MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 5

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

$1+\tan$

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

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

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

$I_n = \int_0^{\pi/4} tan^n \theta d \theta$ , where n is a positive integer, then

$n(I_{n-1} + I_{n +1})$ is equal to

0%
1
0%
n - 1
0%
$\dfrac{1}{n - 1}$
0%
None of these

Explanation

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

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

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

$Let\ \ J=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n-2 } } .{ \left( \sec { x } \right) }^{ 2 }dx$

$Let\ \ z=tanx$

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

$J=\displaystyle \int _{ 0 }^{ 1 }{ { z }^{ n-2 }dz }$

$J=\frac { 1 }{ n-1 }$

${ I }_{ n }=J-{ I }_{ n-2 }^{ }$

${ I }_{ n }=\frac { 1 }{ n-1 } -{ I }_{ n-2 }^{ }$

$\left( n-1 \right) { I }_{ n }+\left( n-1 \right) { I }_{ n-2 }^{ }=1$

$n\rightarrow n+1$

$\Rightarrow n{ I }_{ n+1 }^{ }+n{ I }_{ n-1 }^{ }=1$

The value of

$\displaystyle \int_{1/n}^{(an - 1)/n} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}} dx$ is equal to

0%
$\dfrac {a}{2}$
0%
$\dfrac {n\cdot a + 2}{2n}$
0%
$\dfrac {n\cdot a - 2}{2n}$
0%
$\dfrac {n\cdot a}{2}$

Explanation

Let,

$I =\displaystyle \int_{1/2}^{(an - 1)/n} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}}dx$

$=\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}} dx$ ....(i)

$= \displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\left (\sqrt {\dfrac {1}{n} + a - \dfrac {1}{n} - x}\right )}{\sqrt {a - \left (\dfrac {1}{n} + a - \dfrac {1}{n} - x\right )} + \sqrt {\dfrac {1}{n} + a - \dfrac {1}{n} - x}}dx$

$\Rightarrow I =\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\sqrt {a - x}}{\sqrt {x} + \sqrt {a - x}} dx$ .... (ii)
On adding Eqs. (i) and (ii), we get

$2I =\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} 1dx = [x]_{\dfrac {1}{n}}^{a - \dfrac {1}{n}}$

$= a - \dfrac {1}{n} - \dfrac {1}{n} = \dfrac {na - 2}{n}\Rightarrow I = \dfrac {na - 2}{2n}$ .

The value of

$\displaystyle\int _{ 0 }^{ \pi }{ \dfrac { dx }{ 5+3\cos { x } } }$ is

0%
$\dfrac{ \pi }{ 4 }$
0%
$\dfrac{ \pi }{ 8 }$
0%
$\dfrac{ \pi }{ 2 }$
0%
Zero

Explanation

Consider,

$\dfrac{1}{5+3\cos x}$

$=\dfrac{1}{5+\dfrac{3-3\tan^{2}(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})}}$ ......

$\left[\because \cos 2x=\dfrac{1-\tan^{2}x}{1+\tan^{2}x}\right]$

$=\dfrac{1+\tan^{2}\frac{x}{2}}{5+5\tan^{2}\frac{x}{2}+3-3\tan^{2}\frac{x}{2}}$

$=\dfrac{\sec^{2}\frac{x}{2}}{8+2\tan^{2}\frac{x}{2}}$

Now,

$\displaystyle \int_{0}^{\pi}\dfrac{1}{5+3\cos x}$

$=\displaystyle \int_{0}^{\pi} \dfrac{\sec^{2}\frac{x}{2}}{8+2\tan^{2}\frac{x}{2}}dx$

Take

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

$\implies \dfrac{1}{2}.\sec^{2}\dfrac{x}{2}dx=dt$

Then, we get

$\displaystyle \int \dfrac{dt}{4+t^{2}}dx$

$=\dfrac{1}{2}\left[\tan^{-1}\left(\dfrac{t}{2}\right)\right]$

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

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

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

$\displaystyle \int_{\log 2}^{x}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi }{6}$ , then the value of

$x$ is

0%
$\log 2$
0%
$\log 3$
0%
$\log 4$
0%
None of these

Explanation

Let

$I=\displaystyle \int_{\log 2}^{x}\frac{1}{\sqrt{e^x-1}}dx$
Put

$e^x-1=t^2$

$\Rightarrow e^xdx=2t\, dt \Rightarrow (t^{2}+1)dx=2t\ dt \Rightarrow dx=\dfrac{2t}{1+t^{2}} dt$

When

$x=\log 2, t=\pm 1$

$\therefore I=\displaystyle \int_{1}^{\sqrt{e^{x}-1}}\frac{2t}{t(1+t^2)}dt$

$=2 \displaystyle \int_{1}^{\sqrt{e^{x}-1}}\frac{1}{1+t^2}dt$

$=2|\tan^{-1} t|_{1}^{\sqrt{e^{x}-1}}$

$=2[\tan^{-1}\sqrt{e^x-1}-\tan^{-1}1]$

$=2 \tan^{-1}\sqrt{e^x-1}-\dfrac{\pi}{2}$
But

$I=\displaystyle \int_{\log 2}^{x}\frac{1}{\sqrt{e^x-1}}dx=\frac{\pi}{6}$

$\Rightarrow 2 \tan^{-1}\sqrt{e^x-1}-\dfrac{\pi}{2}=\dfrac{\pi}{6}$

$\Rightarrow 2 \tan^{-1}\sqrt{e^x-1}=\dfrac{2\pi}{3}$

$\Rightarrow \sqrt{e^x-1}=\tan \left(\frac{\pi}{3}\right)=\sqrt{3}$

$\Rightarrow e^x-1=3$

$\Rightarrow e^x=4$

$\therefore x=\log 4$

$\displaystyle\int^{\pi /2}_{-\pi /2}\cos x$ ln

$\left(\displaystyle\frac{1+x}{1-x}\right)dx$ is equal to.

