Integrals Test

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Integrals Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\displaystyle\frac{d}{dx}(F(x))=\frac{e^{\displaystyle\sin{x}}}{x}$$, $$x>0$$. If $$\displaystyle\int_1^4{\frac{2e^{\displaystyle\sin{x^2}}}{x}dx}=F(k)-F(1)$$, then the possible value of $$k$$ is

A
10

B
14

C
16

D
18

Solution

Substitute $${ x }^{ 2 }=z$$
Then $$\displaystyle \int _{ 1 }^{ 4 }{ \frac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =\int _{ 1 }^{ 16 }{ \frac { { e }^{ \sin { z } } }{ z } dz } =\int _{ 1 }^{ 16 }{ \frac { d }{ dz } \left\{ F\left( z \right) \right\} } dz$$
$$={ \left( F\left( z \right) \right) }_{ 1 }^{ 16 }=F\left( 16 \right) -F\left( 1 \right) $$
$$\therefore F\left( 16 \right) -F\left( k \right) =F\left( 16 \right) -F\left( 1 \right) $$
Gives $$k=16$$
Question 2
1 / -0

Let $$\displaystyle \frac{df\left ( x \right )}{dx}=\frac{e^{\sin x}}{x}, x> 0$$. If $$\displaystyle \int_{1}^{4}\displaystyle \frac{3e^{\sin x^{3}}}{x}dx=f\left ( k \right )-f\left ( 1 \right )$$ then one of the possible values of $$k$$ is

A
$$16$$

B
$$63$$

C
$$64$$

D
$$15$$

Solution

Substitute $${ x }^{ 3 }=z$$
Then $$\displaystyle \int _{ 1 }^{ 4 }{ \frac { 3{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =\int _{ 1 }^{ 16 }{ \frac { { e }^{ \sin { z } } }{ z } dz } =\int _{ 1 }^{ 64 }{ \frac { d }{ dz } \left\{ F\left( z \right) \right\} } dz$$
$$={ \left( F\left( z \right) \right) }_{ 1 }^{ 64 }=F\left( 64 \right) -F\left( 1 \right) $$
$$\therefore F\left( 64 \right) -F\left( k \right) =F\left( 64 \right) -F\left( 1 \right) $$
Gives $$k=64$$
Question 3
1 / -0

$$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1$$ is equal to?

A
$$\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx$$

B
$$\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx$$

C
$$\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}$$

D
$$\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}$$

Solution

Let $$\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ 1+x^{ n } } dx=\int _{ \infty }^{ 0 } \frac { -dx }{ 1+x^{ n } } $$

Substitute $$\displaystyle x=\frac { 1 }{ y } \Rightarrow dx=-\frac { 1 }{ { y }^{ 2 } } $$

$$\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ y^{ 2 } } \frac { dy }{ 1+\left( \frac { 1 }{ y } \right) ^{ n } } =\int _{ 0 }^{ \infty } \frac { y^{ n-1 } }{ y(1+y^{ n }) } dy$$

Substitute $$t^{ n }-1+y^{ n }\Rightarrow n{ t }^{ n-1 }dt=n{ y }^{ n-1 }dy$$

$$\displaystyle I=\int _{ 1 }^{ \infty } \frac { dt }{ t(t^{ n }-t)^{ 1/n } } =\int _{ \infty }^{ 1 } -\frac { 1 }{ t^{ 2 } } .\frac { dt }{ \left( 1-\frac { 1 }{ t^{ n } } \right) ^{ 1/n } } $$

Substitute $$\displaystyle z=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } } dt$$

$$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dz }{ (1-z^{ n })^{ 1/n } } =\int _{ 0 }^{ 1 } \frac { dx }{ (1-x^{ n })^{ 1/n } } $$
Question 4
1 / -0

The value of $$\displaystyle \int_{1}^{1/e}f(x)dx+\int_{1}^{e}f(x)dx$$ where $$f(x)$$ is given as $$\displaystyle \frac{log\:x}{1+x}$$ equals

A
$$0$$

B
$$\displaystyle -\frac{1}{2}$$

C
$$\displaystyle \frac{1}{2}$$

D
$$1$$

Solution

Let $$\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { logx }{ 1+x } dx$$ and $$\displaystyle l_{ 2 }=\int _{ 1 }^{ 0 } \frac { log\: x }{ 1+x } dx$$

Now for $$\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { log\: t }{ 1+x } $$

