MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 7

$\displaystyle \int _{ 0 }^{ \pi /4 }{ \log { \left( 1+\tan ^{ 2 }{ \theta } +2\tan { \theta } \right) d\theta = } }$

0%
$\pi \log 2$
0%
$(\pi \log 2)/2$
0%
$(\pi \log 2)/4$
0%
$\log 2$

$\displaystyle\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx = K{{\sin }^{ - 1}}} ({2^x}) + C,$ then K is equal to

0%
$\ell n2$
0%
$\displaystyle{1 \over 2}\ell n2$
0%
$\displaystyle{1 \over 2}$
0%
$\displaystyle{1 \over {\ell n2}}$

Explanation

$I=\displaystyle\int \dfrac{2^x}{\sqrt{1-4^x}}dx$

Let

$2^x=z$

$\Rightarrow 2^x\log 2dx=dz$

$\Rightarrow I=\displaystyle\int \dfrac{\dfrac{1}{\log2}}{\sqrt{1-z^2}}dz$

$=\dfrac{1}{\log 2}\cdot \sin^{-1}z+c$

$=\dfrac{1}{\log 2}\cdot \sin^{-1}(e^x)+c$

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

then

$k=\dfrac{1}{\log 2}$ .

The value of the defined integral

$\displaystyle \int^{\pi/2}_{0}(\sin x+\cos x)\sqrt {\dfrac {e^{x}}{\sin x}}dx$ equals

0%
$2\sqrt {e^{\pi/2}}$
0%
$\sqrt {e^{\pi/2}}$
0%
$2\sqrt {e^{\pi/2}}.\cos 1$
0%
$\dfrac {1}{2}e^{\pi/4}$

Evaluate

$\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$
0%
$\pi$
0%
None of these

Explanation

Let

$I=\displaystyle\int^{\pi/2}_0\dfrac{\cos x}{(1+\sin^2x)}dx$

Put

$\sin x=t$ and

$\cos xdx=dt$ .

$[x=0\Rightarrow t=0]$ and

$\left[x=\dfrac{\pi}{2}\Rightarrow t=1\right]$ .

$\therefore I=\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$

$=[\tan^{-1}t]^1_0$ .......

$\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C$

$=\dfrac{\pi}{4}-0\\$ .

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

Evaluate

$\displaystyle\int^{\sqrt{8}}_{\sqrt{3}}x\sqrt{1+x^2}dx$

0%
$\dfrac{19}{3}$
0%
$\dfrac{19}{6}$
0%
$\dfrac{38}{3}$
0%
$\dfrac{9}{4}$

Evaluate

$\displaystyle\int^{\pi/2}_0\cos^3xdx$

0%
$1$
0%
$\dfrac{3}{4}$
0%
$\dfrac{2}{3}$
0%
None of these

Evaluate :

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

0%
$\dfrac{\pi}{2}$
0%
$\left(\dfrac{\pi}{2}-1\right)$
0%
$\left(\dfrac{\pi}{2}+1\right)$
0%
None of these

Evaluate:

$\int _{ 1/3 }^{ 1 }{ \cfrac { { \left( x-{ x }^{ 3 } \right) }^{ 1/3 } }{ { x }^{ 4 } } } dx=$

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

Explanation

$I=\displaystyle\int^1_{1/3}\dfrac{x^{1/3}(1-x^2)^{1/3}}{x^4}dx$

Let

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

$I=\displaystyle\int^{\pi/2}_{\sin^{-1}1/3}\dfrac{(\sin\theta)^{1/3}(\cos^{2/3}\theta)}{\sin^4\theta}\times \cos\theta d\theta$

$=\displaystyle\int \dfrac{\cos^{5/3}\theta}{\sin^{11/3}\theta}d\theta \times \dfrac{\cos^2\theta}{\cos^2\theta}$

$=\displaystyle\int \dfrac{\cos^{11/3}\theta}{\sin^{11/3}\theta\cos^2\theta}d\theta$