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

Explanation

Let the given integral be

$I$

We have

$I=\displaystyle \int _{ -\tfrac { \pi }{ 2 } }^{ \tfrac { \pi }{ 2 } }{\cos x\ln { \left(\dfrac { 1+x }{ 1-x }\right)dx }}$

In the above expression ,

$x$ can be replaced with

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

$\Rightarrow I=\displaystyle \int _{ -\tfrac { \pi }{ 2 } }^{ \tfrac { \pi }{ 2 } }{ \cos x\ln { \left(\dfrac { 1+x }{ 1-x } \right)dx } =\int _{ -\tfrac { \pi }{ 2 } }^{ \tfrac { \pi }{ 2 } }{\cos x\ln { \left(\dfrac { 1-x }{ 1+x }\right)dx } \quad } \quad \quad }$

By adding both, we get

$2I=\displaystyle \int _{ -\tfrac { \pi }{ 2 } }^{ \tfrac { \pi }{ 2 } } \cos x\left(\ln \left(\dfrac { 1+x }{ 1-x } \right)+\ln { \left(\dfrac { 1-x }{ 1+x } \right) } \right)dx$

$=\displaystyle \int _{ -\tfrac { \pi }{ 2 } }^{ \tfrac { \pi }{ 2 } } \cos x\left(\ln \left(\dfrac { 1+x }{ 1-x } \times \dfrac { 1-x }{ 1+x } \right)\right)dx =\int _{-\tfrac {\pi}{2}}^{\tfrac{\pi}{2}} \cos x(\ln (1))dx =0$

$\Rightarrow I=0$

Therefore the correct option is

$A$

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

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

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

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

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

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

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

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

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

$f\left( x \right)$ is defined

$\left[ -2,2 \right]$ by

$f\left( x \right) =4{ x }^{ 2 }-3x+1$ and

$g\left( x \right) =\dfrac { f\left( -x \right) -f\left( x \right) }{ { x }^{ 2 }+3 }$ , then

$\displaystyle\int _{ -2 }^{ 2 }{ g\left( x \right) dx }$ is equal to

0%
$64$
0%
$-48$
0%
$0$
0%
$24$

Explanation

Given that

$f\left( x \right) =4{ x }^{ 2 }-3x+1,$

$g\left( x \right) =\dfrac { f\left( -x \right) -f\left( x \right) }{ { x }^{ 2 }+3 }$
Therefore,

$g\left( x \right) =\dfrac { \left( 4{ x }^{ 2 }+3x+1 \right) -\left( 4{ x }^{ 2 }-3x+1 \right) }{ { x }^{ 2 }+3 }$

$=\dfrac { 6x }{ { x }^{ 2 }+3 }$
Now,

$g\left( -x \right) =-\dfrac { 6x }{ { x }^{ 2 }+3 } =-g\left( x \right)$
Which is an odd function
Thus

$\displaystyle\int _{ -2 }^{ 2 }{ g\left( x \right) dx } =0$

$\displaystyle \int_0^{\pi} \dfrac{x^2}{(1+\sin \,x)^2} dx = A$ the

$\displaystyle \int_0^{\pi} \dfrac{2x^2cos^2(x/2)}{(1+\sin x)^2} dx=$ ?

0%
$A+\pi - \pi^2$
0%
$A-\pi+ \pi^2$
0%
$A-\pi - \pi^2$
0%
$A+2\pi - \pi^2$

Explanation

$\displaystyle \int_{0}^{\pi} \frac{2x^2 \, cos^2 (x/2)}{(1+sinx )^2} dx = \int_{0}^{\pi} \frac{x^2 (1+ cosx)}{(1+sinx)^2} dx$

$\displaystyle = \int_{0}^{\pi}\frac{x^2 }{(1+sinx)^2}dx + \int_{0}^{\pi} \frac{cosx}{(1+sinx)^2} dx$

On solving

$\displaystyle \int_{0}^{\pi} \frac{x^2 \,cos x}{(1+sinx)^2}dx$ through by parts we get;

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

1180999_685257_ans_e396c5685061449b90cd75d4e680c5cb.jpg

$I=\displaystyle\int _{ { 1 }/{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } \cdot \sin { \left( x-\dfrac { 1 }{ x } \right) } dx }$ , then

$I$ is equal to

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

Explanation

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

Let $t=\dfrac { 1 }{ x }$

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

Substituting in $(i)$

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

using $\sin { (-x)=-\sin { x } }$

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

Using $\displaystyle{\int _{ a }^{ b }{ f\left( x \right) dx=-\int _{ b }^{ a }{ f\left( x \right) dx } } }$ and changing the variable from $t$ to $x$

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

Adding $(i)$ and $(ii)$

$\displaystyle{2I=\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx-\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx=0\\ \Rightarrow I=0}$

So option $(A)$ is correct.

Evaluate

$\displaystyle\int^4_1x\sqrt{x}dx$

0%
$12.8$
0%
$12.4$
0%
$7$
0%
None of these

Explanation

$I=\displaystyle\int^4_1x^{3/2}dx$

$=\left[\dfrac{2}{3}x^{5/2}\right]^4_1$ .............

$\because\displaystyle\int{{x}^{n}dx}=\dfrac{{x}^{n+1}}{n+1}+c$

$=\dfrac{62}{5}=12.4$ .

