MCQ Questions for Class 12 Commerce Maths Integrals Quiz 10

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%
$none\ of\ these$

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

Evaluate :

$\int { \cfrac { dx }{ x\cos { x } } }$

0%
$\ln { x } +\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { 3x }^{ 3 } }{ 3 } +...$
0%
$\ln { x } -\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 4x }^{ 4 } }{ 16 } +...$
0%
$\ln { x } +\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 5x }^{ 4 } }{ 96 } +...$
0%
$\ln { x } -\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { x }^{ 4 } }{ 9 } +...$

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

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

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

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

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

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

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

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

$\int_{}^{} {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx}$ would be equal to

0%
$\frac{{\sin x + x\cos x}}{{x\sin x + \cos x}} + c$
0%
$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}} + c$
0%
$\frac{{\sin x - x\cos x}}{{x\sin x - \cos x}} + c$
0%
none of these

$\displaystyle \int_{}^{} {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx}$ is equal to :

0%
$- {e^x}\tan \frac{x}{2} + c$
0%
$- {e^x}\cot \frac{x}{2} + c$
0%
$- \frac{1}{2}{e^x}\tan \frac{x}{2} + c$
0%
$- \frac{1}{2}{e^x}\cot \frac{x}{2} + c$

Explanation

Let

$I=\displaystyle\int{{e}^{x}\dfrac{1-\sin{x}}{1-\cos{x}}dx}$

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

$=\displaystyle\int{{e}^{x}\dfrac{1-2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}}{2{\sin}^{2}{\dfrac{x}{2}}}dx}$

$=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$

Let

$f\left(x\right)=-\cot{\dfrac{x}{2}}$

$\Rightarrow\,{f}^{\prime}{\left(x\right)}=\dfrac{1}{2}{\csc}^{2}{\dfrac{x}{2}}$

$\Rightarrow\,I=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$

Using

$\displaystyle\int{{e}^{x}\left(f\left(x\right)+{f}^{\prime}{\left(x\right)}\right)dx}={e}^{x}f\left(x\right)+C$

$\Rightarrow\,I=-{e}^{x}\cot{\dfrac{x}{2}}+c$

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

$\cfrac { \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x } \right) }^{ \sqrt { 3 } +1 } } dx }{ \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x } \right) }^{ \sqrt { 3 } -1 } } }$ is

0%
$\cfrac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 }$
0%
$\cfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 }$
0%
$\cfrac { \sqrt { 3 } +1 }{ \sqrt { 3 } }$
0%
none of these

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%
$0$

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

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

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

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

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

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

$\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^{2}}}dx$ is equal to

0%
$\frac{1}{2}+\frac{\pi }{2\sqrt{3}}$
0%
$\frac{1}{2}-\frac{\pi }{2\sqrt{3}}$
0%
$\frac{1}{2}+\frac{\pi }{4\sqrt{3}}$
0%
$\frac{1}{2}-\frac{\pi }{4\sqrt{3}}$

The integral

$\int { \dfrac { { sin }^{ 2 }\times { cos }^{ 2 }\times }{ { (sin }^{ 5 }\times +{ cos }^{ 3 }\times { sin }^{ 2 }\times +{ sin }^{ 3 }\times { cos }^{ 2 }\times +{ cos }^{ 5 }{ \times ) }^{ 2 } } }$ dx is equal to:

0%
$\dfrac { 1 }{ { 1+cos }^{ 3 }x } +c$
0%
$\dfrac { -1 }{ { 1+cos }^{ 3 }x } +c$
0%
$\dfrac { 1 }{ 3\left( 1+{ tan }^{ 3 }x \right) } +c$
0%
$\dfrac { -1 }{ 3\left( 1+{ tan }^{ 3 }x \right) } +c$

The value of

$\int_{0}^{[x]} (x-[x])dx$ , where

$[x]$ is the greatest integer

$|le x$ is equal to

0%
$4[x]$
0%
$2[x]$
0%
$\dfrac{1}{2} [x]$
0%
$\dfrac{1}{5} [x]$

The value of the definite integral

$\int _{ 2 }^{ 3 }{ \left[ \sqrt { 2x-\sqrt { 5\left( 4x-5 \right) } } +\sqrt { 2x+\sqrt { 5\left( 4x-5 \right) } } \right] } dx=$

0%
$\dfrac {7\sqrt {3}+3\sqrt {5}}{3\sqrt {2}}$
0%
$4\sqrt {2}$
0%
$4\sqrt {3}+\dfrac {4}{3}$
0%
$\dfrac {7\sqrt {7}-2\sqrt {5}}{3\sqrt {2}}$

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%
$5$

The integral

$\displaystyle \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}{\dfrac{8\cos 2x}{(\tan x+\cot x)^{3}}dx}$ equals :

0%
$\dfrac{15}{128}$
0%
$\dfrac{15}{64}$
0%
$\dfrac{13}{32}$
0%
$\dfrac{13}{256}$

Let

$f:R\longrightarrow R,g : R\longrightarrow R$ be continuous functions. then the value of integeral.