Substitute $$\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } } $$

$$l_{ 1 }=\int _{ 1 }^{ e } \dfrac { log\: t }{ t(1+t) } dt=\int _{ 1 }^{ e } \dfrac { log\: x }{ x(1+x) } dx$$ ( Replacing $$t\rightarrow \: x$$)

$$\displaystyle \therefore l_{ 1 }+l_{ 2 }=\int _{ 1 }^{ e } \frac { log\: x }{ x(1+x) } dx+\int _{ 1 }^{ e } \frac { log\: x }{ 1+x } dx$$

$$\displaystyle =\int _{ 1 }^{ e } \frac { (log(x))(1+x) }{ x(1+x) } dx=\int _{ 1 }^{ e } \frac { log\: x }{ x } dx$$

$$\displaystyle =\frac { 1 }{ 2 } \left[ (log\: x) \right] _{ 1 }^{ e }=\frac { 1 }{ 2 } [log_{ e }e-log_{ e }1]=\frac { 1 }{ 2 } $$
Question 5
1 / -0

If $$\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$$ then I is equal to

A
$$0$$

B
$$\displaystyle \pi $$

C
$$\displaystyle \pi -\frac{1}{\pi }$$

D
$$\displaystyle \pi +\frac{1}{\pi }$$

Solution

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

Substitute $$\displaystyle x=\dfrac { 1 }{ t } $$

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

$$\displaystyle =-\int _{ \dfrac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ t } \sin { \left( t-\dfrac { 1 }{ t } \right) dt } } -I\\ \Rightarrow 2I=0\Rightarrow I=0$$
Question 6
1 / -0

The value (s) of $$\displaystyle \int_{0}^{1}\frac{x^{4}\left ( 1-x \right )^{4}}{1+x^{2}} dx $$ is (are)

A
$$\displaystyle \frac{22}{7}-\pi $$

B
$$\displaystyle \frac{2}{105}$$

C
$$0$$

D
$$\displaystyle \frac{71}{15}-\frac{3\pi}{2}$$

Solution

Let $$ \displaystyle I=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx $$
$$ \displaystyle =\int _{ 0 }^{ 1 }{ \frac { \left( { x }^{ 4 }-1 \right) { \left( 1-x \right) }^{ 4 }+{ \left( 1-x \right) }^{ 4 } }{ \left( 1+{ x }^{ 2 } \right) } dx } $$
$$ \displaystyle =\int _{ 0 }^{ 1 }{ \left( { x }^{ 2 }-1 \right) } { \left( 1-x \right) }^{ 4 }dx+\int _{ 0 }^{ 1 }{ \frac { { \left( 1+{ x }^{ 2 }-2x \right) }^{ 2 } }{ \left( 1+{ x }^{ 2 } \right) } dx } \\ $$
$$ \displaystyle =\int _{ 0 }^{ 1 }{ \left\{ \left( { x }^{ 2 }-1 \right) { \left( 1-x \right) }^{ 4 }+\left( 1+{ x }^{ 2 } \right) -4x+\frac { { 4x }^{ 2 } }{ \left( 1+{ x }^{ 2 } \right) } \right\} dx } $$
$$ \displaystyle =\int _{ 0 }^{ 1 }{ \left( \left( { x }^{ 2 }-1 \right) { \left( 1-x \right) }^{ 4 }+\left( 1+{ x }^{ 2 } \right) -4x+4\frac { 4 }{ 1-{ x }^{ 2 } } \right) dx } $$
$$ \displaystyle =\int _{ 0 }^{ 1 }{ \left( { x }^{ 6 }-{ 4x }^{ 5 }+{ 5x }^{ 4 }-{ 4x }^{ 2 }+4-\frac { 4 }{ 1+{ x }^{ 2 } } \right) dx } $$
$$ \displaystyle =\left[ \frac { { x }^{ 7 } }{ 7 } -\frac { { 4x }^{ 6 } }{ 6 } +\frac { { 5x }^{ 5 } }{ 5 } -\frac { { 4x }^{ 3 } }{ 3 } +4x-4\tan ^{ -1 }{ x } \right] _{ 0 }^{ 1 } $$
$$ \displaystyle =\frac { 1 }{ 7 } -\frac { 4 }{ 6 } +\frac { 5 }{ 5 } -\frac { 4 }{ 3 } +4-4\left( \frac { \pi }{ 4 } -0 \right) =\frac { 22 }{ 7 } -\pi $$
Question 7
1 / -0

If $$\displaystyle I= \int_{0}^{1}\frac{x dx}{8+x^{3}}$$ then the smallest interval in which $$I$$ lies is