The value of the definite integral

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

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

Explanation

$I=\displaystyle \int_{1}^{\infty}(e^{x+1}+e^{3-x})^{-1}dx=\int_{1}^{\infty}\dfrac{1}{e^{x+1}+e^{3-x}}dx$

$I=\displaystyle \int_{1}^{\infty}\dfrac{e^x}{e^{2x+1}+e^{3}}dx=\dfrac{1}{e}\int_{1}^{\infty}\dfrac{e^x}{e^{2x}+e^{2}}dx$

Substitute

$e^x=t\Rightarrow e^xdx=dt$ and limits will be

Lower limit

$x=1\Rightarrow t=e$

Upper limit

$x=\infty\Rightarrow t=\infty$

$I=\displaystyle \dfrac{1}{e}\int_{e}^{\infty}\dfrac{1}{t^{2}+e^{2}}dt$

$=\dfrac{1}{e}\left| \dfrac{1}{e}\tan^{-1}\left | \dfrac{t}{e}\right |\right |_{e}^{\infty}=\dfrac{1}{e^2}\left | \dfrac{\pi}{2}-\dfrac{\pi}{4}\right |=\dfrac{\pi}{4e^2}$

Hence,

$(A)$

$\displaystyle \int _{ 0 }^{ \pi/2}{ \frac{1}{a + b \cos x}} dx, a > |b| =$ .

0%
$\displaystyle \frac{2}{\sqrt{a^2 - b^2}} \tan^{-1} \sqrt{\frac{a + b}{a - b}}$
0%
$\displaystyle \frac{2}{\sqrt{a^2 - b^2}} \cot^{-1} \sqrt{ \frac{a - b}{a + b} }$
0%
$\displaystyle \frac{2}{\sqrt{a^2 - b^2}} \tan^{-1} \sqrt{ \frac{a - b}{a + b}}$
0%
$\displaystyle \frac{\pi}{\sqrt{a^2 - b^2}}$ .

Explanation

$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ a+b\cos { x } } } =\int _{ 0 }^{ \pi /2 }{ \dfrac { dx }{ a+b\left(\dfrac { 1-\tan ^{ 2 }{ \dfrac { x }{ 2 } } }{ 1+\tan ^{ 2 }{ \dfrac { x }{ 2 } } } \right) } } =\int _{ 0 }^{ \pi /2 }{ \dfrac { (1+\tan ^{ 2 }{ \dfrac { x }{ 2 } } )dx }{ (a+b)+(a-b)\tan ^{ 2 }{ \dfrac { x }{ 2 } } } } =\int _{ 0 }^{ \pi /2 }{ \dfrac { \sec ^{ 2 }{ \dfrac { x }{ 2 } } dx }{ (a+b)+(a-b)\tan ^{ 2 }{ \dfrac { x }{ 2 } } } }$

Let

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

i.e.

$dt=\dfrac { 1 }{ 2 } \sec ^{ 2 }{ \dfrac { x }{ 2 }} dx$

So, by substituting the above equation and changing the limits,

$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { 2dt }{ (a+b)+(a-b){ t }^{ 2 } } } =\left[\dfrac { 2 }{ \sqrt { (a+b)(a-b) } } \tan ^{ -1 }{ \sqrt { \dfrac { a-b }{ a+b } } t } \right] _{ 0 }^{ 1 }=\dfrac { 2 }{ \sqrt { { a }^{ 2 }-{ b }^{ 2 } } } \tan ^{ -1 }{ \sqrt { \dfrac { a-b }{ a+b } } }$

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

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

$\int _0^1 xdx = \dfrac {\pi}{4} - \dfrac {1}{2} ln 2$ 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%
$2 ln 2$

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

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

Explanation

$1+{ x }^{ 2 }=t$

$2xdx=dt$

$xdx=\cfrac { dt }{ 2 }$

$x=0,t=1$

$x=\infty ,t=\infty$

$x=\sqrt { t-1 }$

$\ln { x } =\cfrac { 1 }{ 2 } \ln { \left( t-1 \right) }$

$\Rightarrow \int _{ 1 }^{ \infty }{ \cfrac { 1 }{ 2 } \cfrac { \ln { \left( t-1 \right) } }{ { t }^{ 2 } } } \cfrac { dt }{ 2 }$

$=\cfrac { 1 }{ 4 } \int _{ 1 }^{ \infty }{ \cfrac { 1 }{ { t }^{ 2 } } \ln { \left( t-1 \right) } } dt$
Integrating by parts

$=\cfrac { 1 }{ 4 } { \left[ -\cfrac { 1 }{ t } \ln { \left( t-1 \right) } \right] }_{ 1 }^{ \infty }-\cfrac { 1 }{ 4 } \int _{ 1 }^{ \infty }{ -\cfrac { 1 }{ t\left( t-1 \right) } } dt$

$=\cfrac { 1 }{ 4 } { \left[ -\cfrac { 1 }{ \left( 1+{ x }^{ 2 } \right) } \ln { \left( { x }^{ 2 } \right) } \right] }_{ 0 }^{ \infty }+\cfrac { 1 }{ 4 } \left[ \int _{ 1 }^{ \infty }{ \cfrac { 1 }{ t-1 } } dt-\int _{ 1 }^{ \infty }{ \cfrac { 1 }{ t-1 } } dt \right]$

$=\cfrac { 1 }{ 4 } \times 0+\cfrac { 1 }{ 4 } \times 0=0$

Given

$\int_{1}^{2} e^{x^{2}} dx = a$ the the value of

$\int_{e}^{e^{4}}\sqrt {ln x} dx$ is

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

Explanation

$\int_{1}^{2} e^{x^{2}} dx = a$ (given)
Let

$I = \int_{e}^{e^{4}}\sqrt {\log x} dx$ Putting

$\log x = t^{2} \ dx = 2te^{t^{2}} dt$

$= \int_{1}^{2} t(2t)e^{t^{2}} dt = \int_{1}^{2} t(2t e^{t^{2}}) dt$

$= [te^{t^{2}}]_{1}^{2} - \int_{1}^{2} e^{t{2}} dt = 2e^{4} - e - a$

$\left \{As \int_{1}^{2} e^{t^{2}} dt = \int_{1}6{2} e^{x^{2}} dx = a\right )$
Hence (a) is correct choice.

