MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 4

$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$ for

$x> 0,$

and

$\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$

then

$\displaystyle I$ is?

0%
$\displaystyle f\left ( 2 \right )-f\left ( 1/2 \right )$
0%
$\displaystyle f\left ( 1/2 \right )-f\left ( 2 \right )$
0%
$0$
0%
None of these

Explanation

Substitute

$\displaystyle z=\dfrac { 1 }{ t }$

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

$\displaystyle =\int _{ x }^{ \dfrac { 1 }{ x } }{ f\left( t \right) } dt=-\int _{ \dfrac { 1 }{ x } }^{ x }{ f\left( t \right) } dt=-I\\ \therefore 2I=0$

$\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx$ and

$\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx$

then

$I_{1}+I_{2}$ equals?

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle -\frac{1}{3}$
0%
$0$
0%
None of these

Explanation

$I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx$

$\displaystyle x=\frac { -y-3 }{ 3 }$

$\displaystyle dx=-\frac { dy }{ 3 }$

$\displaystyle \Rightarrow I_{ 2 }=3\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } } e^{ \left( 3x-2 \right) ^{ 2 } }dx=-\int _{ -4 }^{ -5 } e^{ \left( y+5 \right) ^{ 2 } }dx=-I_{ 1 }$

$\displaystyle \Rightarrow I_{ 1 }+I_{ 2 }=0$

$f\left ( a+x \right )= f\left ( x \right )$ , then

$\forall$

$a> 0, n\epsilon N$ the value of

$\displaystyle \int_{0}^{n a}f\left ( x \right )dx$ equals ?

0%
$\displaystyle \left ( n-1 \right )\int_{0}^{a}f\left ( x \right )dx$
0%
$\displaystyle \left ( 1-n \right )\int_{0}^{a}f\left ( x \right )dx$
0%
$\displaystyle n\int_{0}^{a}f\left ( x \right )dx$
0%
None of the above

Explanation

$I=\displaystyle \int _{ 0}^{ na }{ f\left( x \right) dx }$

Substitute

$x=a+t\Rightarrow dx=dt$

$I=\displaystyle \int _{ 0 }^{ \left( n-1 \right) a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+t \right) dt } \\ =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( a+x \right) dx } =\left( n-1 \right) \displaystyle \int _{ 0 }^{ a }{ f\left( x \right) dx }$

$\displaystyle f\left ( x \right )=\int_{-1}^{1}\frac{\sin x}{1+t^{2}}dt$ then

$\displaystyle {f}'\left ( \frac{\pi }{3} \right )$ is

0%
nonexistent
0%
$\displaystyle \pi /4$
0%
$\displaystyle \pi \sqrt{3/4}$
0%
none of these

Explanation

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

$\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right]$

$\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 }$

$\displaystyle \int_{0}^{1}\frac{2^{x+1}-3^{x-1}}{6^{x}}dx$

0%
$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{6}\log _{2}e.$
0%
$\displaystyle -\frac{4}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$
0%
$\displaystyle \frac{4}{3}\log _{3}e-\frac{1}{3}\log _{2}e.$
0%
$\displaystyle -\frac{2}{3}\log _{3}e+\frac{1}{6}\log _{2}e.$

Explanation

$\displaystyle \int _{ 0 }^{ 1 } \frac { 2^{ x+1 }-3^{ x-1 } }{ 6^{ x } } dx=\int _{ 0 }^{ 1 } \left( 2.3^{ -x }-\frac { 1 }{ 3 } 2^{ -x } \right) dx$

$\displaystyle =\left[ -2\frac { 3^{ -x } }{ \log 3 } +\frac { 1 }{ 3 } \frac { 2^{ -x } }{ \log 2 } \right] ^{ 1 }$

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

$\displaystyle =\frac { 4 }{ 3 } \log _{ 3 } e-\frac { 1 }{ 6 } \log _{ 2 } e$
Hence, option 'A' is correct.

