MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 11

$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ 2+\cos { x } } = }$

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

Evaluate:

$\displaystyle \overset{\pi/3}{\underset{\pi/6}{\int}} \dfrac{dx}{1 + \sqrt{\tan x}}$

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

Explanation

Let,

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{dx}{1+\sqrt{\tan x}}}$

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{dx}{1+\sqrt{\dfrac{\sin x}{\cos x}}}}$

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}$ …… (1)

We know that,

$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$

Therefore,

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}}{\sqrt{\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}+\sqrt{\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}}dx}$

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}dx}$

$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$ …… (2)

Add equations (1) and (2).

$2I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

$2I=\int\limits_{\pi /6}^{\pi /3}{1\cdot dx}$

$2I=\left[ x \right]_{\pi /6}^{\pi /3}$

$2I=\dfrac{\pi }{3}-\dfrac{\pi }{6}$

$2I=\dfrac{\pi }{6}$

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

Hence, this is the required value of the integral.

$\displaystyle \int^{\pi/2}_{-\pi/2}\sqrt {\cos x-\cos^{3}x}dx=$

0%
$1$
0%
$4/3$
0%
$-1/3$
0%
$0$

${ I }_{ n }=\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 50 } \right) }^{ n }dx }$ , then

$\dfrac { { I }_{ 100 } }{ { I }_{ 101 } }$

0%
$\cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5001}{5500}$ .
0%
$\cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5051}{5050}$ .
0%
$\cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5050}{5550}$ .
0%
$\cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5510}{5055}$ .

Explanation

Given that:-

${I}_{n} = \int_{0}^{1}{{\left( 1 - {x}^{50} \right)}^{n} dx}$

To find:-

$\cfrac{{I}_{100}}{{I}_{101}} = ?$

${I}_{n+1} = \int_{0}^{1}{{\left( 1 - {x}^{50} \right)}^{n+1} dx}$

$\Rightarrow {I}_{n+1} = \int_{0}^{1}{\left( 1 - {x}^{50} \right) {\left( 1 - {x}^{50} \right)}^{n} dx}$

$\Rightarrow {I}_{n+1} = \int_{0}^{1}{{\left( 1 - {x}^{50} \right)}^{n} dx} - \int_{0}^{1}{{x}^{50} {\left( 1 - {x}^{50} \right)}^{n} dx}$

$\Rightarrow {I}_{n+1} = {I}_{n} + \int_{0}^{1}{x.\left(- {x}^{49} {\left( 1 - {x}^{50} \right)}^{n} \right) dx}$

Using integration by parts, we hve

${I}_{n+1} = {I}_{n} + \left[ \cfrac{x {\left( 1 - {x}^{50} \right)}^{n+1}}{50 \times \left( n + 1 \right)} \right]_{0}^{1} - \int{\cfrac{{\left( 1 - {x}^{50} \right)}^{n + 1}}{50 \left( n + 1 \right)} dx}$

$\Rightarrow {I}_{n+1} = {I}_{n} - \cfrac{\int_{0}^{1}{{\left( 1 - {x}^{50} \right)}^{n} dx}}{50 \left( n + 1 \right)}$

$\Rightarrow {I}_{n+1} = {I}_{n} - \cfrac{{I}_{n+1}}{50 \left( n + 1 \right)}$

$\Rightarrow {I}_{n+1} \left( 1 + \cfrac{1}{50 \left( n + 1 \right)} \right) = {I}_{n}$

As we need to find

$\cfrac{{I}_{100}}{{I}_{101}} \Rightarrow n = 100$

$\therefore {I}_{100 + 1} = \left( 1 + \cfrac{1}{50 \times 101} \right) = {I}_{100}$

$\Rightarrow \cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5051}{5050}$

Hence the required answer is

$\cfrac{{I}_{100}}{{I}_{101}} = \cfrac{5051}{5050}$ .

$\int_\limits{0}^1 \dfrac{x^3}{\sqrt{1-x^2}}dx$

0%
$2\dfrac{\pi}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{3}{2}$
0%
None of these

$\int_{0}^{1}\dfrac{\sin t}{1+t}dt=\alpha$ , then the value of

$\int_{4\pi-2}^{4\pi}\dfrac{\sin (t/2)}{4\pi+2-t}dt=$

0%
$\alpha$
0%
$-\alpha$
0%
$2\alpha$
0%
$4\pi-2\alpha$

$(-1, 2)$ and

$(2, 4)$ are two points on the curve

$y=f(x)$ and if

$g(x)$ is the gradient of the curve at point (x, y), then the value of the integral

$\displaystyle\int^{2}_{-1}g(x)dx$ , is?