$\cfrac { \pi }{ 2 } <t<\cfrac { 2\pi }{ 3 }$ and

$=\int _{ 0 }^{ \sin { 2t } }{ \cfrac { dx }{ \sqrt { 4\cos ^{ 2 }{ t-{ x }^{ 2 } } } } }$ then the value of

$\cfrac { 2005\left( I+t \right) }{ \pi }$ equals

0%
$2004$
0%
$2006$
0%
$2005$
0%
None of these

Explanation

$I=\int _{ 0 }^{ \sin { 2t } }{ \cfrac { dx }{ \sqrt { { \left( 2\cos { t } \right) }^{ 2 }-{ x }^{ 2 } } } } ={ \left\{ \sin ^{ -1 }{ \left( \cfrac { x }{ 2\cos { t } } \right) } \right\} }_{ 0 }^{ \sin { 2t } }$

$=\sin ^{ -1 }{ \left( \cfrac { \sin { 2t } }{ 2\cos { t } } \right) } =\sin ^{ -1 }{ \left( \sin { t } \right) } =\sin ^{ -1 }{ \left( \sin { (\pi -t) } \right) }$

$\left( \because \cfrac { \pi }{ 2 } <t<\cfrac { 2\pi }{ 3 } \right)$

$I=\pi -t$

$I+t=\pi \left( \because \cfrac { \pi }{ 3 } <\pi -t<\cfrac { \pi }{ 2 } \right)$

$\therefore \cfrac { I+t }{ \pi } =1\quad$

$\quad \therefore \cfrac { 2005(I+t) }{ \pi } =2005\quad \quad$

The function

$f(x) = \int_1^x [ 2 (t-1) (t-2)^3+3(t-1)^2 (t-2)^2] dt$ has :

0%
Maximum at $x=1$
0%
Minimum at $x = \dfrac {7}{5}$
0%
Neither maximum nor minimum at $x=2$
0%
All of these

Explanation

We have

$f(x) = \int_1^x [ 2 (t-1) (t-2)^3 + 3 (t-1)^2 (t-2)^2 ] dt$

$= \int_1^x (t-1)(t-2)^2 [ 5t-7 ] dt$

$\Rightarrow f'(x) = (x-1)(x-2)^2 (5x-7)$
for max. or min.

$f'(x) = 0 \Rightarrow x = 1,2, \dfrac {7}{5}$
Now

$f''(x) = (x-2)^2 (5x-7) + 5(x-1)(x-2)^2 + 2 (x-2) (x-1) (5x-7)$

$\Rightarrow f''(0) < 0 , f'' (7/5) > 0$ and

$f''(2) = 0$
Hence

$f(x)$ attains its maximum at

$x=1$

Let

$f(x)$ and

$g(x)$ be two function satisfying

$f(x^2)+g (4-x)=4x^3, g(4-x)+g(x)=0$ , then the value of

$\int_{-4}^{4} f(x^2)dx$ is:

0%
512
0%
64
0%
256
0%
0

Explanation

Let

$I=\int_{-4}^{4}f(x^2)dx$

$\therefore I=\int_{-4}^{0}f(x^2)dx+\int_{0}^{4}f(x^2)dx$

$\therefore I=\int_{4}^{0}f((-x)^2)dx+\int_{0}^{4}f(x^2)dx$

$\therefore I=2\int_{0}^{4}f(x^2)dx$

$\therefore I=2\int_{0}^{4}(4x^3-g(4-x))dx$ (from given equation)

$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{4}g(4-x)dx$

$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{2}^{4}g(4-x)dx$

$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{2}^{4}g(4-x)dx$ ... (1)

Let

$I_1=\int_{2}^{4}g(4-x)dx$

Now, consider

$(4-x)=y$ which is substituted in

$I_1$

$\therefore I_1=\int_{0}^{2}g(y)dy$ (inverting limits and

$dy$ both provide negative signs)

Substituting

$I_1$ in (1), we have

$I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}g(4-x)dx-2\int_{0}^{2}g(y)dy$

Also,

$g(4-x)=-g(x)$ from given equation and

$g(y)=g(x)$ from variable substitution

$\therefore I=2\int_{0}^{4}4x^3dx-2\int_{0}^{2}-g(x)dx-2\int_{0}^{2}g(x)dx$

$\therefore I=2\int_{0}^{4}4x^3dx$

$\therefore I=512$

This is the required answer.

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

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

Explanation

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

The value of

$\displaystyle \int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \cos ^{ 3 }{ 2x } dx }$ is:

0%
$\cfrac { 2 }{ 3 }$
0%
$\cfrac { 1 }{ 3 }$
0%
$0$
0%
$\cfrac { 2\pi }{ 3 }$

Explanation

We have to find the value of

$\displaystyle \int_{0}^{\pi/4}\cos^3 (2x)\:dx$

Consider

$\displaystyle \int_{0}^{\pi/4}\cos^3 (2x)\:dx$

Substitute

$u=2x$

$\Rightarrow du=2dx$

When

$x=0, u=0$ and when

$x=\pi/4, u=\pi/2$

$\therefore \displaystyle \int_{0}^{\pi/4}\cos ^3 (2x)\:dx=\int_{0}^{\pi/2}\cos ^3 (u) \: \dfrac{du}{2}$

