Integrals Test

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Integrals Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

The value of integral $$\displaystyle \int_{0}^{\infty }\frac{x\log x}{(1+x^2)^2} \: dx$$ is

A
$$0$$

B
$$\log 7$$

C
$$ 5\log 13$$

D
$$none\ of\ these$$

Solution

$$ \displaystyle I=\int_{0}^{\infty }\frac{x\log xdx}{(1+x^2)^2}$$
Let $$x=\frac{1}{t}$$
$$\displaystyle I=\int_{0}^{\infty }\frac{(\frac{1}{t})\log (\frac{1}{t}) (-\frac{1}{t^2})dt}{(1+(\frac{1}{t^2}))^2}$$
$$\displaystyle I=-\int_{0}^{\infty }\frac{t\log t}{(1+t^2)^2}dt=-I$$
or $$I=0$$
Question 2
1 / -0

Let $$\displaystyle \frac{{df(x)}}{{dx}} = \frac{{{e^{\sin x}}}}{x}, x>0$$. If $$\displaystyle \int_1^4 {\frac{{3{e^{\sin {x^3}}}}}{x}dx = f(k) - f(1)} $$ then one of the possible values of k is

A
$$16$$

B
$$63$$

C
$$64$$

D
$$15$$

Solution

$$\displaystyle \dfrac{df(x)}{dx}=\displaystyle \dfrac{e^{sin x}}{x} , x > 0, \int_1^4 \dfrac{3e^{sin x^3}}{x}dx=f(k)-f(1)$$
substitute $$x^3=p$$
$$dp=3x^2 dx$$
limit of p will be rom $$1 \rightarrow 64$$ as of x was $$1 \rightarrow 4$$
So, $$\displaystyle \int_1^{64} \dfrac{3e^{sin p}}{x} \cdot \dfrac{dp}{3x^2}=\int_1^{64} \dfrac{e^{sin p}}{p}dp=\int_1^{64}df(p)=f(64)-f(1)$$
Question 3
1 / -0

Let $$f(0) = 0$$ and $$\displaystyle \int_{0}^{2}{f}'(2t)e^{f(2t)} \:dt=5$$.
Then the value of $$f (4)$$ is?

A
$$\log 2$$

B
$$\log 7$$

C
$$\log 11$$

D
$$\log 13$$

Solution

We have $$\displaystyle \int_{0}^{2 }{f}'(2t)e^{f(2t)}dt=5$$

Substitute $$e^{f(2t)}=y$$

$$\therefore 2{f}'(2t)e^{f(2t)}dt=dy$$

$$\therefore \dfrac{1}{2} \displaystyle \int_{e^{f(0)}}^{e^{f(4)}} dy=5$$

or $$\displaystyle \int_{e^{f(0)}}^{e^{f(4)}} dy=10$$

or $$e^{f(4)}-e^{f(0)}=10$$

or $$e^{f(4)}=10+1=11$$

or $$f(4)=\log 11$$
Question 4
1 / -0

$$\displaystyle \int_{-3\pi/2}^{-\pi/2}[(x+\pi)^{3}+\cos^{2}(x+3\pi)]dx=$$

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

B
$$\displaystyle \frac{\pi}{4}-1$$

C
$$\displaystyle \frac{\pi^{4}}{32}$$

D
$$\displaystyle \frac{\pi^{4}}{32}+\frac{\pi}{2}$$

Solution

Let $$I=\int { \left( x+\pi \right) ^{ 3 }dx } =\cfrac { 1 }{ 4 } \left( x+\pi \right) ^{ 4 }$$
And $$J=\int { cos^{ 2 }\left( x+3\pi \right) dx } =\int { cos^{ 2 }xdx } \\ =\int { \left( \cfrac { 1 }{ 2 } cos2x+\cfrac { 1 }{ 2 } \right) dx } =\cfrac { 1 }{ 2 } \int { cos2xdx } +\cfrac { 1 }{ 2 } \int { dx } \\ =\cfrac { 1 }{ 4 } sin2x+\cfrac { x }{ 2 } $$
Therefore,
$$\int _{ -\cfrac { 3\pi }{ 2 } }^{ -\cfrac { \pi }{ 2 } }{ \left[ \left( x+\pi \right) ^{ 3 }+cos^{ 2 }\left( x+3\pi \right) \right] dx } \\ =\left[ \cfrac { 1 }{ 4 } \left( x+\pi \right) ^{ 4 }+\cfrac { 1 }{ 4 } sin2x+\cfrac { x }{ 2 } \right] _{ -\cfrac { 3\pi }{ 2 } }^{ -\cfrac { \pi }{ 2 } }=\cfrac { \pi }{ 2 } $$
Question 5
1 / -0