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

What is the value of

$\int_{0}^{\pi}\dfrac {dx}{5-4\cos x}$ ?

0%
$\dfrac {\pi \log 2}{32}$
0%
$dfrac {4\pi}{7}$
0%
$dfrac {\pi}{3}$
0%
$None\ of\ these$

Solve

$\displaystyle\int_{0}^{\infty}\dfrac {x \tan^{-1}x}{(1+x^{2})^{2}}dx$

0%
$\pi/2$
0%
$\pi/6$
0%
$\pi/4$
0%
$\pi/8$

Explanation

Let,

$\tan^{-1} x = t \implies \dfrac{1}{1+x^2} dx = dt$

$x\rightarrow 0 \implies t \rightarrow 0$

$x\rightarrow \infty \implies t \rightarrow \pi/2$

Now,

$\int_{0}^{\infty} \dfrac{x\tan^{-1} x}{(1+x^2)^2} dx$

$= \int_{0}^{\pi/2} \dfrac{\tan t . (t)}{1+\tan^2 t} dt$

$= \int_{0}^{\pi/2} \dfrac{\tan t . (t)}{\sec^2 t} dt$

$= \int_{0}^{\pi/2} \sin t \cos t .(t) dt$

$= \dfrac{1}{2}\int_{0}^{\pi/2} 2\sin t \cos t .(t) dt$

$= \dfrac{1}{2}\int_{0}^{\pi/2} \sin 2t .(t) dt$

$= \dfrac{1}{2} \left[-t. \dfrac{\cos 2t}{2} + \dfrac{\sin 2t}{4}\right] _{0}^{\pi/2}$

$= \dfrac{1}{2} \left[ (\dfrac{-\pi}{4})(-1) + 0\right]$

$= \dfrac{\pi}{8}$ (Ans)

The value of

$\int_{-1}^{1}\dfrac {dx}{(2-x)\sqrt {1-x^{2}}}$ is

0%
$0$
0%
$\dfrac {\pi}{\sqrt {3}}$
0%
$\dfrac {2\pi}{\sqrt {3}}$
0%
$cannot\ be\ evaluated$

$\displaystyle \int_{1}^{x}\dfrac{dt}{|t|\sqrt{t^{2}-1}}=\dfrac{\pi}{6}$ , then

$x$ can be equal to :

0%
$\dfrac{2}{\sqrt{3}}$
0%
$\sqrt{3}$
0%
$2$
0%
$\dfrac{4}{\sqrt{3}}$

Explanation

$\int _{ 1 }^{ x }{ \cfrac { dt }{ \left| t \right| \sqrt { { \left| t \right| }^{ 2 }-1 } } } =\cfrac { \pi }{ 6 } \\ put\quad \left| t \right| =\sec { \theta } \Rightarrow dt\sec { \theta } \tan { \theta } d\theta \\ changing\quad limit,\\ t=x\Rightarrow \sec { \theta } =x\Rightarrow \sec ^{ -1 }{ x } \\ t=1\Rightarrow \sec { \theta } =1\Rightarrow \sec ^{ -1 }{ 1 } \\ \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \cfrac { \sec { \theta \tan { \theta } d\theta } }{ \sec { \theta } \sqrt { \sec ^{ 2 }{ \theta } -1 } } =\cfrac { \pi }{ 6 } } \\ (\because \sec ^{ 2 }{ \theta } -1=\tan ^{ 2 }{ \theta } )\\ \Rightarrow \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \cfrac { \sec { \theta \tan { \theta } d\theta } }{ \sec { \theta } \tan { \theta } } =\cfrac { \pi }{ 6 } } \\ \Rightarrow \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ d\theta =\cfrac { \pi }{ 6 } } \Rightarrow _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \left[ \theta \right] }=\cfrac { \pi }{ 6 } \\ \therefore \sec ^{ -1 }{ x } -\sec ^{ -1 }{ 1 } =\cfrac { \pi }{ 6 } \\ \therefore \sec ^{ -1 }{ x } =\cfrac { \pi }{ 6 } +\sec ^{ -1 }{ 1 } =\cfrac { \pi }{ 6 } +0\\ \therefore x=\sec { \cfrac { \pi }{ 6 } } =\cfrac { 2 }{ \sqrt { 3 } }$

