
0%
$\dfrac {\pi}{4} - ln2$
0%
$2 ln 2$

The value of

$\int_{0}^{1} \dfrac {8\log (1 + x)}{1 + x^{2}} dx$ is

0%
$\dfrac {\pi}{2}\log 2$
0%
$\pi\log 2$
0%
$2\pi\log 2$
0%
None of these

Explanation

Let

$\begin{matrix} I=\int _{ 0 }^{ 1 }{ \frac { { 8\log \left( { 1+x } \right) } }{ { 1+{ x^{ 2 } } } } } dx \\ \, \, \, \, =8\int _{ 0 }^{ 1 }{ \frac { { \log \left( { 1+x } \right) } }{ { 1+{ x^{ 2 } } } } } dx \\ put\, \, \, x=\tan \theta \\ \, \, \, \, \, \Rightarrow dx={ \sin ^{ 2 } }\theta d\theta \\ x,\, \frac { \pi }{ 4 } ,\, \, \, \, x=0 \\ \quad \quad \quad \quad I=8\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { { \log \left( { 1+x } \right) } }{ { { { \sin }^{ 2 } }\theta } } } { \sin ^{ 2 } }\theta d\theta \\ \, \, \, \, \, \quad \quad \quad \quad =8\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \log \left( { 1+\tan x } \right) } dx \\ \quad \quad =8\times \frac { \pi }{ 8 } \log 2 \\ \, \, \, =\pi \log 2 \\ \end{matrix}$

Hence option (A) is true.

$\displaystyle\int { \sqrt { 1+\sin { x } } \cdot f\left( x \right) dx } =\dfrac { 2 }{ 3 } { \left( 1+\sin { x } \right) }^{ { 3 }/{ 2 } }+C$ , then

$f\left( x \right)$ is equal to

0%
$\cos { x }$
0%
$\sin { x }$
0%
$\tan { x }$
0%
$1$

Explanation

$\int { \sqrt { 1+\sin { x } } } .f\left( x \right) dx=\dfrac { 2 }{ 3 } { (1+\sin { x } ) }^{ { 3 }/{ 2 } }+c$

On differentiating both sides, we get

$\sqrt { 1+\sin { x } } .f\left( x \right) =\dfrac { 2 }{ 3 } .\dfrac { 3 }{ 2 } { (1+\sin { x } ) }^{ { 1 }/{ 2 } }.\cos { x } +0\\ \Rightarrow f\left( x \right) =\cos { x }$

$\displaystyle \int _0^{\pi /2} f(\sin 2x)\sin x\, dx = K\int_0^{\pi/2} f(\cos 2x) \cos x\,dx$ where

$k$ equals to

0%
$2$
0%
$4$
0%
$\sqrt{2}$
0%
$2\sqrt{2}$

Explanation

$\displaystyle \int_0^{\dfrac{\pi}{2}} \sin 2x \sin x dx = K \int_0^{\dfrac{\pi}{2}} \cos 2x \cos x dx$

$\displaystyle \int_0^{\dfrac{\pi}{2}} 2 \sin x \cos x \sin x dx = K \int_0^{\dfrac{\pi}{2}} (1 - 2 \sin^2 x) \cos x dx$

$\displaystyle \int_0^{\dfrac{\pi}{2}} 2 \sin^2 x \cos x dx = K \left[\int_0^{\dfrac{\pi}{2}} \cos x dx - \int_0^{\dfrac{\pi}{2}} 2 \sin^2 x \cos x dx \right]$

$\left\{\dfrac{2 \sin^3 x}{3} \right\}_0^{\dfrac{\pi}{2}} = K \left[\{\sin x\}_0^{\dfrac{\pi}{2}} - \left\{\dfrac{2 \sin^3 x}{3} \right\}_0^{\dfrac{\pi}{2}}\right]$