$\int _{ \ell n\lambda }^{ \ell n/\lambda }{ \frac { f\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ f\left( x \right) -f\left( -x \right) \right] }{ g\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ g\left( x \right) +g\left( -x \right) \right] } } dx$ is:

0%
depend on $\lambda$
0%
a non-zero constant
0%
zero
0%
1

$P=\displaystyle \lim _{ n\rightarrow \infty }{ \frac { { \left( \prod _{ r=1 }^{ n }{ \left( { n }^{ 3 }+{ r }^{ 3 } \right) } \right) }^{ 1/n } }{ { n }^{ 3 } } }$ and

$\lambda =\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 3 } } }$ then

$In P$ is equal to

0%
$In 2-1+\lambda$
0%
$In 2-3+3\lambda$
0%
$2In 2-\lambda$
0%
$In 4-3+3\lambda$

The value of

$\displaystyle \int _{0}^{\infty}\dfrac{\sqrt{x^2+1}}{(x+\sqrt{x^{2}+1})^{n+1}} .dx \ \forall \ n\ \in \ N- \{ \pm 1 \}$ is

0%
$0$
0%
$n(n^{2}-1)$
0%
$\dfrac{n}{(n^{2}-1)}$
0%
$n^{2}$

$\displaystyle \int_{-3\pi}^{3\pi}{\sin^{2}\theta\sin^{2} 2\theta d \theta}$ is equal to-

0%
$\pi$
0%
$\dfrac{3\pi}{2}$
0%
$\dfrac{5\pi}{2}$
0%
$6\pi$

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

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

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

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

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

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

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

Value of the definite integral

$\displaystyle \int_{-1}^{1}\dfrac {dx}{(1+x^{3}+\sqrt {1+x^{6}})}$

0%
$is\dfrac {1}{2}$
0%
$is\ 1$
0%
$is\ 2$
0%
$is\ zero$

The value of

$\displaystyle\int _{ 5 }^{ 10 }{ \left( \sqrt { x+\sqrt { 20x-100 } } +\sqrt { x-\sqrt { 20x-100 } } \right) } dx$ is

0%
$10\sqrt{5}$
0%
$5\sqrt{5}$
0%
$10\sqrt{2}$
0%
$8\sqrt{2}$

Let

$f(x)= \left| \begin{matrix} \sec { x } & \cos { x } & \sec { ^{ 2 }x+\cot { x\csc { x } } } \\ \cos { ^{ 2 }x } & \cos { ^{ 2 }x } & \csc { ^{ 2 }x } \\ 1 & \cos { ^{ 2 }x } & \cos { ^{ 2 }x } \end{matrix} \right|$ . Then

${\int}_{0}^{\pi/2}f(x)dx$ is equal to

0%
$\dfrac{15\pi}{60}$
0%
$\dfrac{15\pi+32}{60}$
0%
$\dfrac{15\pi-32}{60}$
0%
$none\ of\ these$

$\displaystyle \int_{0}^{\pi/2}{(\sin x-\cos x).\log(\sin x+\cos x)dx}$ =

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

Let

$f\left(x\right)+f\left(\dfrac{1}{x}\right)=F\left(x\right)$ where

$f\left(x\right)=\displaystyle\int_{1}^{x}{\dfrac{\ln{t}}{1+t}dx}$ .Then

$F\left(e\right)=$

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

Evaluate:

$\displaystyle\int_{-2}^{3}{\left|1-{x}^{2}\right|dx}$

0%
$\dfrac{28}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{5}{3}$

${ I }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ \cos(\sin x)dx, { I }_{ 2 }= } \int _{ 0 }^{ \pi /2 }{ \sin(\cos x)dx }$ and

${ I }_{ 3 }=\int _{ 0 }^{ \pi /2 }{ \cos xdx }$ , then

0%
${ I }_{ 1 }>{ I }_{ 2 }>{ I }_{ 3 }$
0%
${ I }_{ 3 }>{ I }_{ 2 }>{ I }_{ 1 }$
0%
${ I }_{ 3 }>{ I }_{ 1 }>{ I }_{ 2 }$
0%
${ I }_{ 1 }>{ I }_{ 3 }>{ I }_{ 2 }$