$\displaystyle =\dfrac{1}{2}\int_{0}^{\pi/2}\cos ^3(u)\:du$

$\displaystyle =\dfrac{1}{2}\int_{0}^{\pi/2}\cos ^2(u)\:\cos (u)du$

$\displaystyle =\dfrac{1}{2}\int_{0}^{\pi/2}(1-\sin^2u)\:\cos \:u\:du \quad [\because \cos ^2u=1-\sin^2u]$

Substitute

$\sin\:u=v$

$\Rightarrow \cos \:u\:du=dv$

When

$u=0, v=0$ and when

$u=\pi/2, v=\sin (\pi/2)=1$

$\therefore \displaystyle \dfrac{1}{2}\int_{0}^{\pi/2}(1-\sin^2u)\:\cos \:u\:du = \dfrac{1}{2} \int_{0}^{1}(1-v^2) dv$

$\displaystyle =\dfrac{1}{2}\left[v-\dfrac{v^3}{3}\right]_{0}^{1}$

$=\dfrac{1}{2}\left[\left(1-\dfrac{1}{3}\right)-(0-0)\right]$

$=\dfrac{1}{2} \left(\dfrac{2}{3}\right)$

$=\dfrac{1}{3}$

$\therefore \displaystyle \int_{0}^{\pi/4}\cos ^3 (2x)\:dx=\dfrac{1}{3}$

$\int _{ 0 }^{ 1 }{ x{ \left[ 1-x \right] }^{ 11 } } dx=........$

0%
$-\cfrac { 1 }{ 132 }$
0%
$-\cfrac { 1 }{ 156 }$
0%
$-\cfrac { 1 }{ 121 }$
0%
$-\cfrac { 1 }{ 12 }$

Explanation

$\rightarrow I=\int _{ 0 }^{ 1 }{ x{ (1-x) }^{ 11 } } dx$
Take

$1-x$ is place of

$x$

$=\int _{ 0 }^{ 1 }{ (1-x){ \left[ 1-(1-x) \right] }^{ 11 }dx }$

$=-\int _{ 0 }^{ 1 }{ \left( 1-x \right) { x }^{ 11 } } dx=-\int _{ 0 }^{ 1 }{ \left( { x }^{ 11 }-{ x }^{ 12 } \right) } dx=-{ \left[ \cfrac { { x }^{ 12 } }{ 12 } -\cfrac { { x }^{ 13 } }{ 13 } \right] }_{ 0 }^{ 1 }=\cfrac { 1 }{ 13 } -\cfrac { 1 }{ 12 }$

$\therefore I=-\cfrac { 1 }{ 156 }$

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

The value of

$\displaystyle \int_{0}^{\dfrac {\pi}{4}} (\sqrt {\tan x} + \sqrt {\cot x})\,\, dx$ is equal to

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

Explanation

$\begin{align} y & =\int_{0}^{\dfrac{\pi}{4}} \left( \sqrt { \tan x } +\sqrt { \cot x } \right) \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { \sin x+\cos x }{ \sqrt { \sin 2x } } \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { (\sin x+\cos x)' }{ \sqrt { 1-(\sin x-\cos x)^{ 2 } } } \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { 1 }{ \sqrt { 1-u^{ 2 } } } \, du \\ & =\sqrt { 2 } \sin ^{ -1 } u \\ & =\sqrt { 2 } \sin ^{ -1 } (\sin x-\cos x) \end{align}$

Now applying limits, we get

$y=\sqrt { 2 } \sin ^{ -1 }{ \left(\sin { \dfrac { \pi }{ 4 } } -\cos { \dfrac { \pi }{ 4 } } \right) } -\sqrt{2}\sin ^{ -1 }{ (\sin { 0 } -\cos { 0 } ) } \\ =\sqrt { 2 } \times \dfrac { \pi }{ 2 } =\dfrac { \pi }{ \sqrt { 2 } }$

Hence, the answer is

$\dfrac{\pi}{\sqrt 2}$

$\int _{ a }^{ b }{ \cfrac { \log { x } }{ x } } dx=.......\quad$ (where

$a,b\in { R }^{ + }$ )

0%
$\cfrac { 1 }{ 2 } \log { \left( ab \right) } \log { \left( \cfrac { b }{ a } \right) }$
0%
$\cfrac { 1 }{ 2 } \log { \left( ab \right) }$
0%
$\log { \left( \cfrac { b }{ a } \right) }$
0%
$2\log { \left( \cfrac { b }{ a } \right) }$

Explanation

$I=\int _{ a }^{ b }{ \cfrac { \log { x } }{ x } } dx\quad (a,b+{ R }^{ + })=\int _{ a }^{ b }{ \log { x } \left( \cfrac { d }{ dx } \log { x } \right) } dx$

$=\cfrac { 1 }{ 2}{ { \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b } } \left( \therefore \int { { \left( f(x) \right) }^{ n }f'(x) } dx=\cfrac { { \left( f(x) \right) }^{ n+1 } }{ n+1 } \right)$

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

$\therefore I=\cfrac { 1 }{ 2 } \log { \left( ab \right) } \log { \left( \cfrac { b }{ a } \right) }$

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

0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 2 }$
0%
$0$
0%
$-\cfrac { \pi }{ 4 }$

Explanation

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

$=\int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ 1+2\sin { x } \cos { x } } } dx (\therefore cos^{2}{x} + sin^{2}{x} = 1)$

$= \int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ 1+\sin { 2x } } } dx$

$\cfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \cfrac { \cfrac { d }{ dx } \left( 1+\sin { 2x } \right) }{ 1+\sin { 2x } } } dx (\therefore substitute\ cos{2x} = 1+sin{2x})$