Solve

$\displaystyle\int_{0}^{\pi}\dfrac {1}{3+2\sin x+\cos x}dx$

0%
$\dfrac{5\pi}{4}$
0%
$\dfrac{\pi}{4}$
0%
$-\dfrac{\pi}{4}$
0%
None of these

Explanation

Let

${ I }_{ 1 }=\displaystyle \int _{ 0 }^{ \pi }{ \dfrac { 1 }{ 3+2\sin { x } +\cos { x } } }$

Let

$t=\tan { \dfrac { x }{ 2 } }$

$at=\dfrac { 1 }{ 2 } \sec ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } dx$

$\Rightarrow \sec ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } dx=2dt$

$=\displaystyle \int { \dfrac { 1 }{ \dfrac { 2.2\tan { \dfrac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \dfrac { x }{ 2 } } } +\dfrac { 1+\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } }{ 1-\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } } +3 } } dx$

$=\displaystyle\int { \dfrac { 1+\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } }{ 4\tan { \left( \dfrac { x }{ 2 } \right) + } 1-\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } +3+3\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } } } dx$

$=\displaystyle\int { \dfrac { \sec ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } }{ 2\tan ^{ 2 }{ \left( \dfrac { x }{ 2 } \right) } +4\tan { \left( \dfrac { x }{ 2 } \right) + } 4 } } dx$

$=\displaystyle \int { \dfrac { 2dt }{ 2\left( { t }^{ 2 }+2t+2 \right) } }$

$=\displaystyle \int { \dfrac { dt }{ { \left( 1+1 \right) }^{ 2 }+1 } }$

$=\dfrac { 1 }{ 1 } \tan ^{ -1 }{ \left( \dfrac { t+1 }{ 1 } \right) }$

$={ \left[ \left( \tan ^{ -1 }{ \left( \tan { \dfrac { x }{ 2 } } +1 \right) } \right) \right] }_{ 0 }^{ \pi }$

$=\tan ^{ -1 }{ \left(\infty \right) } -\tan ^{ -1 }{ \left( 1 \right) }$

$=\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 4 }$

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

The integral

$\int _{ 0 }^{ \pi }{ \sqrt { 1+4\sin { ^{ 2 }\frac { x }{ 2 } -4 } \sin { \frac { x }{ 2 } } } dx }$ is equal to

0%
$\pi-4$
0%
$\frac{2\pi}{3}-4-4\sqrt{3}$
0%
$4sqrt{3}-4$
0%
$4sqrt{3}-4-\frac{\pi}{3}$

$\int _{ 1/2 }^{ 2 }{ \dfrac { 1 }{ x } \csc { ^{ 101 }\left( x-\dfrac { 1 }{ x } \right) dx } \\ }$ is equal to

0%
$1/4$
0%
$1$
0%
$0$
0%
$\dfrac{101}{2}$

Explanation

Let

$I=\displaystyle \int _{1/2}^2 \dfrac {1}{x} \csc ^{101} \left (x-\dfrac {1}{x} \right)dx$

Let

$x= 1/t \ \Rightarrow \ dx =-\dfrac {1}{t^2} dt$

$\therefore \ I=-\displaystyle \int _2^{1/2} \dfrac {t}{t^2} \csc ^{101} \left (\dfrac {1}{t} -t \right) dt$

$=-\displaystyle \int _2^{1/2} \csc^{101} \left (\dfrac {1}{t} - t \right) dt$

$=-I$

or,

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

The value of

$\int _{ 1/e }^{ \tan { x } }{ \dfrac { t }{ 1+{ t }^{ 2 } } dt+ } \int _{ 1/e }^{ \cot { x } }{ \dfrac { t }{ t\left( 1+{ t }^{ 2 } \right) } dt }$ , where

$x\in \left( \pi /6,\ \pi /3 \right)$ , is equal to :

0%
$0$
0%
$2$
0%
$1$
0%
$cannot\ be\ determined$

$G(x) = \begin{vmatrix}f(x)f(-x) & 0 & x^{4}\\ 3 & f(x) - f(-x) & \cos x\\ x^{4} & 2x & f(x)f(-x)\end{vmatrix}$ , then

