MCQ Questions for Class 12 Commerce Maths Integrals Quiz 7

Evaluate

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

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

Explanation

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

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

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

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

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

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

$\displaystyle\int^2_0\dfrac{dx}{\sqrt{4-x^2}}$

0%
$1$
0%
$\sin^{-1}\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}$
0%
None of these

Evaluate

$\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{8}$
0%
$\dfrac{\pi}{16}$

Explanation

Let

$I=\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$

Put

$x^4=t$ and

$4x^3dx=dt$ .

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

$[x=1\Rightarrow t=1]$

$\therefore I=\dfrac{1}{4}\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$

$=\dfrac{1}{4}[\tan^{-1}t]^1_0$ ...........

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

$=\dfrac{1}{4}[\tan^{-1}1-\tan^{-1}0]\\$

$=\left(\dfrac{1}{4}\times \dfrac{\pi}{4}\right)=\dfrac{\pi}{16}$ .

Evaluate

$\displaystyle\int^{\pi/2}_{\pi/3} cosec xdx$

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

Evaluate

$\displaystyle\int^{\pi/2}_{\pi/4}\cot xdx$

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

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 :

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

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

Evaluate :

$\displaystyle\int^a_{-a}\sqrt{\dfrac{a-x}{a+x}}dx$

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

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

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

Explanation

Let

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

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

$=\displaystyle\int^{\pi}_0\dfrac{(1-\sin x)}{\cos^2x}dx$ .

$=\displaystyle\int^{\pi}_0[\sec^2x-\sec x\tan x]dx$

$=[\tan x-\sec x]^{\pi}_0$

$=(0+1)-(0-1)$

$=2$ .

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

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

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

Explanation

Let

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

$=\displaystyle\int^1_0e^x\left\{\dfrac{(1+x)-1}{(1+x)^2}\right\}dx$

$=\displaystyle\int^1_0e^x\left\{\dfrac{1}{(1+x)}-\dfrac{1}{(1+x)^2}\right\}dx$

$=\displaystyle\int^1_0e^x\{f(x)+f'(x)dx,$ where

$f(x)=\dfrac{1}{(1+x)}$

$=[e^xf(x)]^1_0$

$=\left[e^x\cdot \dfrac{1}{(1+x)}\right]^1_0$

$=\left(\dfrac{e}{2}-1\right)$ .

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

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%
$\dfrac{2}{3}$

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

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

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

$x,x>0$

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

Let

$\displaystyle\dfrac{x^{1/2}}{\sqrt{1-x^{3}}}dx=\dfrac{2}{3}gof(x)+c$ then

0%
$f(x)=\sqrt{x}$
0%
$f(x)=x^{3/2}$
0%
$f(x)=x^{2/3}$
0%
$g(x)=\sin^{-1}x$

Evaluate :

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

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

Evaluate :

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

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

The value of

$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$ is

0%
$e$
0%
$\ln (1+e)$
0%
$e+\ln (1+e)$
0%
$e-\ln (1+e)$

Explanation

$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$

$=\int_{1}^{e} \dfrac{1+x-x+x^{2} \ln x}{x(1+x \ln x)} d x$

Splitting them we get,

$I =\int_{1}^{e}\left(\dfrac{1}{x}+1\right) d x-\int_{1}^{e} \dfrac{1+\ln x}{1+x \ln x} d x \\$

$=[\ln x+x]_{1}^{e}-[\ln (1+x \ln x)]_{1}^{e} \\$

$=e-\ln (1+e)$

$I_{1}=\int_{0}^{\frac{\pi}{2}} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x, I_{2}=\int_{0}^{2 \pi} \cos ^{6} x d x$

$I_{3}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{3} x d x, I_{4}=\int_{0}^{1} \ln \left(\dfrac{1}{x}-1\right) d x,$ then

0%
$I_{2}=I_{3}=I_{4}=0, I_{1} \neq 0$
0%
$I_{1}=I_{2}=I_{3}=0, I_{4} \neq 0$
0%
$I_{1}=I_{3}=I_{4}=0, I_{2} \neq 0$
0%
$I_{1}=I_{2}=I_{3}=0, I_{4} \neq 0$