$=\displaystyle\cot^{11/3}\theta .\sec^2\theta d\theta$

Let

$\tan\theta =t$

$\Rightarrow \sec^2\theta d\theta =dt$

$=\displaystyle\int t^{-11/3}dt$

$=\dfrac{t^{-8/3}}{(-8/3)}=\dfrac{-3}{8}t^{-8/3}$

$=\left.\dfrac{-3}{8}(\tan\theta)^{-8/3}\right|^{\pi/2}_{\sin^{-11}1/3}$

$\sin^{-11/3}=\tan^{-1}\dfrac{1}{2\sqrt{2}}$

$\Rightarrow I=\dfrac{-3}{8}\left(\dfrac{1}{(\infty)^{8/3}}-\left(\dfrac{1}{2\sqrt{2}}\right)^{-8/3}\right)$

$=\dfrac{3}{8}\times (2\sqrt{2})^{8/3}=\dfrac{3}{8}(2^{3/2})^{8/3}$

$=\dfrac{3}{8}\times 16=6\Rightarrow (C)$ .

Evaluate

$\displaystyle \int _ { 0 } ^ { a } \dfrac { x d x } { \sqrt { a ^ { 2 } + x ^ { 2 } } }$

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

Explanation

$\displaystyle \int _{ 0 }^{ a }{ \cfrac { xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } }$

Let

$t=\left( { a }^{ 2 }+{ x }^{ 2 } \right) \Rightarrow dt=2xdx$

when

$x=0,t={ a }^{ 2 }\quad$

when

$x=a,t=2{ a }^{ 2 }$

$\displaystyle =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { 2xdx }{ \sqrt { { a }^{ 2 }+{ x }^{ 2 } } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ \cfrac { dt }{ \sqrt { t } } } =\cfrac { 1 }{ 2 } \int _{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }{ { t }^{ -1/2 }dt }$

$=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { { t }^{ -1/2+1 } }{ -\cfrac { 1 }{ 2 } +1 } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 1 }{ 2 } \quad { \left[ \cfrac { \sqrt { t } }{ \cfrac { 1 }{ 2 } } \right] }_{ { a }^{ 2 } }^{ 2{ a }^{ 2 } }=\cfrac { 2 }{ 2 } \left[ \sqrt { 2{ a }^{ 2 } } -\sqrt { { a }^{ 2 } } \right]$

$=a\left( \sqrt { 2 } -1 \right)$

The correct evaluation of

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

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

Explanation

$I=\displaystyle \int_0^{\pi/2} \sin x \sin 2x \, dx$

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

$u = \sin x \Rightarrow du = \cos x\ dx$

$= \left.\begin{matrix}\displaystyle \int_0^1 2u^2 du = 2 \dfrac{u^3}{3}\end{matrix}\right|^1_0= \dfrac{2}{3}$

$\int { { e }^{ x^{ 3 } }+{ x }^{ 2-1 }(3{ x }^{ 4 }+{ 2x }^{ 3 }+{ 2x }^{ 2 }\quad x=h(x)+c }$ then the value of

$h(1)h(-1)$ .

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

$\int _{ 0 }^{ \pi }{ \cfrac { { x }^{ 2 } }{ { \left( 1+sinx \right) }^{ 2 } } } dx$ equals

0%
$\pi (\pi -2)$
0%
$\pi ^2 (\pi -2)$
0%
$\pi (2-\pi )$
0%
none of these

The value of the integral

$\displaystyle \int_{-\pi/2}^{\pi/2} \left(x^{2}+\log \dfrac{\pi-x}{\pi+x}\right) \cos x dx$ is

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

Explanation

$I=\displaystyle \int_{-\pi/2}^{\pi/2}\left[x^2+log\left( \dfrac{\pi-x}{\pi+x}\right)\right]cosxdx$