$\int_{-2}^{2} x^{4}G(x)dx$ is equal to

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

The value of

$\int\limits_0^\pi {\frac{{{e^{\cos x}}}}{{{e^{\cos x}} + {e^{ - \cos x}}}}dx}$

0%
$\pi$
0%
$\frac{\pi }{2}$
0%
$\frac{\pi }{4}$
0%
$\frac{\pi }{5}$

The value of

$\displaystyle \int\limits_0^1 {\dfrac{{x{{\tan }^{ - 1}}x}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}} dx$ is

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

Explanation

Let say

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

Put

$t=\tan^{-1}x \Rightarrow$ if

$x=0,$ then

$t=0$

$dt=\dfrac {1}{1+x^2}dx \quad$ if

$x=1,$ then

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

So,

$I=\displaystyle \int_0^{\pi /4} \dfrac {x\ t}{\sqrt {1+x^2}}dt\quad$ if

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

$I=\displaystyle \int_0^{\pi /4}\dfrac {t(\tan t)}{\sqrt {1+\tan^2 t}}dt=\displaystyle \int_0^{\pi /4}\dfrac {t\ \sin t}{\cos t.\sec t} dt$

$=\displaystyle \int_0^{\pi /4}\dfrac {t\ \sin t}{1} dt$

Integral by parts

$I=[t (-\cos t)]_0^{\pi /4}-\displaystyle \int_0^{\pi /4} (1).(-\cos t)dt$

$I=\left [\dfrac {\pi}{4}\left (-\dfrac {1}{\sqrt 2}\right)-0 \right]+\displaystyle \int_0^{\pi /4}\cos t \ dt$

$I=\dfrac {-\pi}{4\sqrt 2}+[\sin t]_0^{\pi /4}$

$I=-\dfrac {\pi}{4\sqrt 2}+\dfrac {1}{\sqrt 2}=\dfrac {-\pi +4}{4\sqrt 2}$

So,

$\boxed {I=\dfrac {4-\pi}{4\sqrt 2}}\to option \to B$

$\int _{ 0 }^{ 1 }{ \frac { x }{ 1+\sqrt { x } } dx= }$

0%
$\frac {5}{3}-\log {4}$
0%
$\frac {5}{3}+\log {4}$
0%
$\frac {5}{3}\log {4}$
0%
$\frac {3}{5}-\log {4}$

Solve

$\int\limits_0^\pi {\log {{\sin }^2}x} dx$

0%
$2\pi \log \left( {1/2} \right)$
0%
$\pi \log 2$
0%
$\pi /2\log \left( {1/2} \right)$
0%
None of these

Explanation

$\displaystyle \int^{\pi}_0\log \sin^2 xdx$

$I=\displaystyle \int^{\pi}_0\log \sin^2 xdx=2\displaystyle \int^{\pi}_0\log \sin xdx$ .............(A)

$I=2.2\displaystyle \int^{\pi/2}_0\log \sin(\pi-x)dx$

$\left\{\displaystyle \int^{\pi}_0 f(x)dx=\int^{2a}_0f(x)dx=2\int^{a}_0 f (x)dx\,\,when f(2a-x)=f(x)\right\}$

$I=4 \displaystyle \int^{\pi/2}_0\log \sin x dx$ .......(1)

$I=4\displaystyle \int^{\pi/2}_0\log \sin\left(\dfrac{\pi}{2}-x\right) dx=4\displaystyle \int^{\pi/2}_0\log \cos x dx$ .........(2)

$I+I=4\displaystyle \int^{\pi/2}_0\log \sin x+\log \cos x dx=4\displaystyle \int^{\pi/2}_0\log \left(\dfrac{\sin 2x}{2}\right)$

$2I=4\left[\displaystyle \int^{\pi/2}_0\log (\sin 2x)dx-\displaystyle \int^{\pi/2}_0\log 2 dx]=4[\displaystyle \int^{\pi/2}_0\log (\sin 2 x)dx-\dfrac{\pi}{2}\log\right]$

$I=2\displaystyle \int^{\pi/2}_0\log \sin 2 xdx-\pi \log 2\Rightarrow I=2I_1=\pi \log 2$

Now

$I_1=\displaystyle \int^{\pi/2}_0\log (\sin 2 x)dx$ ;let

$2x=t\Rightarrow 2dx=dt$

$\displaystyle I_1=\int^{\pi}_0\log (\sin t)\dfrac{dt}{2}=\dfrac{1}{2}\int^{\pi}_0\log (\sin t)dt=\dfrac{1}{2}\int^{\pi }_0\log (\sin x)dx$