Explanation

$\text { } I_{1} =\int_{0}^{\pi / 2} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x \\$

$=\int_{0}^{\pi / 2} \dfrac{\sin \left(\dfrac{\pi}{2}-x\right)-\cos \left(\dfrac{\pi}{2}-x\right)}{1+\sin \left(\dfrac{\pi}{2}-x\right) \cos \left(\dfrac{\pi}{2}-x\right)} d x\\$

$=\int_{0}^{\pi / 2} \dfrac{\cos x-\sin x}{1+\sin x \cos x} d x=-I_{1}\\$

$\Rightarrow I_{1}=0\\$

$I_{3}=0$ as

$\sin ^{3} x\\$ is odd

$I_{4}=\int_{0}^{1} \ln \left(\dfrac{1-x}{x}\right) d x\\$

$=\int_{0}^{1} \ln \left(\dfrac{1-(1-x)}{1-x}\right) d x\\$

$=\int_{0}^{1} \ln \dfrac{x}{1-x} d x=-I_{4}\\$

$\Rightarrow I_{4}=0\\$

$I_{2}=\int_{0}^{2 \pi} \cos ^{6} x d x=2 \int_{0}^{\pi} \cos ^{6} x d x \neq 0$

Evaluate :

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

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

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]

$\Rightarrow$

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

$\therefore$

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

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

The value of the definite integral

$\displaystyle \int_{0}^{\pi/2}sinx\space sin2x\space sin3xdx$ is equal to

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

Explanation

$=\displaystyle \int_{0}^{\pi/2}\sin x\sin2x\sin3xdx$

$=\displaystyle \int_{0}^{\pi/2}sin\left(\dfrac{\pi}{2}-x\right) sin2\left(\dfrac{\pi}{2}-x\right) sin3\left(\dfrac{\pi}{2}-x\right)dx$

$\Rightarrow I=\displaystyle \int_{0}^{\pi/2}-cosxsin2xcos3xdx$

$\therefore 2I=\displaystyle \int_{0}^{\pi/2}-sin2x(cosxcos3x-sin3xsinx)dx$

$=\displaystyle \int_{0}^{\pi/2}-sin2xcos(4x)dx$

$=-\displaystyle \int_{0}^{\pi/2}sin2x(2cos^22x-1)dx$
Put

$cos2x=t \Rightarrow -sin2x\times 2dx=dt$

$\therefore 2I=\displaystyle \int_{1}^{-1}\dfrac{1}{2}(2t^2-1)dt=\dfrac{1}{2}\left[ \dfrac{2t^3}{3}-t \right]^{-1}$

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

$=\dfrac{1}{2}\left[ -\dfrac{2}{3}+1-\dfrac{2}{3}+1\right]=\dfrac{1}{2}\left[\dfrac{2}{3}\right]\qquad I\Rightarrow \dfrac{1}{6}$

$\displaystyle \int f(x) dx = F(x)$ , then

$\displaystyle \int x^3 f(x^2) dx$ is equal to

0%
$\frac{1}{2} [x^2$ { $F(x)$ } $^2 - \displaystyle \int$ { $F(x)$ } $^2 dx$ ]
0%
$\frac{1}{2} [x^2 F(x^2) - \displaystyle \int F(x^2) d (x^2)]$
0%
$\frac{1}{2} [x^2 F(x^2) - \frac{1}{2} \displaystyle \int$ { $F(x)$ } $^2 dx$ ]
0%
None of the above

Explanation

We have,

$\displaystyle \int f(x) dx = F(x)$

$\therefore \displaystyle \int x^3 f(x^2) dx = \frac{1}{2} \displaystyle \int \underbrace{x^2} \underbrace{f(x^2) d(x^2)}$
I II