$\displaystyle \int_{1}^{2}\left ( x+\frac{1}{x} \right )^{3/2}\frac{x^{2}-1}{x^{2}}dx$

0%
$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}+\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \frac{5}{2}\sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \sqrt{\left ( \frac{5}{2} \right )}-\frac{8}{5}\sqrt{2}$
0%
$\displaystyle \frac{3}{2}\sqrt{\left ( \frac{3}{2} \right )}-\frac{8}{5}\sqrt{2}$

Explanation

Let

$\displaystyle I=\int _{ 1 }^{ 2 } \left( x+\frac { 1 }{ x } \right) ^{ 3/2 }\frac { x^{ 2 }-1 }{ x^{ 2 } } dx$

Put

$\displaystyle x+\frac { 1 }{ x } =t\Rightarrow \left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx\Rightarrow dt$
Therefore

$\displaystyle I=\int _{ 2 }^{ 5/2 } t^{ 3/2 }dt=\frac { 2 }{ 5 } t^{ 5/2 }=\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 5/2 }-2^{ 5/2 } \right]$

$\displaystyle =\frac { 2 }{ 5 } \left[ \left( \frac { 5 }{ 2 } \right) ^{ 2 }\sqrt { \left( \frac { 5 }{ 2 } \right) } -2^{ 2 }\sqrt { 2 } \right] =\frac { 5 }{ 2 } \sqrt { \left( \frac { 5 }{ 2 } \right) } -\frac { 8 }{ 5 } \sqrt { 2 }$
Hence, option 'B' is correct.

Evaluate the integrals

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

0%
$\displaystyle \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}$
0%
$\displaystyle \frac{1}{4} \log\left ( ab\right )\cdot \log \frac{b}{a}$
0%
$\displaystyle \log\left ( ab\right )\cdot \log \frac{b}{a}$
0%
$\displaystyle -\frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}$

Explanation

Substitute

$\displaystyle \log x=t \therefore \left ( 1/x \right )dx=dt$ and adjust the limits

$\displaystyle \therefore I= \int t\:dt=\left [ \frac{1}{2}t^{2} \right ]_{\log a}^{\log b}$

$\displaystyle \therefore I= \frac{1}{2}\left [ \left ( \log b \right )^{2}-\left ( \log a \right )^{2} \right ]$

$\displaystyle = \frac{1}{2}\left [ \left ( \log b+\log a \right )\left ( \log b-\log a \right ) \right ]$

$\displaystyle \therefore I= \frac{1}{2} \log\left ( ab\right )\cdot \log \frac{b}{a}.$

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

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

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

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

0%
$0$
0%
$\displaystyle \frac{1}{2}$
0%
$1$
0%
$3/2$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { 1 }{ sin{ x } } dx } =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \csc { x } } dx$
Multiply numerator and denominator by

$cot{ x }+\csc { x }$ , we get

$\displaystyle \therefore I=-\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { -cot{ x }\csc { x } -\csc ^{ 2 }{ x } dx }{ cot{ x }+\csc { x } } dx }$
Substitute

$t=cot{ x }+\csc { x } \Rightarrow dt=\left( -\csc ^{ 2 }{ x } -cot{ x }\csc { x } \right) dx$

$\displaystyle \therefore I=-\int _{ 0 }^{ 1 }{ \dfrac { 1 }{ t } } dt=\left[ -\log { t } \right] _{ 0 }^{ 1 }=0$

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

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

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

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

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

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

$f\left ( x \right )= A\sin \left ( \dfrac {\pi x}{2} \right )\: +\: B,f{}'\left ( \dfrac 12 \right )= \sqrt{2}$ and

$\displaystyle \int_{0}^{1}f\left ( x \right )dx= \displaystyle \frac{2A}{\pi }$ ,

then the constants

$A$ and

$B$ are

0%
$A=\dfrac {\pi}{2}, B=\dfrac {\pi}{2}$
0%
$A=\dfrac {2}{\pi}, B=3\pi$
0%
$A=0, B=-4\pi$
0%
$A=\dfrac {4}{\pi}, B=0$

Explanation

$f(x)=A\sin{(\pi x/2)}+B$

$\implies f^{'}(x)=\dfrac{A\pi}{2}\cos{(\pi x/2)}$

$\implies f^{'}(1/2)=\dfrac{A\pi}{2}\cos{(\pi/4)=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}}$

$\implies\sqrt{2}=\dfrac{A\pi}{2}\dfrac{1}{\sqrt{2}}$

Hence,

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

$\displaystyle\int_{0}^{1} f(x)dx=\left[-\dfrac{2A}{\pi}\cos{(\pi x/2)}+Bx\right]_{0}^{1}$

$\implies\dfrac{2A}{\pi}=\dfrac{2A}{\pi}+B$

Hence,

$B=0$

Hence, answer is option-(D).