$I_1=\dfrac{1}{2}\dfrac{I}{2}$ from eq(A)

putting

$I_1$ in eq (3)

$I=2\left(\dfrac{I}{4}\right)-\pi \log 2$

$I=-2\pi \log 2=2\pi \log (1/2)$

The value of the integral

$\int _{ \dfrac { 1 }{ 3 } }^{ 1 }{ \dfrac { \left( x-{ x }^{ 3 } \right) ^{ \dfrac { 1 }{ 3 } } }{ { x }^{ 4 } } }dx$ is

0%
6
0%
0
0%
3
0%
4

$\overset { -5 }{ \underset { -4 }{ \int } } e^(x + 5)^2 dx + 3 \overset { 2/3}{ \underset { 1/3 }{ \int } } e^9(9(x-2/3)^2$ dx is equal toi

0%
$e^5$
0%
$e^4$
0%
$3e^2$
0%
0

$I=\displaystyle\int^1_0\cos\left(2\cot^{-1}\sqrt{\left(\dfrac{1-x}{1+x}\right)}\right)dx$ then?

0%
$I > \dfrac{1}{2}$
0%
$I=-\dfrac{1}{2}$
0%
$0 < I < \dfrac{1}{2}$
0%
None of these

$\overset { 1 }{ \underset { 0 }{\int } } 2^{x^2}dx,I_2=\overset { 1 }{ \underset { 0 }{\int } }2^{x^3}dx,I_3 =\overset { 2 }{ \underset { 1 }{\int } }2^{x^2}dx,$ and

$I_4=\overset { 1 }{ \underset { 0 }{\int } }2^{x^3}dx$ then-

0%
$I_2>I_1$
0%
$I_1>I_2$
0%
$I_3=I_4$
0%
$I_3>I_4$

$\overset { 1 }{ \underset { -1 }{ \int } } \dfrac{x^3+|x|+1}{x^2+2|x|+1}$ dx is equal to

0%
$In 3$
0%
$2In 3$
0%
$\dfrac{1}{3}In 3$
0%
none of these

The value of

$\int\limits_1^2 {\frac{1}{{{x^2}}}{e^{ - 1/x}}} dx$ is

0%
$\frac{1}{{\sqrt e }} + \frac{1}{e}$
0%
$\frac{1}{e} - \frac{1}{{\sqrt e }}$
0%
$\frac{1}{{\sqrt e }} - \frac{1}{e}$
0%
0

Explanation

Consider

$\int_{1}^{2}\dfrac{1}{x^2}e^{-1/x}dx$

Let

$u=-\dfrac{1}{x}$

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

When

$x=1$ ,

$u=-1$ and when

$x=2, u=-\dfrac{1}{2}$

$\int_{1}^{2}\dfrac{1}{x^2}e^{-1/x}dx=\int_{-1}^{-1/2}e^u du$

$=-\int_{-1/2}^{-1}e^u du$

$=-[e^u]_{-1/2}^{-1}$

$=-[e^{-1}-e^{-1/2}]$

$=-\left[\dfrac{1}{e}-\dfrac{1}{\sqrt{e}}\right]$

$I=\int_{1}^{2}\dfrac{1}{x^2}e^{-1/x}dx$

$=\dfrac{1}{\sqrt{e}}-\dfrac{1}{e}$

The value of

$\int_{-1}^{3}[tan^{-1}(\frac{x}{x^{2}+1})+tan^{-1}(\frac{x^{2}+1}{x})]dx$

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

The value of the integral

$\underset{0}{\overset{\pi / 4}{\int}} \dfrac{\sin \, x + \cos \, x}{3 + \sin \, 2x} dx$ , is

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

Explanation

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

$=\int_{0}^{\frac{\pi}{4}}\dfrac{\sin x+\cos x}{4-(\sin x-\cos x)^2}dx$

substitute

$u=\sin x-\cos x\Rightarrow (\sin x+\cos x)dx=dt$

$x(0,\dfrac{\pi}{4})\rightarrow t(-1,0)$

$=\int_{-1}^{0}\dfrac{dt}{4-t^2}$

$=\int_{-1}^{0}\dfrac{dt}{2^2-t^2}$

$=\dfrac{1}{2\times 2}\left [ \ln \dfrac{2+t}{2-t} \right ]_{-1}^0$

$=\dfrac{1}{4}\left \{ \ln 1 -\ln \dfrac{1}{3} \right \}$

$=\dfrac{1}{4}\ln 3$

$\int^{\pi/4}_{0}\dfrac {\sin^{2}x\cos^{2}x}{(\sin^{3}x+\cos^{3}x)^{2}}dx$ is

0%
$1/3$
0%
$1/2$
0%
$1/6$
0%
$1/4$

$I = \int _ { 0 } ^ { 1 } \frac { \tan x } { \sqrt { x } } d x$ then ?