$= \frac{1}{2} [x^2 F(x^2) - \displaystyle \int F(x^2) d (x^2)]$

Observe the following Lists

List-I	List-II
A: $\displaystyle \int_{-2}^{2}\frac{1}{4+x^{2}}dx$	1) $\displaystyle \frac{\pi}{3}$
B: $\displaystyle \int_{1}^{2}\frac{1}{x\sqrt{x^{2}-1}}dx$	2) 0
C: $\displaystyle \int_{0}^{\pi}\cos 3x.\cos 2xdx$	3) $\displaystyle \frac{\pi}{4}$
	4) $\displaystyle \frac{\pi}{2}$

0%
A-3, B-1, C-4
0%
A-3, B-1, C-2
0%
A-1, B-3, C-2
0%
A-4, B-1, C-2

Explanation

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

$\displaystyle =\left[ \frac { 1 }{ 2 } \tan ^{ -1 }{ 1 } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( -1 \right) } \right] =\frac { \pi }{ 8 } +\frac { \pi }{ 8 } =\frac { \pi }{ 4 }$

B)

$\displaystyle \int _{ 1 }^{ 2 }{ \frac { 1 }{ x\sqrt { { x }^{ 2 }-1 } } } dx=\left[ \sec ^{ -1 }{ x } \right] _{ 1 }^{ 2 }$

$\displaystyle =\left[ \sec ^{ -1 }{ 2 } -\sec ^{ -1 }{ 1 } \right] =\left[ \frac { \pi }{ 3 } -0 \right] =\frac { \pi }{ 3 }$

C)

$\displaystyle \int _{ 0 }^{ \pi }{ \cos { 3x } \cos { 2x } } dx=\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \left( \cos { x } +\cos { 5x } \right) } dx$

$\displaystyle =\frac { 1 }{ 10 } \left[ 5\sin { x } +\sin { 5x } \right] _{ 0 }^{ \pi }=0$

$\int_{0}^{log 5} \displaystyle \frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+3} dx =$

0%
$3+ 2 \pi$
0%
$4 - \pi$
0%
$2 + \pi$
0%
$\pi -4$

Explanation

Let

$I=\int { \cfrac { { e }^{ x }\sqrt { { e }^{ x }-1 } }{ { e }^{ x }+3 } }$
Substitute

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

$I=\int { \cfrac { \sqrt { t-1 } }{ t+3 } dt }$
Substitute

$u=\sqrt { t-1 } \Rightarrow du=\cfrac { 1 }{ 2\sqrt { t-1 } } dt$ , we get

$I=2\int { \cfrac { { u }^{ 2 } }{ { u }^{ 2 }+4 } du } =2\int { \left( 1-\cfrac { 4 }{ { u }^{ 2 }+4 } \right) du } =2\int { du } -2\int { \cfrac { 4 }{ { u }^{ 2 }+4 } } du\\ =2u-4{ tan }^{ -1 }\left( \cfrac { u }{ 2 } \right) =2\sqrt { t-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { t-1 } }{ 2 } \right) \\ =2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right)$
Therefore,

$\int _{ 0 }^{ \log { 5 } }{ Idx } ={ \left[ 2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right) \right] }_{ 0 }^{ \log { 5 } }=4-\pi$

$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$ then

$I_1, I_2, I_3..$ are in

0%
A.P.
0%
G.P.
0%
H.P.
0%
A.G.P.

Explanation

$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$

Substituting

$t=\tan x \Rightarrow dt =\sec^2xdx$

$I_n=\int_{0}^{1}t^ndt$ =

$\dfrac{1}{n\;+\;1}$

$\Rightarrow I_n\;=\;\dfrac{1}{n\;+\;1}$

Reciprocal of the terms

$I_1,I_2,I_3..$ are in A.P, therefore

$\dfrac{1}{n+1}, \dfrac{1}{n+2}, \dfrac{1}{n+3}$ are in H.P.

Hence, option 'C' is correct.