A
$$\displaystyle \left ( 0,\frac{1}{8} \right )$$

B
$$\displaystyle \left ( 0,\frac{1}{9} \right )$$

C
$$\displaystyle \left ( 0,\frac{1}{10} \right )$$

D
$$\displaystyle \left ( 0,\frac{1}{7} \right )$$

Solution

$$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { x }{ 8+{ x }^{ 3 } } } dx=\int _{ 0 }^{ 1 }{ \frac { x }{ \left( x+2 \right) \left( { x }^{ 2 }-2x+4 \right) } } dx$$

$$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \frac { x+2 }{ 6\left( { x }^{ 2 }-2x+4 \right) } -\frac { 1 }{ 6\left( x+2 \right) } \right) } dx$$

$$\displaystyle =\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \left( \frac { 2x-2 }{ 2\left( { x }^{ 2 }-2x+4 \right) } +\frac { 3 }{ 6{ x }^{ 2 }-2x+4 } \right) dx } -\frac { 1 }{ 6 } \int { \frac { 1 }{ x+2 } } dx$$

$$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { 2x-2 }{ { x }^{ 2 }-2x+4 } } dx+\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ 2 }-2x+4 } } -\frac { 1 }{ 6 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ x+2 } } dx$$

$$\displaystyle =\left[ \frac { 1 }{ 12 } \log { \left( { x }^{ 2 }-2x+4 \right) } -\frac { 1 }{ 6 } \log { \left( x+2 \right) } \right] +\frac { \tan ^{ -1 }{ \left( \frac { x-1 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } } $$

$$\displaystyle =\frac { 1 }{ 12 } \log { \left( 3 \right) } -\frac { 1 }{ 6 } \log { \left( 3 \right) } +\frac { \tan ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } } $$

$$\displaystyle -\frac { 1 }{ 12 } \log { \left( 4 \right) -\frac { 1 }{ 6 } \log { \left( 2 \right) } } +\frac { \tan ^{ -1 }{ \left( \frac { -1 }{ \sqrt { 3 } } \right) } }{ 2\sqrt { 3 } } $$

And this lies between $$\displaystyle \left( 0,\frac { 1 }{ 9 } \right) $$
Question 8
1 / -0

$$ \displaystyle \int_{1}^{\infty }\frac{\log \left ( t-1 \right )}{t^2\log t+\log \left ( \frac{t}{t-1} \right )}\:dt $$ equals

A
$$ \displaystyle \frac{1}{2} $$

B
$$ \displaystyle \frac{1}{3} $$

C
$$ \displaystyle \frac{2}{3} $$

D
None of these

Solution

$$\displaystyle\int_{1}^{\infty}\frac{\log(t-1)}{t^{2}\log

t\log\displaystyle \frac{t}{(t-1)}}dt=\int_{0}^{1}\displaystyle

\frac{\log\left (\displaystyle \frac{1}{x}-1 \right )}{\log

x\log(1-x)}dx$$
(By substituting $$t=\displaystyle\frac{1}{x} dt=\frac{1}{x^{2}dx} \mbox{and} t=1\Leftrightarrow x=1 \mbox{and} t=\infty \Leftrightarrow x=0$$)
$$\therefore I = \displaystyle\int_{0}^{1}\frac{\log(1-x)-\log x}{\log x\log(1-x)}dx$$
$$=\displaystyle\int_{0}^{1}\left (\displaystyle\frac{1}{\log x}-\frac{1}{\log(1-x)}\right )dx$$
$$=\displaystyle\int_{0}^{1}\left (\displaystyle\frac{1}{\log(1-x)}-\frac{1}{\log x}\right )dx = -I\therefore I = 0$$
$$\left [ \displaystyle\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx \right ]$$
Question 9
1 / -0

The value of $$ \displaystyle \int_{0}^{\pi /2}\sin \theta \log \left ( \sin \theta \right )\:d\theta $$ equals