The value of the integral $$\displaystyle \int_{0}^{1}\frac{1}{(x^{2}+1)^{3/2}} dx$$ is

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

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

C
$$1$$

D
$${\sqrt{2}}$$

Solution

Let $$I=\int { \cfrac { 1 }{ \left( x^{ 2 }+1 \right) ^{ \cfrac { 3 }{ 2 } } } dx } $$
Substituting $$x=tant\Rightarrow dx=sec^{ 2 }tdt$$, we get
$$I=\int { cot\left( t \right) dt } =sin\left( t \right) =sin\left( tan^{ -1 }x \right) =\cfrac { x }{ \sqrt { x^{ 2 }+1 } } $$
Therefore,
$$\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ \left( x^{ 2 }+1 \right) ^{ \cfrac { 3 }{ 2 } } } dx } =\left[ \cfrac { x }{ \sqrt { x^{ 2 }+1 } } \right] _{ 0 }^{ 1 }=\cfrac { 1 }{ \sqrt { 2 } } $$
Question 6
1 / -0

If $$\displaystyle \int^{x}_{\log 2} \dfrac{1}{\sqrt{e^{x}-1}}dx = \dfrac{\pi}{6}$$

then $$x $$ is equal to?

A
$$e^{2}$$

B
$$1/e$$

C
$$\log 4$$

D
$$\log 2$$

Solution

Let $$\displaystyle I=\int { \cfrac { 1 }{ \sqrt { e^{ x }-1 } } dx } $$
Substituting $$t=e^{ x }\Rightarrow dt=e^{ x }dx$$, we get
$$\displaystyle I=\int { \cfrac { 1 }{ t\sqrt { t-1 } } dt } $$
Again substituting $$u=t-1\Rightarrow du=dt$$, we get
$$\displaystyle I=\int { \cfrac { 1 }{ \sqrt { u } \left( u+1 \right) } du } $$
now substituting $$s=\sqrt { u } \Rightarrow ds=\cfrac { 1 }{ 2\sqrt { u } } du$$, we get
$$\displaystyle I=2\int { \cfrac { 1 }{ s^{ 2 }+1 } ds } \\ =2tan^{ -1 }s=2tan^{ -1 }\sqrt { u } =2tan^{ -1 }\sqrt { t-1 } =2tan^{ -1 }\sqrt { e^{ x }-1 } $$
Therefore,
$$\displaystyle\int _{ \log { 2 } }^{ x }{ 2tan^{ -1 }\left( \sqrt { e^{ x }-1 } \right) } =\cfrac { \pi }{ 6 } \Rightarrow \left[ 2tan^{ -1 }\left( \sqrt { e^{ x }-1 } \right) \right] _{ \log { 2 } }^{ x }=\cfrac { \pi }{ 6 } $$
$$\therefore x=\log{4}$$
Question 7
1 / -0

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

A
$$4$$

B
$$\displaystyle -4$$

C
$$16$$

D
none of these

Solution

$$\int _{ 1 }^{ 4 }{ \cfrac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =F\left( k \right) -F\left( 1 \right) $$
Substitute $${ x }^{ 2 }=t\Rightarrow 2xdx=dt$$
$$\int _{ 1 }^{ 16 }{ 2\cfrac { { e }^{ \sin { t } } }{ t } \cfrac { dt }{ 2 } } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow \int _{ 1 }^{ 16 }{ \cfrac { { e }^{ \sin { t } } }{ t } dt } =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow { \left[ F\left( t \right) \right] }_{ 1 }^{ 16 }=F\left( k \right) -F\left( 1 \right) \\ \Rightarrow F\left( 16 \right) -F\left( 1 \right) =F\left( k \right) -F\left( 1 \right) \\ \Rightarrow k=16$$
Question 8
1 / -0

If $$\displaystyle \int_{0}^{1}xe^{x^{2}}dx= \lambda \int_{0}^{1}e^{x^{2}}dx$$ then $$\lambda $$