Evaluate the integral

$\displaystyle \int_{0}^{\pi/4}\frac{ {s}i {n}\theta+ {c} {o} {s}\theta}{9+16 {s}i {n}2\theta} \ {d}\theta$

0%
$\displaystyle \frac{1}{20} \log 2$
0%
$\displaystyle \frac{1}{20} \log 3$
0%
$\log 3$
0%
$\log 2$

Explanation

$\displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16\sin 2\theta} d\theta$

Let $\sin \theta - \cos \theta = t$

$\therefore (\cos \theta + \sin \theta) d\theta = dt$

$= \displaystyle \int_{-1}^{0} \dfrac {dt}{9 + 16(1 – t^{2})} (\because t^{2} = 1 – 2\cdot \sin \theta \cdot \cos \theta)$

$= \displaystyle \int_{-1}^{0} \dfrac {dt}{25 – 16t^{2}}$

$= \dfrac {1}{16} \displaystyle \int_{-1}^{0} \dfrac {dt}{\left (\dfrac {5}{4}\right )^{2} – t^{2}}$

$= \dfrac {1}{16}\times \dfrac {1}{2\times \dfrac {5}{4}}\left [ln \left |\dfrac {\dfrac {5}{4} +t}{\dfrac {5}{4} – t}\right |\right ]_{-1}^{0}$

$\left (\because \displaystyle \int \dfrac {1}{a^{2} – x^{2}} dx = \dfrac {1}{2a} ln \left |\dfrac {a + x}{a – x}\right | + c\right )$

$= \dfrac {1}{40} \left [ln |1| - ln \left |\dfrac {\dfrac {1}{4}}{\dfrac {9}{4}}\right |\right ]$

$= \dfrac {1}{40} ln 9 = \dfrac {ln 3}{20}$

$\therefore \displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16 \sin 2\theta} d\theta = \dfrac {ln 3}{20}$ .

Evaluate:

$\displaystyle \int_{0}^{\pi}\frac{dx}{5+4\cos x}$

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

Explanation

Putting

$\cos{x}=\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}$ ,

$\displaystyle\int_{0}^{\pi}\dfrac{dx}{5+4\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}}$

$=\displaystyle\int_{0}^{\pi} \dfrac{\{1+\tan^2{(x/2)\}}dx}{9+\tan^2{(x/2)}}$

$=\displaystyle\int_{0}^{\pi} \dfrac{\sec^2{(x/2)}dx}{\tan^2{(x/2)}+3^2}$

Let ,

$z=\tan{(x/2)}\implies dz=\dfrac{1}{2}\sec^2{(x/2)}dx\implies \sec^2{(x/2)}dx=2dz$

And when,

$x=0, z=0$ and when

$x=\pi, z=\infty$

Hence, integration becomes:-

$\displaystyle\int_{0}^{\infty} \dfrac{2dz}{z^2+3^2}$

$=\dfrac{2}{3}\left[\tan^{-1} {\dfrac{z}{3}}\right]_{0}^{\infty}$

$=\dfrac{2}{3}\left[\tan^{-1} {\infty}-\tan^{-1} {0}\right]$

$=\dfrac{2}{3}\left(\dfrac{\pi}{2}-0\right)$

$=\dfrac{\pi}{3}$

$\displaystyle \int_{0}^{a}\frac{dx}{x+\sqrt{a^{2}-x^{2}}}=$

0%
$\pi$
0%
$\dfrac{\pi}{3}$
0%
$-\pi$
0%
$\dfrac{\pi}{4}$

Explanation

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

Substituting

$x=a\sin t\Rightarrow dx=a\cos tdt$

$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \cos tdt }{ \sin t+\cos t } }$ ...(1)

Using property

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

$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \sin tdt }{ \cos t+\sin t } }$ ...(2)

Adding (1) and (2), we get

$\displaystyle 2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ dt } =\cfrac { \pi }{ 2 } \Rightarrow I=\cfrac { \pi }{ 4 }$

$\displaystyle \int_{0}^{\pi/2}\sqrt{\cos x}\sin^{5}xdx=$

0%
$\displaystyle \frac{34}{231}$
0%
$\displaystyle \frac{64}{231}$
0%
$\displaystyle \frac{30}{321}$
0%
$\displaystyle \frac{128}{231}$