The value of the integral

$\displaystyle \int_{0}^{\pi /4}\displaystyle \frac{\sin x+\cos x}{3+\sin 2x}dx$ is

0%
$\log 2$
0%
$\log 3$
0%
$\left ( 1/4 \right )\log 3$
0%
$\left ( 1/8 \right )\log 3$

Explanation

Let

$\displaystyle I=\int _{ 0 }^{ \pi /4 } \frac { \sin x+\cos x }{ 3+\sin 2x } dx=\int _{ 0 }^{ \pi /4 } \frac { \sin { x } +\cos { x } }{ 4-{ \left( \sin { x } -\cos { x } \right) }^{ 2 } } dx$

Substitute

$\sin { x } -\cos { x } =t$

$\displaystyle I=\int _{ -1 }^{ 0 }{ \frac { dt }{ 4-{ t }^{ 2 } } } =\frac { 1 }{ 4 } \log { 3 }$

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

$\alpha >1$

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

$I$ is equal to

0%
$1$
0%
$1/2$
0%
$1/\sqrt{2}$
0%
$1/\sqrt{3}$

Explanation

As form of integrand is

$\displaystyle \dfrac { 1 }{ L\sqrt { Q } }$
Substitute

$\displaystyle L=\dfrac { 1 }{ t } \Rightarrow x+1=\dfrac { 1 }{ t }$

$\displaystyle I=\int _{ \dfrac { 1 }{ 2 } }^{ \dfrac { 1 }{ 3 } }{ \dfrac { \left( \dfrac { -1 }{ { t }^{ 2 } } \right) dt }{ \dfrac { 1 }{ t } \sqrt { { \left( \dfrac { 1 }{ t } -1 \right) }^{ 2 }-1 } } } =\int _{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ \dfrac { dt }{ \sqrt { 1-2t } } }$

$\displaystyle ={ \left[ \dfrac { { \left( 1-2t \right) }^{ \dfrac { 1 }{ 2 } } }{ \left( -2 \right) \left( \dfrac { 1 }{ 2 } \right) } \right] }_{ \dfrac { 1 }{ 3 } }^{ \dfrac { 1 }{ 2 } }{ = }\dfrac { 1 }{ \sqrt { 3 } }$

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

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

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

The value of

$\displaystyle \int_{8}^{15}\displaystyle \frac{dx}{\left ( x-3 \right )\sqrt{x+1}}$ is

0%
$\log \displaystyle \frac{5}{3}$
0%
$\displaystyle \frac{1}{2}\log \displaystyle \frac{5}{3}$
0%
$2\log \displaystyle \frac{5}{3}$
0%
$\sqrt{2}\log \left ( \sqrt{2}-1 \right )$

Explanation

$\displaystyle I=\int _{ 8 }^{ 15 } \frac { dx }{ \left( x-3 \right) \sqrt { x+1 } }$

Substitute

$\sqrt { x+1 } =t\Rightarrow x+1={ t }^{ 2 }$

$\displaystyle I=\int _{ 3 }^{ 4 }{ \frac { 2t }{ \left( { t }^{ 2 }-4 \right) t } dt } ={ \left[ \frac { 2 }{ 2.2 } \log { \left( \frac { t-2 }{ t+2 } \right) } \right] }_{ 3 }^{ 4 }$

$\displaystyle =\frac { 1 }{ 2 } \left[ \log { \frac { 1 }{ 3 } } -\log { \frac { 1 }{ 5 } } \right] =\frac { 1 }{ 2 } \log { \frac { 5 }{ 3 } }$

$b > a,$ and

$\displaystyle I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,$ then

$I$ equals

0%
$\displaystyle \frac{\pi}{2} (b -a)$
0%
$\pi (b -a)$
0%
$\pi/2$
0%
$2\pi (b-a)$

Explanation

Substitute

$b -x = t^{2}$ so that

$\displaystyle I = \int _{\sqrt{b-a}}^{0}\sqrt{\frac{b-t^{2}-a}{t^{2}}} (-2t) dt$