0%
$I < \frac { 2 } { 3 }$
0%
$I > \frac { 2 } { 3 }$
0%
$I < \frac { 5 } { 9 }$
0%
$I < \frac { 1 } { 3 }$

$\int _ { 0 } ^ { \pi ^ { 2 } / 4 } \sin \sqrt { x } d x$ is

0%
0
0%
1
0%
2
0%
4

$\displaystyle \int_{0}^{\pi/4}{\dfrac{x.\sin x}{\cos^{3}x}dx}$ equals to :

0%
$\dfrac{\pi}{4}+\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}-\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{4}+1$

Explanation

$\displaystyle\int_{0}^{\frac{\pi}{4}}{\dfrac{x\sin{x}}{{\cos}^{3}{x}}dx}$

$=\displaystyle\int_{0}^{\frac{\pi}{4}}{x\tan{x}{\sec}^{2}{x}dx}$

Let

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

$\Rightarrow\,x={\tan}^{-1}{t}$

When

$x=0\Rightarrow\,t=0$

When

$x=\dfrac{\pi}{4}\Rightarrow\,t=1$

$=\displaystyle\int_{0}^{1}{{\tan}^{-1}{t}\,dt}$

Let

$u={\tan}^{-1}{t}\Rightarrow\,du=\dfrac{1}{1+{t}^{2}}dt$

$dv=t\Rightarrow\,v=\dfrac{{t}^{2}}{2}$

$=\left[{\tan}^{-1}{t}\dfrac{{t}^{2}}{2}\right]_{0}^{1}-\displaystyle\int_{0}^{1}{\dfrac{{t}^{2}}{2}\times\dfrac{1}{1+{t}^{2}}dt}$

$=\dfrac{1}{2}\left[{t}^{2}{\tan}^{-1}{t}\right]_{0}^{1}-\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{{t}^{2}}{1+{t}^{2}}dt}$

$=\dfrac{1}{2}\left[{t}^{2}{\tan}^{-1}{t}\right]_{0}^{1}-\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{{t}^{2}+1-1}{1+{t}^{2}}dt}$

$=\dfrac{1}{2}\left[{t}^{2}{\tan}^{-1}{t}\right]_{0}^{1}-\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{{t}^{2}+1}{1+{t}^{2}}dt}+\dfrac{1}{2}\displaystyle\int_{0}^{1}{\dfrac{dt}{1+{t}^{2}}}$

$=\dfrac{1}{2}\left[{\tan}^{-1}{1}-0\right]-\dfrac{1}{2}\left[t\right]_{0}^{1}+\dfrac{1}{2}\left[{\tan}^{-1}{t}\right]_{0}^{1}$

$=\dfrac{1}{2}\left[\dfrac{\pi}{4}-0\right]-\dfrac{1}{2}\left[1-0\right]_{0}^{1}+\dfrac{1}{2}\left[\dfrac{\pi}{4}-0\right]_{0}^{1}$

$=\dfrac{\pi}{8}-\dfrac{1}{2}+\dfrac{\pi}{8}$

$=\dfrac{2\pi}{8}-\dfrac{1}{2}$

$=\dfrac{\pi}{4}-\dfrac{1}{2}$

The value of

$\int_{a}^{b}{(x-a)^{3}(b-x)^{4}dx}$ is

0%
$\dfrac{(b-a)^{4}}{6^{4}}$
0%
$\dfrac{(b-a)^{8}}{280}$
0%
$\dfrac{(b-a)^{7}}{7^{3}}$
0%
$none\ of\ these$

$\displaystyle\int^1_0\tan^{-1}\left[\dfrac{2x-1}{1+x-x^2}\right]dx=?$

0%
$0$
0%
$1/2$
0%
$1$
0%
$\pi/6$

Consider

$f ( x ) = \int _ { 0 } ^ { \pi } \frac { \ln ( 1 + x \cos \theta ) } { \cos \theta } d \theta$
Range of