A
$$ \displaystyle \log_{e}\left ( \frac{1}{e} \right ) $$

B
$$ \displaystyle \log _{2}e $$

C
$$ \displaystyle \log_{e}{2}-1 $$

D
$$ \displaystyle \log_{e}\left ( \frac{e}{2} \right ) $$

Solution

Let $$I=\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\sin \theta \log(\sin \theta )d\theta =\displaystyle \frac{1}{2}\int_{0}^{\displaystyle \frac{\pi}{2}}\sin \theta \log(\sin^{2} \theta )d\theta \left ( \because \log_{e}a^{m}=m\log_{e}a \right )$$
$$=\displaystyle\frac{1}{2}\int_{0}^{\displaystyle \frac{\pi}{2}}\sin \theta \log(1-\cos^{2}\theta)d\theta$$
Take $$t=\cos\theta$$ $$\Rightarrow \sin \theta d\theta=-dt$$
$$ \theta=0,t=1$$ and $$\theta =\displaystyle\frac{\pi}{2},t=0$$
Then, $$I=\displaystyle\frac{1}{2}\int_{0}^{1}\log(1-t^{2})dt$$
$$=-\displaystyle \frac{1}{2}\left [ \int_{0}^{1}\left ( t^{2}+\frac{t^{4}}{2}+\frac{t^{6}}{3}+...\infty \right )dt \right ]$$
$$=-\displaystyle\frac{1}{2}\left [\displaystyle\frac{1}{3}+\frac{1}{2\cdot5}+\frac{1}{7\cdot3}+...\infty \right ]$$
$$=-\left [\displaystyle\frac{1}{2.3}+\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+....\infty \right ]$$
$$=-\left [\left (\displaystyle\frac{1}{2}-\frac{1}{3}\right )+\left (\displaystyle\frac{1}{4}-\displaystyle\frac{1}{5} \right )+\left (\displaystyle\frac{1}{6}-\displaystyle \frac{1}{7} \right )+....\infty\right ]$$
$$=-\displaystyle\frac{1}{2}+\frac{1}{3}-\displaystyle\frac{1}{4}+\frac{1}{5}-\displaystyle \frac{1}{6}+\frac{1}{7}+....\infty$$
$$=\left(1-\displaystyle\frac{1}{2}+\frac{1}{3}-\displaystyle\frac{1}{4}+\frac{1}{5}-\displaystyle \frac{1}{6}+\frac{1}{7}+....\infty\right)-1$$
$$=\log_{e}2-1$$
$$=\log_{e}2-\log e=\log_{e}\left (\displaystyle\frac{2}{e} \right )$$

Ans: C
Question 10
1 / -0

If $$\displaystyle I_{n} =\int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx$$

then $$\displaystyle \frac{1}{I_{2}+I_{4}},\frac{1}{I_{3}+I_{5}},\frac{1}{I_{4}+I_{6}}$$ are in?

A
$$A.P$$

B
$$H.P$$

C
$$G.P$$

D
None of these

Solution

$$\displaystyle I_{n}= \int_{0}^{\pi /4}\tan ^{n-2}x\tan ^{2}xdx$$
$$\displaystyle= \int_{0}^{\pi /4}\tan ^{n-2}x\left ( \sec ^{2}x-1 \right

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

^{n-2}xdx$$
$$\displaystyle I_{n}= \left [ \frac{\tan ^{n-1}x}{n-1} \right ]_{0}^{\pi /4}-I_{n-2}$$
$$\displaystyle \therefore I_{n}+I_{n-2}= \frac{1}{n-1}$$
Substitute $$n=4,5,6, ....$$ we get
$$\displaystyle I_{4}+I_{2}, I_{5}+I_{3}, I_{6}+I_{4},....$$ are respectively $$\displaystyle \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\cdots
$$ which are in H.P. and hence their reciprocals are in A.P.

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Re-Attempt Weekly Quiz Competition

Integrals Test - 50

Result

Integrals Test - 50

Score

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SHARING IS CARING

Question 1

10

14

16

18

Solution

Question 2

$$16$$

$$63$$

$$64$$

$$15$$

Solution

Question 3

$$\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx$$

$$\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx$$

$$\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}$$

$$\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}$$

Solution

Question 4

$$0$$

$$\displaystyle -\frac{1}{2}$$

$$\displaystyle \frac{1}{2}$$

$$1$$

Solution

Question 5

$$0$$

$$\displaystyle \pi $$

$$\displaystyle \pi -\frac{1}{\pi }$$

$$\displaystyle \pi +\frac{1}{\pi }$$

Solution

Question 6

$$\displaystyle \frac{22}{7}-\pi $$

$$\displaystyle \frac{2}{105}$$

$$0$$

$$\displaystyle \frac{71}{15}-\frac{3\pi}{2}$$

Solution

Question 7

$$\displaystyle \left ( 0,\frac{1}{8} \right )$$

$$\displaystyle \left ( 0,\frac{1}{9} \right )$$

$$\displaystyle \left ( 0,\frac{1}{10} \right )$$

$$\displaystyle \left ( 0,\frac{1}{7} \right )$$

Solution

Question 8

$$ \displaystyle \frac{1}{2} $$

$$ \displaystyle \frac{1}{3} $$

$$ \displaystyle \frac{2}{3} $$

None of these

Solution

Question 9

$$ \displaystyle \log_{e}\left ( \frac{1}{e} \right ) $$

$$ \displaystyle \log _{2}e $$

$$ \displaystyle \log_{e}{2}-1 $$

$$ \displaystyle \log_{e}\left ( \frac{e}{2} \right ) $$

Solution

Question 10

$$A.P$$

$$H.P$$

$$G.P$$

None of these

Solution

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