As,

$\displaystyle \int_{-a}^{a}f(x)dx=0,$ when

$f(-x)=-f(x)$

$\therefore I=\displaystyle \int_{-\pi/2}^{\pi/2}x^2cosxdx+0=2\displaystyle \int_{0}^{\pi/2}(x^2cosx)dx$

$=2\left\{ (x^2sinx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}(x^3cosx) \right\}$

$2\left[ \dfrac{\pi^2}{4}-2\left\{(-xcosx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}1.(-cosx)dx \right\}\right]$

$=2\left[ \dfrac{\pi^2}{4}-2(sinx)_{0}^{\pi/2}\right]=2\left[\dfrac{\pi^2}{4}-2\right]=\left( \dfrac{\pi^2}{2}-4\right)$

Let

$\theta$ be the angle between the lines

${ L }_{ { 1 } }:\left[ \begin{array}{l} { { x }=2{ t }+{ 1 } } \\ { { y }={ t }+{ 1 } } \\ { { z }=3{ t }+{ 1 } } \end{array} \right.$ and

${ L }_{ { 2 } }:\left[ \begin{array}{l} { { x }=3{ s }+2 } \\ { { y }=6{ s }-1 } \\ { { z }=4 } \end{array} \right.$ where

$s , t \in { R }.$ Then the value of

$\int _ { 0 } ^ { \theta } \dfrac { 1 } { 1 + \tan x } d x =$

0%
$\pi / 6$
0%
$\pi / 4$
0%
$\pi / 2$
0%
$\pi / 3$

Let

${ I }_{ 1 }=\displaystyle \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 50 } \right) }^{ 100 }dx }$ and

${ I }_{ 2 }=\displaystyle \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 50 } \right) }^{ 101 }dx }$ , then

$\dfrac { { I }_{ 1 } }{ { I }_{ 2 } } =$

0%
$\dfrac {5051}{5050}$
0%
$\dfrac {5051}{5049}$
0%
$\dfrac {51}{50}$
0%
$\dfrac {101}{100}$

$\displaystyle\int _{ -9 }^{ 9 }{ \log { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } dx }$ equal

0%
$2\log (9^{2}+1)$
0%
$2\log (\sqrt{9^{2}+1}-9)$
0%
$0$
0%
$2\log (9+\sqrt{9^{2}+1})$

$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\tan \theta}{\sqrt{2k \sec \theta}}d\theta=1-\dfrac{1}{\sqrt{2}},(k>0),$ then the value of k is :

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

Explanation

$LHS= \dfrac{1}{\sqrt{2k}}\displaystyle\int_0^{\dfrac{\pi}{3}} \dfrac{\tan \theta}{\sqrt{\sec \theta}}d\theta$

$=\dfrac{1}{\sqrt{2k}}\displaystyle \int_0^{\dfrac{\pi}{3}}\dfrac{\sin \theta}{\sqrt{\cos \theta}}d\theta$

$=-\dfrac{1}{\sqrt{2k}}|2\sqrt{\cos \theta}|_0^{\dfrac{\pi}{3}}$

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

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

$\because$ it is given that

$RHS = 1-\dfrac{1}{\sqrt{2}}$

$\therefore$ by comparing we get

$k=2$

The value of

$\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$ is equal to?

0%
$2\pi$
0%
$\pi^2$
0%
$2\pi^2$
0%
$4\pi$

Explanation

Let

$I = \displaystyle \int_0^{2 \pi} \dfrac{x \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(1)

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 (2\pi - x)}{\sin^8 (2 \pi - x) + \cos^8 (2 \pi - x)}$

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(2)

Add (1) + (2) we get,

$2I = \displaystyle \int_0^{2 \pi} \dfrac{2 \pi \sin^8 x}{\sin^8 x + \cos^8 x}$

$2I = 2 \times 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x}$

$I = 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ __(3)

$I = 4\pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 \left(\dfrac{\pi}{2} - x \right)}{\sin^8 \left(\dfrac{\pi}{2} - x \right) + \cos^8 \left(\dfrac{\pi}{2} - x \right)}$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\cos^8 (x)}{\sin^8 x + \cos^8 x}$ __(4)

add (3) + (4) we get

$I = 2 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx = 2\pi \displaystyle \int_0^{\pi/2} 1 dx = 2 \pi \times \dfrac{\pi}{2} = \pi^2$

$f(a-x)=-f(x)$ , then

$\displaystyle \int_{0}^{a}f(x)dx=0$ .