$=\cfrac { 1 }{ 2 } { \left[ \log { \left| 1+\sin { 2x } \right| } \right] }_{ 0 }^{ \pi /2 }=\cfrac { 1 }{ 2 } \left[ \log { 1 } -\log { 1 } \right] \quad$

$\therefore I=0$

Evaluate :

$\displaystyle\int^1_0\dfrac{dx}{\sqrt{5x+3}}$

0%
$\dfrac{2}{5}(\sqrt{8}-\sqrt{3})$
0%
$\dfrac{2}{5}(\sqrt{8}+\sqrt{3})$
0%
$\dfrac{2}{5}\sqrt{8}$
0%
None of these

Explanation

$I=\displaystyle\int^1_0(5x+3)^{-1/2}dx=\left[2\cdot \dfrac{(5x+3)^{1/2}}{5}\right]^1_0=\dfrac{2}{5}(\sqrt{8}-\sqrt{3})$ .

Evaluate :

$\displaystyle\int^2_0\sqrt{6x+4}dx$

0%
$\dfrac{64}{9}$
0%
$7$
0%
$\dfrac{56}{9}$
0%
$\dfrac{60}{9}$

Evaluate

$\displaystyle\int^{\pi/4}_0\tan^2xdx$

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

Solve:

$\displaystyle \int _{ 0 }^{ \pi /4 }{ \tan ^{ 100 }{ x } } dx+\int _{ 0 }^{ \pi /4 }{ \tan ^{ 102 }{ x } } dx=....$

0%
$\cfrac { 1 }{ 101 }$
0%
$\cfrac { 1 }{ 102 }$
0%
$\cfrac { 1 }{ 100 }$
0%
$101$

Explanation

$\displaystyle I=\int _{ 0 }^{ \pi /4 }{ \tan ^{ 100 }{ x } } dx+\int _{ 0 }^{ \pi /4 }{ \tan ^{ 102 }{ x } } dx$

$\displaystyle I=\int _{ 0 }^{ \pi /4 }{ \tan ^{ 100 }{ x } +\tan ^{ 102 }{ x } } dx=\int _{ 0 }^{ \pi /4 }{ \tan ^{ 100 }{ x } \left[ 1+\tan ^{ 2 }{ x } \right] } dx$

$\displaystyle I=\int _{ 0 }^{ \pi /4 }{ \tan ^{ 100 }{ x } \sec ^{ 2 }{ x } } dx=\int _{ 0 }^{ \pi /4 }{ { \left[ f(x) \right] }^{ n }f'(x) } dx$

where

$f(x)=\tan { x } \Rightarrow f'(x)=\sec ^{ 2 }{ x }$

$n=100$

$\therefore I={ \left[ \cfrac { { [f(x)] }^{ n-1 } }{ n+1 } \right] }_{ 0 }^{ \pi /4 }=\cfrac { 1 }{ n+1 } { \left\{ { \left( \tan { x } \right) }^{ n+1 } \right\} }_{ 0 }^{ \pi /4 }=\cfrac { 1 }{ n+1 } \left\{ { \left( \tan { \cfrac { \pi }{ 4 } } \right) }^{ n+1 }-{ \left( \tan { 0 } \right) }^{ n+1 } \right\}$

$=\cfrac { 1 }{ 100+1 } \left\{ { (1) }^{ 100+1 }-0 \right\} =\cfrac { 1 }{ 101 } \left( { 1 }^{ 101 }-0 \right) \quad$

$\therefore I=\cfrac { 1 }{ 101 }$

$I=\displaystyle\overset{1}{\underset{0}{\displaystyle\int}}x(1-x)^{1/2}dx$ 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%
$\dfrac{5}{2}(\sqrt{17}+\sqrt{2})$

$\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } } dx=.........$

0%
$25\pi$
0%
$\cfrac { 25\pi }{ 4 }$
0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { 25 }{ 4 }$

Explanation

$\rightarrow I=\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } } dx$

$=\int _{ 0 }^{ 5 }{ \sqrt { { (5) }^{ 2 }-{ x }^{ 2 } } } dx\quad$
Now take

$a=5$ in the

$\int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx$

$\therefore I={ \left\{ \cfrac { x\sqrt { { 5 }^{ 2 }-{ x }^{ 2 } } }{ 2 } +\cfrac { { 5 }^{ 2 } }{ 2 } \sin ^{ -1 }{ \left( \cfrac { x }{ 5 } \right) } \right\} }_{ 0 }^{ 5 }$

$=\left\{ \cfrac { 5\sqrt { { 5 }^{ 2 }-{ 5 }^{ 2 } } }{ 2 } +\cfrac { 25 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { 5 }{ 5 } \right) } \right\} -\left\{ \cfrac { 0\sqrt { { 5 }^{ 2 }-0 } }{ 2 } +\cfrac { 25 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { 0 }{ 5 } \right) } \right\}$

$=0+\cfrac { 25 }{ 2 } \left( \cfrac { \pi }{ 2 } \right) -\left\{ 0+\cfrac { 25 }{ 2 } (0) \right\} =\cfrac { 25\pi }{ 4 } -0$

$\quad I=\cfrac { 25\pi }{ 4 }$

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

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;

$\Rightarrow 2I=0$

$\Rightarrow I=0$

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

$\int _{ 0 }^{ \pi /3 }{ \dfrac { \cos { x } }{ 3+4\sin { x } } dx } =k\log { \left( \dfrac { 3+2\sqrt { 3 } }{ 3 } \right) }$ , 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%
$-{ e }^{ -x }dx\rightarrow dt$
0%
$-\int _{ 0 }^{ 1 }{ \dfrac { dt }{ t } }$
0%
$-\int _{ 0 }^{ 1 }{ ln\left| t \right| }$