Explanation

$\int_{0}^{\dfrac{\pi}{2}} \sqrt{cos x}sin^5 x   dx$
$\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x}(sin^4x)sin x   dx$
$-\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos   x}(1-cos^2 x)^2d cos x$
$-\left [ \int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} dcos x+\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^4 x dx - 2\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^2 x   dx \right ]$
$-\left [ \dfrac{2}{3} cos^{\dfrac{3}{2}} x \int_{0}^{\dfrac{\pi}{2}}+\dfrac{2}{11} cos^{\dfrac{11}{2}}x\int_{0}^{\dfrac{\pi}{2}}-2(\dfrac{2}{7}) cos^{\dfrac{7}{2}}x \int_{0}^{\dfrac{\pi}{2}} \right ]$
$=-\left [\dfrac{2}{3} (-1)+\dfrac{2}{11}(-1)-2(\dfrac{2}{7})(-1)\right ]$
$=\dfrac{64}{231}$

$\displaystyle \int_{0}^{\pi/2}\frac{1}{a+bcosx}dx=$ , where

$a>|b|$

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

Let

$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { 1 }{ a+b\cos x } dx }$

$\tan\left( \cfrac { x }{ 2 } \right) =t\Rightarrow \cfrac { 1 }{ 2 } { \sec }^{ 2 }\left( \cfrac { x }{ 2 } \right) dx=dt$
Gives

$\cos x=\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } }$

$\displaystyle I=\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ a+b\left(\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }\right) } \cfrac { 2dt }{ 1+{ t }^{ 2 } } } =\cfrac { 2 }{ a+b } \int _{ 0 }^{ 1 }{ \cfrac { 1 }{ \cfrac { { t }^{ 2 }\left( a-b \right) }{ a+b } +1 } dt }$
Substitute

$\cfrac { t\sqrt { a-b } }{ \sqrt { a+b } } =u\Rightarrow \cfrac { dt\sqrt { a-b } }{ \sqrt { a+b } } =du$

$\displaystyle I=\cfrac { 2 }{ \sqrt { a-b } \sqrt { a+b } } \int _{ 0 }^{ \cfrac { \sqrt { a-b } }{ \sqrt { a+b } } }{ \cfrac { 1 }{ 1+{ u }^{ 2 } } du }$

$={ \left[ \cfrac { 2{ \tan }^{ -1 }u }{ \sqrt { a-b } \sqrt { a+b } } \right] }_{ 0 }^{ \cfrac { \sqrt { a-b } }{ \sqrt { a+b } } } =\cfrac { 2{ \tan }^{ -1 }\cfrac { \sqrt { a-b } }{ \sqrt { a+b } } }{ \sqrt { a-b } \sqrt { a+b } } -0\\ =\cfrac { 2 }{ \sqrt { { a }^{ 2 }-{ b }^{ 2 } } } { \tan }^{ -1 }\sqrt{\cfrac {a-b}{a+b}}$

$\displaystyle \int_{1}^{2}\mathrm{x}^{2\mathrm{x}}[1+\log \mathrm{x}]\mathrm{d}\mathrm{x}=$

0%
$\displaystyle \frac{9}{2}$
0%
$\displaystyle \frac{11}{2}$
0%
$\displaystyle \frac{13}{2}$
0%
$\displaystyle \frac{15}{2}$

Explanation

Let

$\displaystyle {x}^{x} = t$
Diiferentiating both sides,

$\displaystyle {x}^{x}(1 + \log x) dx = dt$
Substituting the above obtained values back into the expression,

$\displaystyle \int {x}^{2x}(1 + log x) dx = \int t dt$

$\displaystyle = \frac{{t}^{2}}{2} + c$
Putting the value of t back, we get the integral as

$\displaystyle \frac{{({x}^{x})}^{2}}{2} + c$
Substituting the limits from 1 to 2, we get

$= \displaystyle \left( \frac{{({2}^{2})}^{2}}{2} \right) - \left( \frac{{({1}^{1})}^{2}}{2}\right)$
.

$\displaystyle = 8 - \frac{1}{2}$
.