0%
True
0%
False

Explanation

$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$\displaystyle\int_{a}^{a}{f\left(x\right)dx}=I+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

Now

$I=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}$

Put

$x=-t$

$\Rightarrow dx=-dt$

When

$x=-a,\,t=a$ and when

$x=0,\,t=0$

$I=\displaystyle\int_{a}^{0}{f\left(-t\right)\left(-dt\right)}$

$=-\displaystyle\int_{a}^{0}{f\left(-t\right)dt}$

$=\displaystyle\int_{0}^{a}{f\left(-t\right)dt}...........$ since

$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(x\right)dx}$

$=\displaystyle\int_{0}^{a}{f\left(-x\right)dt}..........$ since

$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(t\right)dt}$

Thus,

$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{f\left(-x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$=\displaystyle\int_{0}^{a}{\left[f\left(-x\right)+f\left(x\right)\right]dx}$

$=2\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$f\left(x\right)$ is an odd function, then

$f\left(-x\right)=-f\left(x\right)$

$\Rightarrow \displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{\left[-f\left(x\right)+f\left(x\right)\right]dx}=0$

Hence proved.

Thus, the given statement is true.

$\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos { 2x } } }$

0%
$200\sqrt 2$
0%
$400\sqrt 2$
0%
$800\sqrt 2$
0%
$none$

Explanation

$I=\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos x } dx } \\ =400\int _{ 0 }^{ \pi }{ \sqrt { 1-\cos 2x } dx } \\ =800\int _{ 0 }^{ \pi }{ \sqrt { 2 } \sqrt { { { \sin }^{ 2 } }x } dx } \\ =800\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin xdx } \\ =800\sqrt { 2 } \left[ { \cos x } \right] _{ \frac { \pi }{ 2 } }^{ 0 } \\ =800\sqrt { 2 }$

$\\ Hence,\, the\, option\, C\, is\, the\, correct\, answer.$

$\displaystyle \int_{0}^{1}\sin^{-1}x dx=\dfrac {\pi}{2}-1$

0%
True
0%
False

Let a function

$f:R\rightarrow R$ be defined as

$f\left( x \right) =x+\sin { x }$ . The value of

$\int _{ 0 }^{ 2\pi }{ { f }^{ -1 }(x) } dx$ will

0%
$2{ \pi }^{ 2 }$
0%
$2{ \pi }^{ 2 }-2$
0%
$2{ \pi }^{ 2 }+2$
0%
${ \pi }^{ 2 }$

Explanation

We have

$f(x) = x + \sin x$

Let

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

let

$x = f(t) \Rightarrow dx = f'(t) dt$

$I = \displaystyle \int_{f^{-1} (0)}^{f^{-1} (2 \pi)} t . f'(t) dt = \int_0^{2\pi} t. f'(t) dt$

$= t f(t) \displaystyle |_0^{2\pi} - \int_0^{2\pi} f(t) dt$

$= 2 \pi . f(2 \pi) - \displaystyle \int_0^{2\pi} t + \sin t$

$4 \pi^2 - \dfrac{t^2}{2} \displaystyle |_0^{2\pi} + \cos t |_0^{2\pi}$

$= 4 \pi^2 - 2 \pi^2 + 0$

$= 2 \pi^2$

The value of

$\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta=$

0%
$\dfrac{\pi}{4}-1$
0%
$\dfrac{\pi}{4}$
0%
$1-\dfrac{\pi}{4}$
0%
none of these

Explanation

Now,

$\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta$

$=\displaystyle\int\limits_{0}^{\frac{\pi}{4}} (\sec^2 \theta-1)\ d\theta$ [ Since