Explanation

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

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

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

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

$\displaystyle \int _{ 2 }^{ x }{ { \left( { e }^{ x }-1 \right) }^{ -1 }dx=\left( \dfrac { 3 }{ 2 } \right) }$ then

$x=$

0%
$\ln 4$
0%
$1$
0%
${e}^{2}$
0%
$\dfrac{1}{e}$

Explanation

According to,

$\begin{array}{l} { \int _{ \log 2 }^{ x }{ ({ e^{ x } }-1) }^{ -1 } }dx=\, \log \, \left( { \frac { 3 }{ 2 } } \right) \\ \int _{ \log 2 }^{ x }{ \frac { 1 }{ { ({ e^{ x } }-1) } } }dx=\, \log \, \left( { \frac { 3 }{ 2 } } \right) \\ \int _{ \log 2 }^{ x }{ \frac { { { e^{ -x } } } }{ { (1-{ e^{ -x } }) } } }dx=\, \log \, \left( { \frac { 3 }{ 2 } } \right) \\ \left[ { \log (1-{ e^{ -x } }) } \right] _{ \log 2 }^{ x }=\log \, \left( { \frac { 3 }{ 2 } } \right) \\ \log (1-{ e^{ -x } })-log\frac { 1 }{ 2 } =\log \, \left( { \frac { 3 }{ 2 } } \right) \\ \log (1-{ e^{ -x } })=\log \left( { \frac { 3 }{ 4 } } \right) \\ (1-{ e^{ -x } })=\left( { \frac { 3 }{ 4 } } \right) \\ { e^{ -x } }=\frac { 1 }{ 4 } \\ { e^{ x } }=4={ 2^{ 2 } } \\ \therefore \, \, x=2\, \, then\, \, ,\, { e^{ 2 } } \end{array}$

Sothe correct option is C.

Integrate

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

0%
$4$
0%
$6$
0%
$8$
0%
None of these

Explanation

$\int_0^{2\pi } {\sqrt {1 + \sin x} } dx$

$= \int_0^{2\pi } {\sqrt {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos } \frac{x}{2}} dx$ (

${\therefore {{\sin }^2}x + {{\cos }^2}x = 1}$ and

$\sin 2x = 2\sin x\cos x$ )

$= \int_0^{2\pi } {\sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} dx}$

$= \int_0^{2\pi } {\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)} dx$

$= \left[ { - 2\cos \frac{x}{2} + 2\sin \frac{x}{2}} \right]_0^{2\pi }$

$= 2\left[ {\sin \frac{x}{2} - \cos \frac{x}{2}} \right]_0^{2\pi }$

$= 2\left[ {1 + 1} \right]$

$= 4$

Evaluate the following integral

$\int _{ 0 }^{ \infty }{ \dfrac { dx }{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) \left( { x }^{ 2 }+{ b }^{ 2 } \right) } } =$

0%
$\dfrac {\pi (a+b)}{ab}$
0%
$\dfrac {\pi ab}{(a+b)}$
0%
$\dfrac {\pi}{2(a+b)}$
0%
$\dfrac {\pi}{2ab(a+b)}$

Explanation

$\displaystyle \int_{0}^{\infty }\frac{dx}{(x^{2}+a^{2})(x^{2}+b^{2})}$

$\displaystyle\Rightarrow \frac{1}{(a^{2}-b^{2})} \int_{0}^{\infty} \frac{(a^{2}-b^{2})dx}{(x^{2}+a^{2})(x^{2}+b^{2})}$

$\displaystyle \Rightarrow \frac{1}{(a^{2}-b^{2})} \int_{0}^{\infty}\left [ \frac{(x^{2}+a^{2})-(x^{2}+b^{2})}{(x^{2}+a^{2})(x^{2}+b^{2})} \right ]dx$

$\displaystyle = \frac{1}{(a^{2}-b^{2})} \left [ \int_{0}^{\infty} \frac{dx}{(x^{2}+b^{2})} -\int_{0}^{\infty} \dfrac{dx}{(x^{2}+a^{2})}\right ]$

$\displaystyle = \dfrac{1}{(a^{2}-b^{2})} \left [ \frac{1}{b} tan^{-1}\frac{x}{b}-\frac{1}{a} tan^{-1}\frac{x}{a} \right ]_{0}^{\infty }$

$\displaystyle = \frac{1}{(a^{2}-b^{2})} \left [ \frac{1}{b}tan^{-1}\infty - \frac{1}{a}tan^{-1} \infty ] -(0-0) \right ]$

$\displaystyle = \dfrac{1}{(a^{2}-b^{2})} \left [ \dfrac{1}{b} \times \dfrac{\pi}{2}-\dfrac{1}{a} \times \dfrac{\pi}{2} \right ]$

$= \dfrac{\pi}{2(a-b)(a+b)} \left [ \dfrac{(a-b)}{ab} \right ]$

$= \dfrac{\pi}{2ab (a+b)}$

1115615_1036091_ans_a71ae45f1f484418a2558057a3e846f2.jpg

$\displaystyle \int_{0}^{40} \dfrac{dx}{2x+1} =\log a,$ then

$a$ is:

0%
$3$
0%
$9$
0%
$81$
0%
$40$

Explanation

$\displaystyle \int_{0}^{40} \dfrac{dx}{2x+1} =\left[\dfrac{\log(2x+1)}{2}\right]^{40}_0=\dfrac12 \log81=\log81^{1/2}=\log9$

Hence, Value of a is 9

$\displaystyle \int_{2}^{4} \dfrac{\sqrt{x^{2}-4}}{x^{4}}dx=$

0%
$\dfrac{3}{32}$
0%
$\dfrac{\sqrt{3}}{32}$
0%
$\dfrac{3}{8}$
0%
$\dfrac{\sqrt{3}}{8}$