$\displaystyle= 2 \int_{0}^{c} \sqrt{c^{2}-t^{2}} dt$ , where

$c= \sqrt{b-a}$

$=\displaystyle 2 \left [ \frac{1}{2} t \sqrt{c^{2}-t^{2}}+ \frac{c^{2}}{2} \sin ^{-1} \left( \frac{t}{c}\right) \right] _{0}^{c}$

$=0 + c^{2} \sin^{-1} (1) -0$

$\displaystyle =\frac{\pi}{2}(b-a)$

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

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

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

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

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

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

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

$0 < \alpha < 1$ and

$\displaystyle I=\int _{-1}^{1} \frac{dx}{\sqrt{1-2\alpha x+\alpha^{2}}}$ then

$I$ equals

0%
$1/\alpha$
0%
$2/\alpha$
0%
$3/\alpha$
0%
none of these

Explanation

$\displaystyle I =\int_{-1}^{1} (1 -2\alpha x + \alpha^{2} )^{-1/2} dx$

$\displaystyle =\left. \frac{2(1-2\alpha x+\alpha^{2})^{1/2}}{-2\alpha}\right ]_{-1}^{1}$

$\displaystyle =-\frac{1}{\alpha}[([(1 -2\alpha + \alpha^{2})^{1/2} -(1 + 2 \alpha + \alpha ^{2})^{1/2} ]$

$=-\dfrac{1}{\alpha}[\sqrt{(\alpha-1)^2}-\sqrt{(\alpha+1)^2}]$

$=-\dfrac1\alpha\left[(1-\alpha)-(1+\alpha)\right]\ldots(\because0<\alpha<1)$

$=2$

Ans: D

$\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx$ 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

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

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

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

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

The value of the definite integral

$\displaystyle \int_{1}^{\infty}\left ( e^{x+1}+e^{3-x} \right )^{-1}dx$ is

0%
$\displaystyle \frac{\pi }{4e^{2}}$
0%
$\displaystyle \frac{\pi }{4e}$
0%
$\displaystyle \frac{1}{e^{2}}\left ( \frac{\pi }{2}-\tan ^{-1}\frac{1}{e} \right )$
0%
$\displaystyle \frac{\pi }{2e^{2}}$

Explanation

$I=\int _{ 1 }^{ \infty }{ \cfrac { 1 }{ { e }^{ x+1 }+{ e }^{ 3-x } } dx } =\int _{ 1 }^{ \infty }{ \cfrac { { e }^{ x-1 } }{ \left( { e }^{ x }-ie \right) \left( { e }^{ x }+ie \right) } dx }$
Substituting

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

$I=\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( t-ie \right) \left( t+ie \right) } dt } =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \cfrac { 1 }{ \left( e-it \right) \left( e+it \right) } dt } \\ =\cfrac { 1 }{ e } \int _{ e }^{ \infty }{ \left( \cfrac { i }{ 2\left( et+i{ e }^{ 2 } \right) } -\cfrac { i }{ 2\left( et-i{ e }^{ 2 } \right) } \right) } dt\\ ={ \left[ \cfrac { i\log { \left( et+i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } -\cfrac { i\log { \left( et-i{ e }^{ 2 } \right) } }{ 2{ e }^{ 2 } } \right] }_{ e }^{ \infty }\\ ={ \left[ \cfrac { \tan ^{ -1 }{ \left( { e }^{ x-1 } \right) } }{ { e }^{ 2 } } \right] }_{ e }^{ \infty }=\cfrac { \pi }{ 4{ e }^{ 2 } }$

$\displaystyle \int ^{\displaystyle \frac{3 \pi}{10}}_{\displaystyle \frac{\pi}{5}} \frac{sin x}{sin x + cos x}dx$ is equal to

0%
$\pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{20}$

Explanation

Let

$\displaystyle I=\int { \frac { \sin { x } }{ \sin { x } +\cos { x } } dx }$
Multiply numerator and denominator by

$I=\int { \dfrac { \sin { x } }{ \sin { x } +\cos { x } } dx } \\ \csc ^{ 3 }{ x }$

$\displaystyle I=\int { \frac { \csc ^{ 2 }{ x } }{ \csc ^{ 2 }{ x } +\cot { x } \csc ^{ 2 }{ x } } dx } =\int { \frac { \csc ^{ 2 }{ x } }{ \cot ^{ 3 }{ x } +\cot ^{ 2 }{ x } +\cot { x } +1 } dx }$