$\sec^2 \theta-\tan^2 \theta=1$ ]

$=\left[\tan \theta-\theta\right]_0^{\frac{\pi}{4}}$

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

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

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

Explanation

$\displaystyle\int _{ 0 }^{ 1 }{ \dfrac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ \frac { 3 }{ 2 } } } dx } =\dfrac{1}{2}\int ^{1}_0 \dfrac{d(x^2+1)}{(x^2+1)^{3/2}}=\dfrac{1}{2}\bigg[\dfrac{(x^2+1)^{-3/2+1}}{-3/2+1}\bigg]^{1}_{0}=\bigg[\dfrac{1}{2}\times \dfrac{(x^2+1)^{-1/2}}{-1/2}\bigg]^{1}_{0}=-[\dfrac{1}{\sqrt{2}}-1\bigg]=1-\dfrac{1}{\sqrt{2}}$

$\displaystyle\int^{1/2}_0\dfrac{dx}{(1+x^2)\sqrt{1-x^2}}$ is equal to?

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

Explanation

Let

$x=\sin\theta$

$dx=\cos \theta d\theta$

$\displaystyle\int^{\pi/6}_0\dfrac{\cos\theta d\theta}{(1+\sin^2\theta)\cos \theta}=\displaystyle\int^{\pi/6}_0\dfrac{\sec^2\theta}{1+2\tan^2\theta}=d\theta$ .

Multiply numeration and denomination by

$\sec^2\theta$

$=\dfrac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\theta)^{\dfrac{\pi}{6}}_{0}$

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

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

0%
$\dfrac { \sqrt { e } }{ 2 } \left( \sqrt { 3 } +1 \right)$
0%
$\dfrac { \sqrt { 3e } }{ 2 }$
0%
$\sqrt { 3e }$
0%
$\sqrt { \dfrac { e }{ 3 } }$

$\int _{ 1 }^{ 3} \dfrac{4x}{x^2+1} \, dx$

0%
$\log { 5 }$
0%
$\cfrac { 1 }{ 2 } \log { 5 }$
0%
$\log { 25 }$
0%
$\log { 100 }$

Explanation

$\int _{ 1 }^{ 3 }{ \left( \cfrac { 4x }{ { 1+x }^{ 2 } } \right) } dx$

substitute

$x^2+1=t$

$x=1$ then

$t=2$

$x=3$ then

$t=10$

$\int_2^{10}\dfrac2tdt$

$=2{ \left[ \ln { \left( { 1+x }^{ 2 } \right) } \right] }_{ 1 }^{ 3 }$

$=2\left[ \ln { 10 } -\ln { 2 } \right]$

$=\ln { 25 }$

$\int _{ log2 }^{ x }{ \dfrac { dx }{ \sqrt { { e }^{ x }-1 } } } =\dfrac { \pi }{ 6 } ,$ then x is equal to _________.

0%
4
0%
in 8
0%
in 4
0%
None of these

$\displaystyle\int^1_0\dfrac{dx}{e^x+e^{-x}}$ is equal to?

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

Explanation

$\displaystyle\int^1_0\dfrac{dx}{e^x+e^{-1}}=\displaystyle\int^1_0\dfrac{e^x}{(e^x)^2+1}dx$

Put

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

$x\rightarrow 0\Rightarrow t\rightarrow e^0=1$

$x\rightarrow 1\Rightarrow t\rightarrow e^1=e$

$\displaystyle \int^e_1 \dfrac{1}{1+t^2}dt=\tan^{-1}t+c$

$=(\tan^{-1}(e^x))^1_0+c=\tan^{-1}(e)-\dfrac{\pi}{4}$ .