$\displaystyle = \frac{15}{2}$

$\displaystyle \int_{0}^{\infty}e^{-\mathrm{x}^{2}}\mathrm{d}\mathrm{x}=\frac{\sqrt{\pi}}{2}$ , then

$\displaystyle \int_{0}^{\infty}e^{-ax^{2}}dx,\ \mathrm{a}>0$ is

0%
$\displaystyle \frac{\sqrt{\pi}}{2}$
0%
$\displaystyle \frac{\sqrt{\pi}}{2a}$
0%
$2 \displaystyle \frac{\sqrt{\pi}}{a}$
0%
$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$

Explanation

$\int _{ 0 }^{ \infty }{ { e }^{ { -ax }^{ 2 } } } dx\\ =\int _{ 0 }^{ \infty }{ { e }^{ -(\sqrt { a } x)^{ 2 } } } dx$

Let,

$\sqrt { a } .{ x }=1$

$\sqrt { a } .dx=dt\\ dx=\cfrac { dt }{ \sqrt { a } } \\ I=\int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } .\cfrac { dt }{ \sqrt { a } } \\ =\cfrac { 1 }{ \sqrt { a } } \int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } dt\\ =\cfrac { 1 }{ \sqrt { a } } \times \cfrac { \sqrt { \pi } }{ 2 } \\ =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } }$

Hence the correct answer is

$\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } }$

$\displaystyle \int_{0}^{\pi/3}\frac{\cos x}{3+4\sin x}dx=k\log(\frac{3+2\sqrt{3}}{3})$ , then

$\mathrm{k}$ is equal to

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

Explanation

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

Let

$z=3+4\sin{x}\implies dz=4\cos{x}dx\implies \cos{x}dx=\dfrac{dz}{4}$

When

$x=0,z=3$ and when

$x=\dfrac{\pi}{3},z=3+2\sqrt{3}$

Hence, integration becomes:-

$\dfrac{1}{4}\displaystyle\int_{3}^{3+2\sqrt{3}}\dfrac{1}{z} dz$

$=\dfrac{1}{4}\left[\log{z}\right]_{3}^{3+2\sqrt{3}}$

$=\dfrac{1}{4}\log{\left(\dfrac{3+2\sqrt{3}}{3}\right)}$

$\implies k=\dfrac{1}{4}$

Hence, answer is option-(C).

The value of the integral

$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$ is

0%
$2\displaystyle \sec\alpha\log(\frac{1}{2}\text{cosec}\alpha)$
0%
$2\displaystyle \sec\alpha\log(\frac{1}{2}\sec\alpha)$
0%
$2 \text{cosec}\alpha\log(\sec\alpha)$
0%
$2 \displaystyle \text{cosec}\alpha\log(\frac{1}{2}\sec\alpha)$

Explanation

$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$

$=\displaystyle \int_{0}^{3\alpha}\frac{1}{\sin (x-\alpha)\sin (x-2\alpha)}dx$

$\displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha} \displaystyle \frac{\sin [(x-\alpha)-(x-2\alpha)]}{\sin (x-\alpha)\sin (x-2\alpha)}dx$

$\Rightarrow \displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha}[\cot(x-2\alpha)-\cot(x-\alpha)]dx$

$=\displaystyle \frac{1}{\sin \alpha}(\log [\sin (x-2\alpha)]|_{0}^{3\alpha}-\log (\sin (x-\alpha))|_{0}^{3\alpha})$

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

$=2\displaystyle \frac{1}{\sin \alpha}\log \left | \displaystyle \frac{\sin \alpha}{2\sin \alpha \cos \alpha} \right |$

$=2 \text{cosec}\alpha\ \log\left(\dfrac{1}{2} \sec\alpha\right)$

$I_{n}=\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{n} dx,$ then

$\dfrac{1}{I_{2}+I_{4}},\dfrac{1}{I_{3}+I_{5}},\dfrac{1}{I_{4}+I_{6}}\cdots$ form