$\dfrac{2}{3} = K \left[1 - \dfrac{2}{3} \right]$

$K = 2$

$\int_{0}^{1}{\frac{dx}{x\sqrt{x}}}$

0%
$2$
0%
$-2$
0%
$1$
0%
$3$

Explanation

$\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ x\sqrt { x } } dx } \\ \int _{ 0 }^{ 1 }{ { x }^{ \cfrac { -3 }{ 2 } }dx } ={ [\cfrac { { x }^{ \cfrac { -3 }{ 2 } +1 } }{ \cfrac { -3 }{ 2 } +1 } ] }_{ 0 }^{ 1 }\\ ={ [\cfrac { { x }^{ \cfrac { -1 }{ 2 } } }{ \cfrac { -1 }{ 2 } } ] }_{ 0 }^{ 1 }=\cfrac { 1-0 }{ \cfrac { -1 }{ 2 } } =-2$

What is

$\displaystyle \int_{0}^{2\pi}\sqrt {1 + \sin \dfrac {x}{2}}dx$ equal to?

0%
$8$
0%
$4$
0%
$2$
0%
$0$

Explanation

$1+\sin\frac { x }{ 2 } ={ \sin }^{ 2 }\frac { x }{ 4 } +{ \cos }^{ 2 }\frac { x }{ 4 } +2\sin\frac { x }{ 4 } \cos\frac { x }{ 4 } \\ { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 }\\ \sqrt { 1+\sin\frac { x }{ 2 } } =\sqrt { { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 } } =\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 }$

$\displaystyle \int _{ 0 }^{ 2\pi }{ \left(\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 }\right )dx } =4\left(-\cos\frac { x }{ 4 } +\sin\frac { x }{ 4 }\right )+c\\ After\quad putting\quad the\quad limit\quad we\quad get\\\Rightarrow 4(1-(-1))=8\\$

So correct answer will be option A

Let

$I=\displaystyle \int _{ \pi /4 }^{ \pi /3 }{ \cfrac { \sin { x } }{ x } } dx$ . Then?

0%
$\cfrac { 1 }{ 2 } \le I\le 1\quad$
0%
$4\le I\le 2\sqrt { 30 }$
0%
$\cfrac { \sqrt { 3 } }{ 8 } \le I\le \cfrac { \sqrt { 2 } }{ 6 }$
0%
$1\le I\le \cfrac { 2\sqrt { 3 } }{ \sqrt { 2 } }$

Explanation

$I=\displaystyle \int^{\tfrac{\pi}{3}}_{\tfrac{\pi}{4}}\dfrac{sinx}{x}dx$

$\dfrac{sinx}{x}$ is a decreasing function in given interval

difference of limits

$=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$

so,

$\dfrac{\pi}{12}\cdot\dfrac{sin\dfrac{\pi}{3}}{\dfrac{\pi}{3}} \le I \le \dfrac{\pi}{12} \cdot\dfrac{sin\dfrac{\pi}{4}}{\dfrac{\pi}{4}}$

$\dfrac{\sqrt3}{8} \leq I \leq \dfrac{\sqrt2}{6}$

$\int { { \left( ex \right) }^{ x }\left( 2+\log { x } \right) } dx=....+c,x\in { R }^{ + }-\left\{ 1 \right\}$

0%
${ x }^{ x }$
0%
${ \left( ex \right) }^{ x }\quad$
0%
${ e }^{ x }$
0%
$\left( 1+\log { x } \right) { \left( ex \right) }^{ x }$

Explanation

$\int (ex)^x(2+\log x)dx=\int (ex)^xdx+\int e^x.x^x(1+\log x)dx$

Integrating by parts ,

$\Rightarrow \int(ex)^x +e^x(\int x^x(1+\log x)) -\int(\dfrac{d}{dx}(e^x).(\int x^x(1+\log x)dx))dx$

$\Rightarrow \int (ex)^x+ e^xx^x-\int (ex)^x$

$\Rightarrow (ex)^x$

$I=\displaystyle\overset{1}{\underset{0}{\displaystyle\int}}x(1-x)^{1/2}dx$ and

$60I+k=25$ then

$k=$ _________.

$(k\in R)$ .