Explanation

$\int_2^4 {\cfrac{{\sqrt {{x^2} - 4} }}{4}}$

$\int_2^4 {\cfrac{{\sqrt {1 - \cfrac{4}{{{x^2}}}dx} }}{{{x^4}}}}$

$\int_2^4 {\cfrac{{\sqrt {1 - \cfrac{4}{{{x^2}}}dx} }}{{{x^3}}}}$

putting

$1 - \cfrac{4}{{{x^2}}} = t$

$- 4 \times \cfrac{{ - 2}}{{{x^3}}}dx = dt$

$\cfrac{{dx}}{{{x^3}}} = \cfrac{{dt}}{8}$

$\int_0^{\cfrac{3}{4}} {\cfrac{{\sqrt t }}{8} \times dt}$

$= \cfrac{1}{8} \times \cfrac{2}{3}{\left[ {{t^{\cfrac{3}{2}}}} \right]^\cfrac{3}{4}}$

$= \cfrac{1}{{12}}\left[ {{{\left( {\cfrac{3}{4}} \right)}^\cfrac{3}{2}} - 0} \right]$

$= \cfrac{1}{{12}}\left[ {{{\left( {\cfrac{{\sqrt 3 }}{2}} \right)}^3}} \right]$

$= \cfrac{1}{{12}} \times \cfrac{{3\sqrt 3 }}{8}$

$= \cfrac{{\sqrt 3 }}{{32}}$

Integrate

$\int_{0}^{1}\dfrac{\sqrt x}{1+x^3}$

0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {\pi}{2}$

Explanation

$\int_{0}^{1} \dfrac{\sqrt{x}}{1+^{3}} dx$

Let

$x^{3/2} = t$

$\dfrac{3}{2}x^{1/2}dn = dt$

$\sqrt{x}dn = \dfrac{2}{3}dt$

$= \dfrac{2}{3} \int_{0}^{1} \dfrac{dt}{1+t^{2}}$

$= \dfrac{2}{3}\int_{0}^{1}\dfrac{dt}{1+t^{2}}$

$= \dfrac{2}{3}\left [ tan^{-1}t \right ]_{0}^{1} = \dfrac{2}{3}[tan^{-1}-tan^{-1}(0)]$

$= \dfrac{2}{3} \times \dfrac{\pi}{2} = \dfrac{\pi}{6}$

$= \boxed{\dfrac{\pi}{6}}$

So, option (C) is right.

1115609_1036088_ans_5153178945c64810883e731a5c5f2221.jpg

The value of

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

0%
$\dfrac 12$
0%
$\dfrac 34$
0%
$2$
0%
$None\ of\ these$

Explanation

$I=\displaystyle \int_0^{\pi/2}\sin x\cos xdx$

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

$x\to 0\to \dfrac \pi 2\\t\to 0\to 1$

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

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

$I=\dfrac 12-0=\dfrac 12$

$\displaystyle \int _{ 0 }^{ 2x }{ \sqrt { 1+\sin { x } } } dx$

0%
$\sin\dfrac{x}{2}-\cos\dfrac{x}{2}-\dfrac{\pi}{2}+C$
0%
$\sin\dfrac{x}{2}-\cos\dfrac{x}{2}+\dfrac{\pi}{2}+C$
0%
$2\sin x - 2\cos x + 2$
0%
None of these

Explanation

$\int_0^{2x} {\sqrt {1 + \sin x} dx}$

$= \int_0^{2x} {\sqrt {{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} } dx$ (since

${\sin ^2}x + {\cos ^2}x = 1$ )

$= \int_0^{2x} {\sqrt {{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}} } dx$

$= \int_0^{2x} {\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)} dx$

$= \left[ { - 2\cos \dfrac{x}{2} + 2\sin \dfrac{x}{2}} \right]_0^{2x}$

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

$= 2\left[ {\sin x - \cos x + 1} \right]$

$= 2\sin x - 2\cos x + 2$

$\displaystyle \int_{0}^{1}{\dfrac{ln(1+x)}{1+{x}^{2}}}dx=$

0%
$\dfrac{\pi}{4}ln{2}$
0%
$\dfrac{\pi}{2}ln{2}$
0%
$\dfrac{\pi}{8}ln{2}$
0%
$\pi\ ln{2}$

Explanation

$\displaystyle \int_{0}^{1}{\dfrac{\ln(1+x)}{1+{x}^{2}}}dx=$

$\Rightarrow \dfrac{-i\left(-\ln(1-i))\ln(|x+i|)+\ln(i+1)\ln(|x-i|)+li_2\left(\dfrac{x+i}{x-1}\right)-li_2\left(\dfrac{-x+i}{1+i}\right)\right)+c}{2}$

$\therefore arc\tan(x)\ln(x+1)-arc\tan|(x)\ln\left(\dfrac{x^2+2x+1}{2}\right)-\tan 2\left(\dfrac{x+1}{2},\dfrac{x+1}{2}\right)$

$\dfrac{\ln(x^2+1)-ili_2\left(l\dfrac{i(i+1)x-i+1}{2}\right)+ili_2\left(\dfrac{i(i+1)x-i-1}{2}\right)^{+c}}{2}$

$\dfrac{\therefore -ili_2\left(\dfrac{i+1}{2}\right)-ili_2\left(\dfrac{-i-1}{2}\right)+\dfrac{ili2(-1)}{2}-\dfrac{-i(li2-1)}{2}+\dfrac{\pi \ln(2)}{4}}{2}$

$=\dfrac{\pi \ln(2)}{4}$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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