$\displaystyle\int^{\dfrac{\pi}{2}}_0\dfrac{\cot x}{\cot x+cosec x}dx=m(\pi +n)$ , then

$mn$ is equal to?

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

Explanation

$\displaystyle\int^{\pi /2}_0\dfrac{\cot xdx}{\cot x+cosec x}$

$\displaystyle\int^{\pi /2}_0\dfrac{\cos x}{\cos x+1}=\displaystyle\int \dfrac{2\cos^2\dfrac{x}{2}-1}{2\cos^2\dfrac{x}{2}}$

$\displaystyle\int^{\pi/2}_0\left(1-\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)dx$

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

$\Rightarrow$

$\dfrac{1}{2}[\pi -2]$

$\Rightarrow$

$m=\dfrac{1}{2}, n=-2$

$mn=-1$ .

Evaluate :

$\displaystyle\int^1_0\dfrac{dx}{(1+x+x^2)}$

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

Evaluate

$\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$

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

Explanation

Let

$I=\displaystyle\int^{\pi/2}_0e^x\left(\dfrac{1+\sin x}{1+\cos x}\right)dx$

$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{(1+\cos x)}+\dfrac{\sin x}{(1+\cos x)}\right\}dx$

$=\displaystyle\int^{\pi/2}_0e^x\left\{\dfrac{1}{2\cos^2(x/2)}+\dfrac{2\sin(x/2)\cos (x/2)}{2\cos^2(x/2)}\right\}dx$

$=\displaystyle\int^{\pi/2}_0e^x\left\{\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right\}dx$

$=\left[e^x\tan \dfrac{x}{2}\right]^{\pi/2}_0=e^{\pi/2}$ .

Evaluate :

$\displaystyle\int^{\sqrt{2}}_0\sqrt{2-x^2}dx$

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

Evaluate :

$\displaystyle\int^1_0\dfrac{(1-x)}{(1+x)}dx$

0%
$(\log 2+1)$
0%
$(\log 2-1)$
0%
$(2 \log 2-1)$
0%
$(2 \log 2+1)$

Evaluate :

$\displaystyle\int^9_0\dfrac{dx}{(1+\sqrt{x})}$

0%
$(3-2 log 2)$
0%
$(3+2 log 2)$
0%
$(6-2 log 4)$
0%
$(6+2 log 4)$

Evaluate :

$\displaystyle\int^2_{-2}|x|dx$

0%
$4$
0%
$3.5$
0%
$2$
0%
$0$

Evaluate

$\displaystyle\int^{1}_0\dfrac{(1-x)}{(1+x)}dx$

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

Evaluate

$\displaystyle\int^{\pi/6}_0\cos x\cos 2xdx$

0%
$\dfrac{1}{4}$
0%
$\dfrac{5}{12}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{7}{12}$

The value of

$\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx$ is?

0%
$50\sqrt{2}$
0%
$100\sqrt{2}$
0%
$150\sqrt{2}$
0%
$200\sqrt{2}$

Evaluate :

$\displaystyle\int^2_1|x^2-3x+2|dx$

0%
$\dfrac{-1}{6}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$

Given

$I_{m}=\displaystyle \int_{1}^{e}(\log x)^{m} d x .$ If

$\dfrac{I_{m}}{K}+\dfrac{I_{m-2}}{L}=e,$ then the values of

$K$ and

$L$ are

0%
$\dfrac{1}{1-m}, \dfrac{1}{m}$
0%
$(1-m), \dfrac{1}{m}$
0%
$\dfrac{1}{1-m}, \dfrac{m(m-2)}{m-1}$
0%
$\dfrac{m}{m-1}, m-2$

Explanation

$I_{m}= \displaystyle \int_{1}^{e}(\log x)^{m} d x \\$

$\therefore I_{m}=\left[x(\log x)^{m}\right]_{1}^{e}-\int_{1}^{e} x \dfrac{m(\log x)^{m-1}}{x} d x \text { (integrating by parts) }$