Put

$t=\cot { x } \Rightarrow dt=-\csc ^{ 2 }{ x } dx$

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

$\displaystyle =\frac { 1 }{ 2 } \int { \frac { t }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }+1 } dt } -\frac { 1 }{ 2 } \int { \frac { 1 }{ t+1 } dt }$

$\displaystyle =\frac { 1 }{ 4 } \log { \left( { t }^{ 2 }+1 \right) } -\frac { 1 }{ 2 } \tan ^{ -1 }{ t } -\frac { 1 }{ 2 } \log { \left( t+1 \right) }$

$\displaystyle =\frac { 1 }{ 2 } \left( x-\log { \left( \sin { x } +\cos { x } \right) } \right)$

Hence

$\displaystyle \int _{ \frac { \pi }{ 5 } }^{ \frac { 3\pi }{ 10 } } \frac { sinx }{ sinx+cosx } dx=\frac { \pi }{ 20 }$

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

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

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

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

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

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

The derivative of

$F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}(x >0)$ is

0%
$\displaystyle\frac{1}{3log x}$
0%
$\displaystyle\frac{1}{3log x}-\frac{1}{2log x}$
0%
$(log x)^{-1}x(x-1)$
0%
$\displaystyle\frac{3x^2}{log x}$

Explanation

Given,

$F(x)=\displaystyle\int^{x^3}_{x^2}\frac{dt}{log t}$

$\Rightarrow F'(x)=\displaystyle\frac{d}{dx}F(x)$

$=\displaystyle\frac{d}{dx}(x^3)\cdot\frac{1}{log x^3}-\frac{d}{dx}(x^2)\cdot\frac{1}{log x^2}$

$=\displaystyle\frac{3x^2}{3log x}-\frac{2x}{2log x}=\frac{x^2-x}{log x}$

$=x(x-1)(log x)^{-1}$

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

$x\in (0,1)$

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

Trapezoidal rule for evaluation of

$\displaystyle\int _{ a }^{ b }{ f(x)dx }$ requires the interval

$(a,b)$ to be divided into

0%
$2n$ sub-intervals of equal width
0%
any number of sub intervals of not equal width
0%
any number of sub intervals of equal width
0%
$3n$ sub-intervals of equal width

Explanation

Trapezoidal rule for evaluation of

$\int _{ a }^{ b }{ f(x)dx }$ requires the interval

$(a,b)$ to be divided into any number of sub-intervals of equal width.

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

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

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

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

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

The value of the integral

$\int_0^{\dfrac{\pi}{2}}\,sin^5x\,dx$ is

0%
$\dfrac{4}{15}$
0%
$\dfrac{8}{5}$
0%
$\dfrac{8}{15}$
0%
$\dfrac{4}{5}$

Explanation

$I=\int_0^{\dfrac{\pi}{2}}sin^4x.sin\,x\,dx$

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

$cos\,x=t$

$\Rightarrow\;-sin\,x\,dx=dt$

$=-\int_1^0(1-t^2)^2dt=\int_0^(t^4-2t^2+1)dt$

$[\because\,-\int_a^bf(x)dx=\int_a^bf(x)dx]$

$=\dfrac{1}{5}(t^5)_0^1-\dfrac{2}{3}(t^3)_0^1+(t)_0^1$

$=\dfrac{1}{5}-\dfrac{2}{3}+1=\dfrac{3-10+15}{15}=\dfrac{8}{15}$

Using Trapezoidal rule and following table

$\int_0^8 {f(x)}dx$ is equal to

$x$	0	2	4	6	8
$f(x)$	2	5	10	17	26

0%
$184$
0%
$92$
0%
$46$
0%
$-36$

$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}$ equals

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{12}$

Explanation

$\displaystyle \int \frac {d}{1+x^2}=\tan^{-1}x=F(x)$
By second fundamental theorem of calculus, we obtain

$\displaystyle \int_1^{\sqrt 3}\frac {dx}{1+x^2}=F(\sqrt 3)-F(1)$

$=\tan^{-1}\sqrt 3-\tan^{-1}1$

$\displaystyle =\frac {\pi}{3}-\frac {\pi}{4}=\frac {\pi}{12}$

$\displaystyle\int _{ 1 }^{ 3 }{ \dfrac { \cos { \left( \log { x } \right) } }{ x } dx }$ is equal to