0%
$9$
0%
$25$
0%
$60$
0%
$41$

Explanation

We know,

$\displaystyle \int_0^af(x)dx=\int_0^af(a-x)dx$

$\therefore I=\displaystyle \int_0^1x(1-x)^{1/2}dx=\int_0^1(1-x)x^{1/2}dx$

$\Rightarrow I=\displaystyle \int_0^1(x^{1/2}-x^{3/2})dx$

$\Rightarrow I=\left[\dfrac{2}{3}x^{3/2}-\dfrac{2}{5}x^{5/2} \right]_0^1$

$\Rightarrow I=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}$

$60I+k=25$

$k=25-60\times \dfrac{4}{15}=9$

$\displaystyle \int_1^{32}\dfrac{dx}{x^{1/5}\sqrt{1+x^{4/5}}}$

0%
$\dfrac{2}{5}(\sqrt{17}+\sqrt{2})$
0%
$\dfrac{2}{5}(\sqrt{17}-\sqrt{2})$
0%
$\dfrac{5}{2}(\sqrt{17}-\sqrt{2})$
0%
$\dfrac{5}{2}(\sqrt{17}+\sqrt{2})$

$\int _{ 0 }^{ \pi /2 }{ \sin { 2x } .\sin { x } } dx=.....$

0%
$\cfrac{1}{3}$
0%
$\cfrac{2}{3}$
0%
$-\cfrac{2}{3}$
0%
$\cfrac{4}{3}$

Explanation

$I=\int _{ 0 }^{ \pi /2 }{ \sin { 2x } .\sin { x } } dx$

$=\int _{ 0 }^{ \pi /2 }{ 2\sin { x } \cos { x } .\sin { x } } dx=2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2 }{ x } \cfrac { d }{ dx } \sin { x } } dx$

$=\cfrac { 2 }{ 3 } { \left[ \sin ^{ 2 }{ x } \right] }_{ 0 }^{ \pi /2 }=\cfrac { 2 }{ 3 } (1-0)=\cfrac { 2 }{ 3 }$

Evaluate

$\displaystyle \int_{-2\pi}^{5\pi} \cot^{-1} (\tan x) dx$ .

0%
$0$
0%
$-1$
0%
$1$
0%
$2$

Explanation

$I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx$ ......

$(1)$

Applying property of limits of integral formula ;

$\Rightarrow I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan (7\pi - x)) dx$

$\Rightarrow I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(-\tan x)dx$

$\Rightarrow I=\displaystyle -\int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx$ ......

$(2)$

Adding both equations;

$\Rightarrow 2I=0$

$\Rightarrow I=0$

$\Rightarrow \displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx=0$

$\int {\left( 3.{ x }^{ 2 }.\tan ^{ -1 }{ x } +\cfrac { { x }^{ 3 } }{ 1+{ x }^{ 2 } } \right) } dx=....+c$ .

0%
${ x }^{ 3 }\tan ^{ -1 }{ x }$
0%
$\cfrac { { x }^{ 3 } }{ 3 } \tan ^{ -1 }{ x }$
0%
${ x }^{ 2 }\tan ^{ -1 }{ x }$
0%
$\cfrac { { x }^{ 2 } }{ 2 } \tan ^{ -1 }{ x }$

Explanation

Given

$\int (3x^2.\tan^{-1}x+\dfrac{x^3}{1+x^2})dx$ =

$\int (3x^2.\tan^{-1}x.dx)+\int(\dfrac{x^3}{1+x^2})dx$

Integrating by parts

$\int 3x^2tan^{-1}x.dx+x^3\int\dfrac{1}{1+x^2}dx-\int( \dfrac{d}{dx}(x^3)(\int \dfrac{1}{1+x^2}dx)dx)$

$\dfrac{d}{dx}x^3=3x^2;\int \dfrac{1}{1+x^2}=tan^{-1}x$

$\Rightarrow\int 3x^2tan^{-1}x.dx + x^3 \tan^{-1}x-\int 3x^2tan^{-1}x.dx$

$\Rightarrow x^3\tan^{-1}x+c$

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$ is equal to:

0%
$\dfrac 1 2$
0%
$\dfrac 14$
0%
$2$
0%
$1$

Explanation

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$

$\sin x=t\implies \cos x dx=dt$

$x\to 0\to \dfrac \pi 2$

$t\to 0\to 1$

$\Rightarrow \displaystyle \int _0^{\pi/2} t dt$

$\Rightarrow\left.\dfrac {t^2}2\right|^1_0$

$\Rightarrow\dfrac 12-0=\dfrac 12$

$I= \int \frac{x+2}{(x+1)^2}dx;$ then I is equal to

0%
$\log (x+1)+\dfrac{1}{x+1}+c$
0%
$\log (x+2)-\dfrac{1}{x+1}+c$
0%
$\log (1+x)-\dfrac{1}{x+1}+c$
0%
$\log (x+2)+\dfrac{1}{x+1}+c$

Explanation

$\int { \dfrac { x+2 }{ { \left( x+1 \right) }^{ 2 } } } dx$

$=\int { \dfrac { x+1+1 }{ { \left( x+1 \right) }^{ 2 } } } dx$

$=\int { \left( \dfrac { x+1 }{ { \left( x+1 \right) }^{ 2 } } +\dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } \right) } dx$

$=\int { \dfrac { 1 }{ x+1 } } dx+\int { \dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } } dx$

$\left[ \because \int { { x }^{ n }dx=\dfrac { { x }^{ n+1 } }{ n+1 } +c } \right]$

$=\log \left(x+1\right) +\dfrac { { \left( x+1 \right) }^{ -2+1 } }{ -2+1 } +c$

$\left[ \because \int { \dfrac { 1 }{ x } dx=\log { x+c } } \right]$

$=\log\left(x+1\right) +\left(-1\right) \dfrac{1}{\left(x+1\right)}+c$

$=\log\left(x+1\right)-\dfrac{1}{x+1}+c$

Hence, the answer is

$\log\left(x+1\right)-\dfrac{1}{x+1}+c.$

$\int _{ 0 }^{ \pi /3 }{ \dfrac { \cos { x } }{ 3+4\sin { x } } dx } =k\log { \left( \dfrac { 3+2\sqrt { 3 } }{ 3 } \right) }$ , then,

$k$ is equal to ?

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$

Explanation

$\sin x=t \Rightarrow dx.\cos x=dt$

$\displaystyle I=\int_{0}^{\frac{\pi }{3}}\frac{dt}{3+4t}$

$\displaystyle =\frac{1}{4}[ln(3+4t)]^{\frac{\pi }{3}}$

When

$x=0\Rightarrow t=0$

When

$\displaystyle x=\frac{\pi }{3}\Rightarrow t=\frac{\sqrt{3}}{2}$

$\displaystyle =\frac{1}{4}[ln(3+4t)]^{\frac{\sqrt{3}}{2}}-\frac{1}{4}[ln(3+4t)]^{0}$

$\displaystyle =\frac{1}{4}ln(3+2\sqrt{3})-\frac{1}{4}ln(3)$

$\displaystyle I=\frac{1}{4}ln\left[\frac{3+2\sqrt{3}}{3}\right]$

$K=\dfrac{1}{4}$

Find proper substitution

$\int _{ 0 }^{ 1 }{ \dfrac { { e }^{ -x } }{ 1+{ e }^{ -x } } dx }$

0%
$1+{ e }^{ -x }\rightarrow t$
0%
$-{ e }^{ -x }dx\rightarrow dt$
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$-\int _{ 0 }^{ 1 }{ \dfrac { dt }{ t } }$
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$-\int _{ 0 }^{ 1 }{ ln\left| t \right| }$

Explanation

$\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ -x } }{ 1+{ e }^{ -x } } dx }$

$1+{ e }^{ -x }=t$

$\Rightarrow { -e }^{ -x }dx=dt$

$\int _{ 0 }^{ 1 }{ \cfrac { -dt }{ t } } \quad \therefore 1+{ e }^{ -x }=t$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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