$\Rightarrow \dot{I}_{m}=e-m \displaystyle \int_{1}^{e}(\log x)^{m-1} d x=e-m I_{m-1}\\$

Replacing

$m$ by

$m-1\\$

$I_{m-1}=e-(m-1) I_{m-2}\\$
From equations

$(1)$ and

$(2),$ we have

$I_{m}=e-m\left[e-(m-1) I_{m-2}\right]\\$

$\Rightarrow I_{m}-m(m-1) I_{m-1}=e(1-m) \\$

$\Rightarrow \dfrac{I_{m}}{1-m}+m I_{m-2}=e \\$

$\Rightarrow K=1-m \text { and } L=\dfrac{1}{m}$

$\displaystyle\int^a_{-a}x|x|dx=?$

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

Explanation

$|x|=-x,x<0$

$x,x>0$

$I=\displaystyle\int^0_{-a}x(-x)dx+\displaystyle\int^a_0x\cdot xdx$

$=\displaystyle\int^0_{-a}-x^2dx+\displaystyle\int^a_0x^2dx$

$=\left[\dfrac{-x^3}{3}\right]^0_{-a}+\left[\dfrac{x^3}{3}\right]^a_0=+\dfrac{1}{3}(-a)^3+\dfrac{a^3}{3}=0$ .

$\displaystyle\int^1_{-2}\dfrac{|x|}{2}dx=?$

0%
$3$
0%
$2.5$
0%
$1.5$
0%
None of these

Explanation

$|x|=-x,x<0$

$x,x>0$

$I=\displaystyle\int^0_{-2}\dfrac{-x}{x}dx+\displaystyle\int^1_0\dfrac{x}{x}dx=\displaystyle\int^0_{-2}-dx+\displaystyle\int^1_0dx=[-x]^0_{-2}+[x]^1_0=(-2+1)=-1$ .

Evaluate :

$\displaystyle\int^1_0|2x-1|dx$

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

Value of

$\displaystyle\int^3_2\dfrac{dx}{\sqrt{(1+x^3)}}$ is?

0%
Less than $1$
0%
Greater than $2$
0%
Lies between $3$ and $4$
0%
None of these

Explanation

Given,

$\int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}}$

Let us consider the function

$f(x) = \int_{2}^{3} \dfrac{dx}{\sqrt{1+x^3}}$

according to the question putting the lower and upper limit we get,

$f(2) = \dfrac{1}{\sqrt{1+2^3}} = \dfrac{1}{3}$

$f(3) = \dfrac{1}{\sqrt{1+3^3}} = \dfrac{1}{\sqrt{28}}$

$\therefore \dfrac{1}{3} > \dfrac{1}{\sqrt{28}}$ we conclude that the Larger integrating value lies in the point 2

$\therefore \dfrac{1}{\sqrt{28}} < f(x) < \dfrac{1}{3}$

Since, m < f(x) < M

$\Rightarrow \int m dx < \int f(x) dx < \int M dx \, \, \, [ \, \forall \, m , M \in \mathbb{R}$ and be the lower and upper limit of the f(n) ]

we can conclude that,

$\int_2^3 f(x) dx < \dfrac{1}{3}$

Therefore the value is less than 1. (Ans)

The value of the definite integral

$\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x$ is

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

Explanation

$\sin n x-\sin (n-2) x=2 \cos (n-1) x \sin x\\$

$\Rightarrow \displaystyle \int \dfrac{\sin n x}{\sin x} d x=\int 2 \cos (n-1) d x+\int \dfrac{\sin (n-2) x}{\sin x} d x\\$

$\therefore \displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 5 x}{\sin x} d x=\int_{0}^{\pi / 2} 2 \cos 4 x d x+\int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\$

$=0+\displaystyle \int_{0}^{\pi / 2} \dfrac{\sin 3 x}{\sin x} d x\\=\displaystyle\int_{0}^{\pi / 2} 3-4sin^2x\ \ d x=\displaystyle\int_{0}^{\pi / 2} 1+2cos2x\ \ d x=\dfrac{\pi}{2}$