$f ( x )$ is

0%
$( 0 , \pi )$
0%
$\left( 0 , \pi ^ { 2 } \right)$
0%
$\left( \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right)$
0%
$\left( \frac { - \pi ^ { 2 } } { 2 } , \frac { \pi ^ { 2 } } { 2 } \right)$

$\displaystyle\int _{ -1 }^{ 1 }{ x\ell n\left( 1+{ e }^{ x } \right) dx } =$

0%
$0$
0%
$\ell n(1+e)$
0%
$\ell n(1+e)-1$
0%
$1/3$

$\displaystyle \int _{0}^{1}\cot^{-1}(1+x^{2}-x)dx=k\left(\dfrac {\pi}{4}-\log_{e}\sqrt {2}\right)$ , then the value of

$k$ is equal to

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

$\int _{ \pi /4 }^{ 3\pi /4 }{ \dfrac { dx }{ 1+\cos { x } } }$ is equal to

0%
$-2$
0%
$2$
0%
$4$
0%
$-1$

The value of

$\displaystyle \int _{0}^{\pi/2} \log{\sin{x}}dx$ is

0%
$-\pi \log{2}$
0%
$\dfrac{-\pi}{2} \log{2}$
0%
$\pi \log{2}$
0%
0

The value of

$\displaystyle \int _{-1}^{1} \log{\left(\dfrac{2-x}{2+x}\right)}\sin^{2}{x}dx$

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

The value of the definite intergral

$\displaystyle \int_{19}^{37}($ {x}

$^2$ +

$3\sin (2\pi x))dx$ , where {.} denotes the fractional part function

0%
$0$
0%
$6$
0%
$9$
0%
can not determine

The value of the integral

$\displaystyle \int _{ { e }^{ -1 } }^{ { e }^{ 2 } }{ \left| \frac { \log _{ e }{ x } }{ x } \right| dx }$ is

0%
$\dfrac {3}{2}$
0%
$\dfrac {5}{2}$
0%
$3$
0%
$5$

$\displaystyle\int^{2+\sqrt{3}}_{2-\sqrt{3}}\dfrac{xdx}{(1+x)(1+x^2)}=?$

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{12}$
0%
$\dfrac{\pi}{24}$

Let

$I=\overset{1}{\underset{0}{\int}}\dfrac{\sin x}{\sqrt x}dx$ and

$J=\overset{1}{\underset{0}{\int}}\dfrac{\cos x}{\sqrt x}dx$ . Then which one of the following is true?

0%
$1>\dfrac{2}{3}$ and $J>2$
0%
$1<\dfrac{2}{3}$ and $J<2$
0%
$1<\dfrac{2}{3}$ and $J>2$
0%
$1>\dfrac{2}{3}$ and $J<2$

The area of the region bounded by the lines

$x = 1, x = 2$ , and the curves

$x(y - e^x) = \sin x$ and

$2xy = 2 \sin x + x^3$ is

0%
$e^2 - e - \dfrac{1}{6}$
0%
$e^2 - e - \dfrac{7}{6}$
0%
$e^2 - e + \dfrac{1}{6}$
0%
$e^2 - e + \dfrac{7}{6}$

Explanation

Given:

$x\left(y-{e}^{x}\right)=\sin{x}$

$\Rightarrow\,xy-x{e}^{x}=\sin{x}$

$\Rightarrow\,xy=\sin{x}+x{e}^{x}$

$\Rightarrow\,y=\dfrac{\sin{x}}{x}+{e}^{x}$

Given:

$2xy=2\sin{x}+{x}^{3}$

$\Rightarrow\,y=\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}$

$A=\displaystyle\int_{1}^{2}{\left\{\dfrac{\sin{x}}{x}+{e}^{x}-\left(\dfrac{\sin{x}}{x}+\dfrac{{x}^{2}}{2}\right)\right\}dx}$

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

$=\left[{e}^{x}-\dfrac{{x}^{3}}{6}\right]_{1}^{2}$

$={e}^{2}-\dfrac{8}{6}-\left(e-\dfrac{1}{6}\right)$

$={e}^{2}-e-\dfrac{7}{6}$

The value of the integral

$\displaystyle \int _ { 0 } ^ { 1 } \dfrac { d x } { x ^ { 2 } + 2 x \cos \alpha + 1 }$ , where

