Integrals Test

Result

Integrals Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle\int^{\lambda}_0\dfrac{y}{\sqrt{y+\lambda}}dy=?$$

A
$$\dfrac{2}{3}(2-\sqrt{2})\lambda \sqrt{\lambda}$$

B
$$\dfrac{2}{3}(2+\sqrt{2})\lambda\sqrt{\lambda}$$

C
$$\dfrac{1}{3}(2-\sqrt{2})\lambda \sqrt{\lambda}$$

D
$$\dfrac{1}{3}(2+\sqrt{2})\lambda \sqrt{\lambda}$$

Solution
Question 2
1 / -0

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:

A
$$\dfrac { 1 }{ { 1+cos }^{ 3 }x } +c$$

B
$$\dfrac { -1 }{ { 1+cos }^{ 3 }x } +c$$

C
$$\dfrac { 1 }{ 3\left( 1+{ tan }^{ 3 }x \right) } +c$$

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

Solution
Question 3
1 / -0

$$\int_{1}^{\infty }(e^{x+1}+e^{3-x})^{-1}dx$$ is equal to:

A
$$\frac{\pi }{4e^{2}}$$

B
$$\frac{\pi }{4e}$$

C
$$\frac{1}{e^{2}}(\frac{\pi }{2}-tan^{-1}\frac{1}{e})$$

D
$$\frac{\pi }{2e^{2}}$$

Solution
Question 4
1 / -0

The value of the integral $$\displaystyle \int _{ { e }^{ -1 } }^{ { e }^{ 2 } }{ \left| \frac { \log _{ e }{ x } }{ x } \right| dx } $$ is

A
$$\dfrac {3}{2}$$

B
$$\dfrac {5}{2}$$

C
$$3$$

D
$$5$$

Solution
Question 5
1 / -0

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

A
$$\sin{\alpha}$$

B
$$\alpha \sin {\alpha}$$

C
$$\dfrac { \alpha } { 2 \sin{\alpha} }$$

D
$$\dfrac { \alpha } { 2 } \sin {\alpha}$$

Solution

$$ \int_{0}^{1}\frac{dx}{x^{2}+2x\, cos\, \alpha +1} = \int_{0}^{1}\frac{dx}{x^{2}+2x\, cos\, \alpha +cos^{2} \alpha +sin^{2}\alpha }$$
$$ = \int_{0}^{1}\frac{dx}{(x+cos\,\alpha )^{2}+sin^{2}\alpha } $$
$$ = \frac{1}{sin\,\alpha } \left [ tan^{-1} \left ( \frac{x+cos\,\alpha }{sin\,\alpha } \right ) \right ]_{0}^{1}$$
$$ = \frac{1}{sin\,\alpha } \left [ tan^{-1}\left ( \frac{1+cos\alpha }{sin\,\alpha } \right )- tan^{-1}\left ( \frac{cos\, \alpha }{sin\,\alpha } \right ) \right ]$$
$$ = \frac{1}{sin\, \alpha } \left [ tan^{-1}\left ( \frac{2\, cos^{2}\alpha /2}{2\, sin\, \alpha /2 \, cos\,\alpha /2} \right ) - tan^{-1}\left ( cot\alpha \right )\right ]$$
$$ = \frac{1}{sin \, \alpha} \left [ tan^{-1}\left ( cot+\,\alpha/2 \right ) - tan^{-1}\left ( cot\alpha \right ) \right ]$$
$$ = \frac{1}{sin\, \alpha}\left [ tan^{-1}\left ( tan\left ( \pi /2 - \alpha/2 \right ) \right ) - tan^{-1}\left ( tan\left ( \pi /2 - \alpha \right ) \right ) \right ]$$
$$ = \frac{1}{sin \alpha} \left [ \pi/2 - \alpha/2 - \pi/2 + \alpha \right ]$$
$$ = \frac{\alpha}{2 sin\, \alpha}$$
Question 6
1 / -0

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

A
$$\displaystyle \frac { \pi } { 4 } + \frac { 1 } { 2 }$$

B
$$\displaystyle \frac { \pi } { 4 } - \frac { 1 } { 2 }$$

C
$$\dfrac { \pi } { 4 }$$

D
$$\dfrac { \pi } { 4 } + 1$$

Solution

Let $$I=\displaystyle\int_{0}^{\pi/4}\dfrac{x.\sin x}{\cos^3 x}dx$$

Applying by-parts formula, we get

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

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

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

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

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

$$=\dfrac{\pi}{4}-\dfrac{1}{2}$$
Question 7
1 / -0

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

A
$$\frac{1}{2}+\frac{\pi }{2\sqrt{3}}$$

B
$$\frac{1}{2}-\frac{\pi }{2\sqrt{3}}$$

C
$$\frac{1}{2}+\frac{\pi }{4\sqrt{3}}$$

D
$$\frac{1}{2}-\frac{\pi }{4\sqrt{3}}$$

Solution
Question 8
1 / -0

Value of the definite integral $$\displaystyle \int_{-1}^{1}\dfrac {dx}{(1+x^{3}+\sqrt {1+x^{6}})}$$

A
$$is\dfrac {1}{2}$$

B
$$is\ 1$$

C
$$is\ 2$$

D
$$is\ zero$$

Solution
Question 9
1 / -0

If $$f(x)=\begin{vmatrix} \cos { x } & 1 & 0 \\ 1 & 2\cos { x } & 1 \\ 0 & 1 & 2\cos { x } \end{vmatrix}$$ then $$\displaystyle\int _{ 0 }^{ \pi /2 }{ f\left( x \right) } dx$$ is equal to

A
$$1/4$$

B
$$-1/3$$

C
$$1/2$$

D
$$none of these $$

Solution
Question 10
1 / -0

The value of $$\int_{0}^{[x]} (x-[x])dx$$, where $$[x]$$ is the greatest integer $$|le x$$ is equal to

A
$$4[x]$$

B
$$2[x]$$

C
$$\dfrac{1}{2} [x]$$

D
$$\dfrac{1}{5} [x]$$