Evaluate :

$\displaystyle\int^1_{-2}|2x+1|dx$

0%
$\dfrac{5}{2}$
0%
$\dfrac{7}{2}$
0%
$\dfrac{9}{2}$
0%
$4$

The value of the definite integral

$\int_{2}^{4}(x(3-x)(4+x)(6-x)$

$(10-x)+\sin x) d x$ equals

0%
$\cos 2+\cos 4$
0%
$\cos 2-\cos 4$
0%
$\sin 2+\sin 4$
0%
$\sin 2-\sin 4$

Explanation

Let

$\left.I=\int_{2}^{4}(x(3-x)(4+x)(6-x)(10-x)+\sin x) d x\right)$ .........(i)

$=\int_{2}^{4}((6-x)(3-(6-x))(4+(6-x))(6-(6-x))$ ..........

$\because \int_{a}^b f(x) \ dx= \int _{a}^{b} f(a+b-x) \ dx$

$=\int_{2}^{4}((6-x)(x-3)(10-x) x(4+x)+\sin (6-x)) d x$ .......... (ii)

Adding equations (1) and

$(2),$ we get

$2 I=\int_{2}^{4}(\sin x+\sin (6-x)) d x \\$

$\quad=(-\cos x+\cos (6-x))_{2}^{4} \\$

$\quad=-\cos 4+\cos 2+\cos 2-\cos 4 \\$

$\quad=2(\cos 2-\cos 4) \\$

$\Rightarrow \quad I=\cos 2-\cos 4$

Let

$f : R \rightarrow R$ be a function as

$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$ . If

$g(x)$ is a polynomial of degree

$\leq 3$ such that

$\displaystyle \int \frac{g(x)}{f(x)} dx$ does not contain any logarithm function and

$g(-2) = 10$ . Then

$\displaystyle \int \frac{g(x)}{f(x)} dx$ , equals

0%
$\tan^{-1} \left ( \frac{x - 2}{2} \right ) + c$
0%
$\tan^{-1} \left ( \frac{x - 1}{1} \right ) + c$
0%
$\tan^{-1} (x) + c$
0%
None of these

Explanation

Here,

$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$

$= (x^2 + 4x + 3) (x^2 - 4x - 12) - 100$

$= (x^2 - 4x)^2 - 9(x^2 - 4x) - 136$

$= (x^2 - 4x + 8)(x^2 - 4x + 17)$

$\displaystyle \int \frac{g(x)}{f(x)} = \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)}$

$= \frac{Ax + B}{x^2 - 4x - 17} + \frac{Cx + D}{x^2 - 4x + 8}$
Clearly,

$A, B$ and

$C$ must be zero.

$\therefore \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{D}{x^2 - 4x + 8}$

$\therefore$

$g(x) = D (x^2 - 4x - 17)$

$g(-2) = D (4 + 8 - 17) = -10$ [given]

$\Rightarrow$

$\frac{g(x)}{f(x)} = \frac{2 (x^2 - 4x - 17)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{2}{x^2 - 4x + 8}$

$\therefore$

$\displaystyle \int \frac{g(x)}{f(x)} dx = \displaystyle \int \frac{2}{x^2 - 4x + 8}dx = 2 \displaystyle \int \frac{dx}{(x - 2)^2 + (2)^2}$

$= 2. \frac{1}{2} \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C = \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C$

$\displaystyle \int_{-2}^{-1} (ax^2-5)dx$ and

$5+\displaystyle \int_{1}^{2} (bx+c)dx=0,$ then

0%
$ax^2-bx+c=0$ has atleast one root in (1,2)
0%
$ax^2-bx+c=0$ has atleast one root in (-2,-1)
0%
$ax^2+bx+c=0$ has atleast one root in (-2,-1)
0%
None of the above

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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