0%
$1$
0%
$\cos { \left( \log { 3 } \right) }$
0%
$\sin { \left( \log { 3 } \right) }$
0%
$\dfrac { \pi }{ 4 }$

Explanation

$\displaystyle \int_{1}^{3} \frac{cos(logx)}{x}dx$

Substituting

$log\,x = t$

$\displaystyle dt = \frac{1}{x}dx$

$\displaystyle \therefore \int_{t_{1}}^{t_{2}} cos\,t\,dt$

Lim

$: t = log\,1 = 0$ to

$t = log\,3$

$\displaystyle \therefore \int_{0}^{log3} cos\,t\,dt \Rightarrow [sint]_{0}^{log3} = sin\,log3-0$

$\displaystyle \therefore \int_{1}^{3} \frac{cos(log\,x)}{x}dx = sin\,log3$

1117026_460415_ans_636e6031355e4ec899c77037353e7401.JPG

The value of

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \frac { x }{ 2 } \right) } \right) dx }$ is

0%
$\cfrac{1}{n}$
0%
$\cfrac{n+2}{2n+1}$
0%
$\cfrac{2n-1}{n}$
0%
$\cfrac{2n-3}{3n-2}$

Explanation

Given

$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \tan ^{ n+1 }{ x } \right) dx } +\cfrac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \tan ^{ n+1 }{ \left( \cfrac { x }{ 2 } \right) } \right) dx }$

In second integral, put

$t=\cfrac{x}{2} \Rightarrow$

$dx=2dt$

$\Rightarrow$ Also, when

$x=0$ then

$t=0$

When

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

$t=\dfrac{\pi}4$

Then

$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ t } \right) } dt$

$I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } \right) } dx+\int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) } dx\quad (\because \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(y)dy } )$

$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n+1 }{ x } +\tan ^{ n-1 }{ x } \right) } dx$

$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \tan ^{ 2 }{ x+1 } \right) } dx$

$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /4 }{ \left( \tan ^{ n-1 }{ x } \right) .\left( \sec ^{ 2 }{ x } \right) } dx$

Put

$t=\tan{x}$

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

Also when

$x=0$ then

$t=0$
when

$x=\pi /4$ , then

$t=1$

$I=\displaystyle \int _{ 0 }^{ 1 }{ { t }^{ n-1 }dt }$

$={ \left[ \cfrac { { t }^{ n } }{ n } \right] }_{ 0 }^{ 1 }=\cfrac { 1 }{ n }$

What is

$\displaystyle \int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ 1+{ x }^{ 2 } } } dx$ equal to?

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

Explanation

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

Put

$\tan ^{ -1 }{ x } =p\Rightarrow \dfrac { dx }{ 1+{ x }^{ 2 } } =dp$

When

$x=0, p=0$ and

$x=1,p=\dfrac{\pi}{4}$

$\Rightarrow I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ pdp }=\dfrac{p^{2}}{2} =\dfrac { { \pi }^{ 2 } }{ 32}$

Hence, B is correct.

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.

$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$ , then

$b =$

0%
$\tan ^{-1}\, \left (\displaystyle \frac{1}{3} \right )$
0%
$\displaystyle \frac{\sqrt 3}{2}$
0%
$\sqrt 3$
0%
$1$

Explanation

$\displaystyle \int_{0}^{b} \displaystyle \frac{dx}{1\, +\, x^2}\, =\, \displaystyle \int_{b}^{\infty} \displaystyle \frac{dx}{1\, +\, x^2}$

$\Rightarrow [\tan^{-1}x]_0^b=[\tan^{-1}x]_b^{\infty}$

$\Rightarrow \tan^{-1}b-0=\dfrac{\pi}{2}-\tan^{-1}b$

$\Rightarrow \tan^{-1}b=\dfrac{\pi}{4}$

$\Rightarrow b=\tan\dfrac{\pi}{4}=1$

Select and write the most appropriate answer from the given alternatives for question :
If

$\displaystyle \int^k_0 4x^3dx=16$ , then the value of

$k$ is _____.

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

Explanation

Given:

$\displaystyle \int^K_0 4x^3dx=16$

$4\left(\dfrac{x^4}{4}\right)^K_0=16$

$(x^4)^k_0=16$

$k^4=16$

$k^4=2^4$

$\boxed{k=2}$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 4 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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