$0 < \alpha < \dfrac { \pi } { 2 } ,$ is equal to

0%
$\sin{\alpha}$
0%
$\alpha \sin {\alpha}$
0%
$\dfrac { \alpha } { 2 \sin{\alpha} }$
0%
$\dfrac { \alpha } { 2 } \sin {\alpha}$

Explanation

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

$= \int_{0}^{1}\frac{dx}{(x+cos\,\alpha )^{2}+sin^{2}\alpha }$

$= \frac{1}{sin\,\alpha } \left [ tan^{-1} \left ( \frac{x+cos\,\alpha }{sin\,\alpha } \right ) \right ]_{0}^{1}$

$= \frac{1}{sin\,\alpha } \left [ tan^{-1}\left ( \frac{1+cos\alpha }{sin\,\alpha } \right )- tan^{-1}\left ( \frac{cos\, \alpha }{sin\,\alpha } \right ) \right ]$

$= \frac{1}{sin\, \alpha } \left [ tan^{-1}\left ( \frac{2\, cos^{2}\alpha /2}{2\, sin\, \alpha /2 \, cos\,\alpha /2} \right ) - tan^{-1}\left ( cot\alpha \right )\right ]$

$= \frac{1}{sin \, \alpha} \left [ tan^{-1}\left ( cot+\,\alpha/2 \right ) - tan^{-1}\left ( cot\alpha \right ) \right ]$

$= \frac{1}{sin\, \alpha}\left [ tan^{-1}\left ( tan\left ( \pi /2 - \alpha/2 \right ) \right ) - tan^{-1}\left ( tan\left ( \pi /2 - \alpha \right ) \right ) \right ]$

$= \frac{1}{sin \alpha} \left [ \pi/2 - \alpha/2 - \pi/2 + \alpha \right ]$

$= \frac{\alpha}{2 sin\, \alpha}$

1203824_1246621_ans_5f6a4a6e0596462b93d687e05a24214d.jpg

$\displaystyle\int^{\lambda}_0\dfrac{y}{\sqrt{y+\lambda}}dy=?$

0%
$\dfrac{2}{3}(2-\sqrt{2})\lambda \sqrt{\lambda}$
0%
$\dfrac{2}{3}(2+\sqrt{2})\lambda\sqrt{\lambda}$
0%
$\dfrac{1}{3}(2-\sqrt{2})\lambda \sqrt{\lambda}$
0%
$\dfrac{1}{3}(2+\sqrt{2})\lambda \sqrt{\lambda}$

$\int _ { \ln 2 } ^ { x } \frac { d x } { \sqrt { e ^ { x } - 1 } } = \frac { \pi } { 6 } ,$ then

$x$ is equal to

0%
$\ln 8$
0%
$\ln 2$
0%
$\ln 4$
0%
$4$

$\displaystyle\int^{2\pi}_0[\sin x]dx$ .

0%
0
0%
$-\pi$
0%
$2\pi$
0%
$-2\pi$

$\displaystyle \int _ { 0 } ^ { \pi / 4 } \frac { x \cdot \sin x } { \cos ^ { 3 } x } d x$ equals to :

0%
$\displaystyle \frac { \pi } { 4 } + \frac { 1 } { 2 }$
0%
$\displaystyle \frac { \pi } { 4 } - \frac { 1 } { 2 }$
0%
$\dfrac { \pi } { 4 }$
0%
$\dfrac { \pi } { 4 } + 1$

Explanation

Let

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{x.\sin x}{\cos^3 x}dx$

Applying by-parts formula, we get

$=\displaystyle\left[\dfrac{x}{2\cos^2 x}\right]_0^{\pi/4}-\int_{0}^{\pi/4}\dfrac{1}{2\cos^2 x}dx$

$=\displaystyle\dfrac{\dfrac{\pi}{4}}{2\left(\dfrac{1}{\sqrt{2}}\right)^2}-0-\int_{0}^{\pi/4}\dfrac{\sec^2 x}{2}dx$

$=\dfrac{\pi}{4}-\dfrac{1}{2}\left[\tan x\right]_0^{\pi/4}$

$=\dfrac{\pi}{4}-\dfrac{1}{2}\left[\tan \dfrac{\pi}{4}-\tan 0\right ]$

$=\dfrac{\pi}{4}-\dfrac{1}{2}[1-0]$

$=\dfrac{\pi}{4}-\dfrac{1}{2}$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 11 - 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 11 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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