A
$$\displaystyle \lambda = 0$$

B
$$\displaystyle \lambda \epsilon = \left ( 0,1 \right )$$

C
$$\displaystyle \lambda \epsilon = \left (-\infty, 0 \right )$$

D
$$\displaystyle \lambda \epsilon = \left ( 1, 2 \right )$$

Solution

For $$x\in(0,1), x e^{x^2}< e^{x^2}$$
$$\Rightarrow \displaystyle \int_0^1xe^{x^2}<\int_0^1e^{x^2}dx$$
Also $$x e^{x^2} >0 \forall x\in(0,1)$$
$$\Rightarrow \displaystyle \int_0^1xe^{x^2} >0$$
Hence $$\lambda \in(0,1)$$
Question 9
1 / -0

The value of $$\displaystyle \int_{1}^{2}\left [ f\left \{ g\left ( x \right ) \right \} \right ]^{-1}.{f}'\left \{ g\left ( x \right ) \right \}.{g}'\left ( x \right )dx,$$ where $$\displaystyle g\left ( 1 \right )=g\left ( 2 \right ),$$ is equal to?

A
$$1$$

B
$$2$$

C
$$0$$

D
none of these

Solution

Let $$I=\int _{ 1 }^{ 2 } \left[ f\left\{ g\left( x \right) \right\} \right] ^{ -1 }.{ f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx$$

Substitute $$\left[ f\left\{ g\left( x \right) \right\} \right] =t\Rightarrow { f }'\left\{ g\left( x \right) \right\} .{ g }'\left( x \right) dx=dt$$

$$\displaystyle \therefore I=\int _{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }{ tdt } ={ \left[ \frac { { t }^{ 2 } }{ 2 } \right] }_{ f\left( g\left( 1 \right) \right) }^{ f\left( g\left( 2 \right) \right) }=0\\ \because g\left( 1 \right) =g\left( 2 \right) $$
Question 10
1 / -0

The value of $$\displaystyle \int_{0}^{\pi /4}\frac{\sec x}{\left ( \sec x+\tan x \right )^{2}}dx$$ is

A
$$\displaystyle 1+\sqrt{2}$$

B
$$\displaystyle -\left ( 1+\sqrt{2} \right )$$

C
$$\displaystyle -\sqrt{2}$$

D
none of these

Solution

Let $$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } } dx$$

Multiply numerator and denominator by $$\cos ^{ 2 }{ x } $$, we get

$$\displaystyle I=\int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ \dfrac { 1 }{ { \left( u+1 \right) }^{ 2 } } } du=\left[ \dfrac { -1 }{ u+1 } \right] _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }$$

$$\displaystyle =-\dfrac { 1 }{ \dfrac { 1 }{ \sqrt { 2 } } +1 } $$

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

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Re-Attempt Weekly Quiz Competition

Integrals Test - 26

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Integrals Test - 26

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SHARING IS CARING

Question 1

$$0$$

$$\log 7$$

$$ 5\log 13$$

$$none\ of\ these$$

Solution

Question 2

$$16$$

$$63$$

$$64$$

$$15$$

Solution

Question 3

$$\log 2$$

$$\log 7$$

$$\log 11$$

$$\log 13$$

Solution

Question 4

$$\displaystyle \frac{\pi}{2}$$

$$\displaystyle \frac{\pi}{4}-1$$

$$\displaystyle \frac{\pi^{4}}{32}$$

$$\displaystyle \frac{\pi^{4}}{32}+\frac{\pi}{2}$$

Solution

Question 5

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

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

$$1$$

$${\sqrt{2}}$$

Solution

Question 6

$$e^{2}$$

$$1/e$$

$$\log 4$$

$$\log 2$$

Solution

Question 7

$$4$$

$$\displaystyle -4$$

$$16$$

none of these

Solution

Question 8

$$\displaystyle \lambda = 0$$

$$\displaystyle \lambda \epsilon = \left ( 0,1 \right )$$

$$\displaystyle \lambda \epsilon = \left (-\infty, 0 \right )$$

$$\displaystyle \lambda \epsilon = \left ( 1, 2 \right )$$

Solution

Question 9

$$1$$

$$2$$

$$0$$

none of these

Solution

Question 10

$$\displaystyle 1+\sqrt{2}$$

$$\displaystyle -\left ( 1+\sqrt{2} \right )$$

$$\displaystyle -\sqrt{2}$$

none of these

Solution

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