0%
an A.P
0%
a G.P
0%
a H.P
0%
an AGP

Explanation

$I_n = \displaystyle \int_{0}^{\pi/4}\tan^{n}x\ dx$

$I_{n+2}=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\tan^{2}x)dx$

$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\sec^{2}x-1)dx$

$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x\ d(\tan x)-I_n$

$\Rightarrow I_{n+2} + I_n=\dfrac{1}{n+1}$

$I_{2}+I_{4}=\dfrac13$

$I_{3}+I_{5}=\dfrac14$

$I_{4}+I_{6}=\dfrac15$

$So,\dfrac{1}{I_{2}+I_{4}};\dfrac{1}{I_{3}+I_{5}};\dfrac{1}{I_{4}+I_{6}}$ forms an AP

$\displaystyle \int_{0}^{1}\frac{3^{x+1}-4^{x-1}}{12^{x}}dx=$

0%
$\displaystyle \frac{9}{4}log_{4}\, e$
0%
$\displaystyle \frac{9}{4}log_{4}\, e-\frac{1}{6}log_{3}e$
0%
$\displaystyle log_{4}3$
0%
None of these

Explanation

$\int_{0}^{1}\dfrac{3^{x+1}-4^{x-1}}{12^{x}} dx$

$=\displaystyle \int_{0}^{1}(\dfrac{3}{4^{\mathrm{x}}}-\dfrac{1}{4}. \displaystyle \dfrac{1}{3^{\mathrm{x}}}) dx$

$=\left \{ 3\dfrac{\dfrac{1}{4^{x}}}{(log\dfrac{1}{4})}-\dfrac{1}{4} \dfrac{\dfrac{1}{3^{x}}}{log\left ( \dfrac{1}{3} \right )}\right \}_{0}^{1}$

$=\displaystyle \dfrac{3(\dfrac{1}{4}-1)}{-\log_{\mathrm{e}}4}-\dfrac{1}{4}\dfrac{(\dfrac{1}{3}-1)}{-\log_{\mathrm{e}}3}$

$=\displaystyle \dfrac{9}{4}\log_{4}\mathrm{e}-\dfrac{1}{6}\log_{3}\mathrm{e}$

Ans: B

Let f be a function defined for every x, such that f'' = -f ,f(0)=0, f' (0) = 1, then f(x) is equal to

0%
tanx
0%
$e^{x}-1$
0%
sinx
0%
2sinx

Explanation

$\int f^{''}(x)f^{'}(x)=-\int f(x)f^{'}(x)$

$(f^{'}(x))^{2}=f^{2}(x)+c$

$c=1$

$f^{'}(x)=\sqrt{1-f^{2}(x)}$

$\int \frac{df(x)}{\sqrt{1-f^{2}(x)}}=\int dx$

$sin^{-1}f(x)=x+c$

$f(x)=sin x$

Let

$\displaystyle \frac{d}{dx}F(x)=\frac{e^{{s}{m}{x}}}{x}$ ,

$x>0$ . lf

$\displaystyle \int_{1}^{4}\frac{3}{x}e^{{s}{m}{x}^{3}}dx=F(k)-F(1)$ , then one of the possible values of

${k}$ is

0%
$16$
0%
$62$
0%
$64$
0%
$15$

Explanation

Given:

$\displaystyle\int_{1}^{4} \dfrac{3}{x} e^{\sin{x^3}}dx$

Let

$z=x^3\implies dz=3x^2dx=\dfrac{3}{x}x^3dx$

$\implies dz=\dfrac{3}{x}zdx$

$\implies \dfrac{3}{x}dx=\dfrac{dz}{z}$

And when

$x=1,z=1$ and when

$x=4,z=64$

Hence, integration becomes:-

$\displaystyle\int_{1}^{64}e^{\sin{z}}\dfrac{dz}{z}$

Now it is given that:-

$\dfrac{d}{dx}F(x)=\dfrac{e^{\sin{x}}}{x}, x>0$

$\implies \displaystyle\int_{1}^{64} \dfrac{e^{\sin{z}}}{z}dz$

$=\displaystyle\int_{1}^{64} dF(z)$

$=F(64)-F(1)$

Hence,

$k=64$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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