$\displaystyle \int_{2}^{8}\dfrac {\sqrt {10-x}}{\sqrt {x}+\sqrt {10-x}}dx$ is

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

${\int}_{0}^{\pi}\dfrac{\sqrt{1-x}}{1+x}dx=$

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

$\int _ { 0 } ^ { \pi / 4 } \tan ^ { 2 } x d x$ equals -

0%
$\pi / 4$
0%
$1 + ( \pi / 4 )$
0%
$1 - ( \pi / 4 )$
0%
$1 - ( \pi / 2 )$

Evaluate

$\int \dfrac{dx}{\sqrt{1-x}}$

0%
$\sin^{-1} \sqrt{x}$
0%
$-\sin^{-1} \sqrt{x}+c$
0%
$2\sqrt{1-x}+c$
0%
$-2\sqrt{1-x}+c$

$\displaystyle \overset{2\pi}{\underset{0}{\int}} x \,log \left(\dfrac{3 + \cos x}{3 - \cos x}\right)dx$

0%
$\dfrac{\pi}{2} \,log \,3$
0%
$\dfrac{\pi}{12} \,log \,3$
0%
$\dfrac{\pi}{3} \,log \,3$
0%
$0$

The smallest interval in which value of

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

0%
$[0,1]$
0%
$[0,\dfrac{1}{2}]$
0%
$[0,\dfrac{1}{4}]$
0%
$[0,\dfrac{1}{8}]$

$\displaystyle \overset{3}{\underset{0}{\int}} \dfrac{dx}{\sqrt{5 - x^2}}$

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{2}$
0%
$-\dfrac{\pi}{2}$
0%
$-\dfrac{\pi}{6}$

$\displaystyle\int _{ 0 }^{ \infty }{ \dfrac { dx }{ \left( x+\sqrt { { x }^{ 2 }+1 } \right) ^{ 3 } } } =$

0%
$\dfrac{3}{8}$
0%
$\dfrac{1}{8}$
0%
$-\dfrac{3}{8}$
0%
$None\ of\ these$

If I =

$\overset { 2 }{ \underset { -3 }{ \int } } (|x + 1| + |x + 2| +|x -1|) dx$ , then i equals

0%
$\dfrac{31}{2}$
0%
$\dfrac{35}{2}$
0%
$\dfrac{47}{2}$
0%
$\dfrac{39}{2}$

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

$I$ equals

0%
$\dfrac{1}{2}\log\dfrac{5}{3}$
0%
$2\log\dfrac{1}{3}$
0%
$\dfrac{1}{2}\log\dfrac{1}{5}$
0%
$2\log\dfrac{5}{3}$

The floor value of integral

$\displaystyle \int_\dfrac{\pi }{4}^{3\pi }\dfrac{x}{1+4x}dx$ is

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

Solve :-

$\int_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}} = }$

0%
$1$
0%
$\frac{1}{{\sqrt 2 }}$
0%
$\frac{1}{2}$
0%
$\sqrt 2$

Value of

$\displaystyle \int_{1}^{5} \left(\sqrt {x+2\sqrt {x-1}}+\sqrt {x-2(x-1)}\right)dx$ is

0%
$\dfrac {8}{3}$
0%
$\dfrac {16}{3}$
0%
$\dfrac {32}{3}$
0%
$\dfrac {34}{3}$

Let

$[\cdot]$ denote the greatest integer function then the value of

$\displaystyle\int^{1.5}_0x[x^2]dx$ is?

0%
$0$
0%
$\dfrac{3}{2}$
0%
$\dfrac{7}{4}$
0%
$\dfrac{5}{4}$

$\int _ { 0 } ^ { 2 } \cfrac { \sqrt { x } } { \sqrt { x } + \sqrt { 2 - x } }$ is equal to

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

$\int _{ 0 }^{ \infty }{ \frac { { x }^{ 2 }+1 }{ { x }^{ 4 }+{ 7x }^{ 2 }+1 } }$ dx=

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

The solution of

$x$ of the equation

$\displaystyle \int_{\sqrt{2}}^{x}{\dfrac{dt}{t\sqrt{t^{2}-1}}}=\dfrac{\pi}{2}$ is

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

If the value of the definite integral

$\int _{ \dfrac { \pi }{ 6 } }^{ \dfrac { \pi }{ 4 } }{ \dfrac { 1+\cot { x } }{ { e }^{ x }sin{ x } } dx }$ , is equal to

${ ae }^{ -\pi /6 }+{ be }^{ -\pi /4 }$ then

$(a+b)$ equals

0%
$2-\sqrt { 2 }$
0%
$2+\sqrt { 2 }$
0%
$2\sqrt { 2 } -2$
0%
$2\sqrt { 3 } -\sqrt { 2 }$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.