Integrals Test

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Integrals Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

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

A
tanx

B
$$e^{x}-1$$

C
sinx

D
2sinx

Solution

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

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

$$c=1$$

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

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

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

$$f(x)=sin x$$
Question 2
1 / -0

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

A
$$16$$

B
$$62$$

C
$$64$$

D
$$15$$

Solution

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, $$k=64$$
Question 3
1 / -0

$$\displaystyle \int_{0}^{1}\displaystyle \frac{x(2x^{2}+1)}{x^{8}+2x^{6}-x^{2}+1}dx=$$

A
$$\displaystyle \frac{\pi }{3}$$

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

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

D
$$\displaystyle \frac{\pi }{6}$$

Solution

Let $$\displaystyle I=\int_{0}^{1}\displaystyle \frac{x(2x^{2}+1)}{x^{8}+2x^{6}-x^{2}+1}dx$$

$$=\int_{0}^{1}\displaystyle \frac{(2x^{3}+x)}{(x^{4}+x^{2})^{2}-(x^{4}+x^{2})+1}dx$$

Put $$t=x^{4}+x^{2}$$
$$\Rightarrow dt =4x^{3}+2x$$
Also $$t=0 $$ for $$ x=0 $$ and $$ t=2 $$ for $$ x=1 $$

So, $$I=\displaystyle \frac{1}{2} \int_{0}^{2} \frac{dt}{t^{2}-t+1}$$

$$=\displaystyle \frac{1}{2}\int_{0}^{2}\displaystyle \frac{dt}{\left ( t-\displaystyle \frac{1}{2} \right )^{2}+\left ( \displaystyle \frac{\sqrt{3}}{2} \right )^{2}}$$

$$=\displaystyle \frac{1}{2}.\displaystyle \frac{2}{\sqrt{3}}\left \{ tan^{-1}\displaystyle \frac{2\left ( t-\displaystyle \frac{1}{2} \right )}{\sqrt{3}} \right \}_{0}^{2}$$

$$=\displaystyle \frac{1}{\sqrt{3}}\left ( tan^{-1} (\sqrt{3})-tan^{-1}\left ( -\displaystyle \frac{1}{\sqrt{3}} \right )\right )$$

$$=\displaystyle \frac{1}{\sqrt{3}}\left ( \displaystyle \frac{\pi }{3} +\displaystyle \frac{\pi }{6}\right )$$

$$=\displaystyle \frac{\pi }{2\sqrt{3}}$$
Question 4
1 / -0

If $$\mathrm{a}>\mathrm{b}$$ then $$\displaystyle \int_{0}^{\pi}\frac{\mathrm{d}\mathrm{x}}{\mathrm{a}+\mathrm{b}\sin \mathrm{x}}=$$

A
$$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

B
$$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

C
$$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

D
$$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

Solution
Question 5
1 / -0

If $$I_{n}=\displaystyle \int_{0}^{\infty}e^{-x}x^{n-1} dx$$, then $$\displaystyle \int_{0}^{\infty}e^{-\lambda x}x^{n-1} dx$$ is equal to?

A
$$\lambda I_{\mathrm{n}}$$

B
$$\displaystyle \frac{1}{\lambda}I_{\mathrm{n}}$$

C
$$\displaystyle \frac{I_{\mathrm{n}}}{\lambda^{\mathrm{n}}}$$

D
$$\lambda^{n}I_{n}$$

Solution

$$I_n=\displaystyle \int_{0}^{\infty }e^{-x}x^{n-1}dx$$

then $$ I_{1}=\displaystyle \int_{0}^{\infty }e^{-x}x^{n-1}dx$$

$$I_{1}=\displaystyle \int_{0}^{\infty }e^{-x}\dfrac{((\lambda x)^{n-1})}{\lambda }dx$$

$$I_{1}=\dfrac{1}{\lambda ^{n}}\displaystyle \int_{0}^{\infty }e^{-t}t^{n-1}dt$$

$$I_{1}=\dfrac{I_n}{\lambda ^{n}}.$$

$$ So, \displaystyle \int_{0}^{\infty }e^{-t}x^{n-1}dt=\dfrac{I_n}{\lambda^n }.$$
Question 6
1 / -0

The value of $$\displaystyle \int_{0}^{\pi}\dfrac {dx}{1+2sin^2x}$$ is

A
$$0$$

B
$$\dfrac {\pi}{2\sqrt 3}$$

C
$$\dfrac {\pi}{\sqrt 3}$$

D
None of these

Solution

$$\displaystyle I=\displaystyle \int_{0}^{\pi}\frac {dx}{1+2sin^2x} $$

$$\displaystyle =2\displaystyle \int_{0}^{\pi /2}\frac {dx}{1+2sin^2x}$$ $$\because \displaystyle \int_{0}^{2a}f(x)dx=2\displaystyle \int_{0}^{a} f(x)dx $$ if $$f(2a-x)=f(x)$$

$$\displaystyle =2\displaystyle \int_{0}^{\pi /2}\frac {\sec^2xdx}{3 \tan^2x+1}$$

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

$$\displaystyle I=2\displaystyle \int_{0}^{\infty}\frac {dt}{3t^2+1}$$

$$\displaystyle =\frac {2}{\sqrt 3}tan^{-1}\sqrt {3t}\mid_{0}^{\infty}$$

$$\displaystyle =\frac {\pi}{\sqrt 3}$$
Question 7
1 / -0

If $$f(x) = x - x^2 +1$$ & $$g(x)=max\left \{ f(t) ;0\leq t< x \right \}$$, then $$\overset {1}{\underset { 0 }{ \int } } g (x) dx = ?$$

A
$$\dfrac{29}{24}$$

B
$$\dfrac{7}{6}$$

C
$$\dfrac{5}{4}$$

D
none of these

Solution

$$f(x) = x - x^2 +1$$ & $$g(x)=max\left \{ f(t) ;0\leq t< x \right \}$$
y = g(x)
So, $$\overset {1}{\underset { 0 }{ \int } }g (x) dx=\overset {1/2}{\underset { 0 }{ \int } } f(x)dx+\left [ 1-\dfrac{1}{2} \right ]\times\dfrac{5}{4}$$
$$=\left [ -\dfrac{x^3}{3}+\dfrac{x^2}{2}+x\right ]_0^{1/2}+\dfrac{5}{8}=\dfrac{29}{24}$$
Question 8
1 / -0

Directions For Questions

If $$a=\sum_{r=1}^{\infty }\frac{1}{2r^{2}-r},b=\sum_{r=1}^{\infty }\frac{1}{2r^{2}+r},c=\sum_{r=1}^{\infty }\frac{1}{4r^{3}-1}$$, then answer the following

...view full instructions

What is the value of a + b ?

A
2

B
1

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

D
$$\frac{5}{2}$$
Question 9
1 / -0

The value of $$\displaystyle {\int}_0^{\pi}\frac{dx}{1+2 \sin^2 x}$$ is

A
$$0$$

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

C
$$\displaystyle \frac{\pi}{\sqrt{3}}$$

D
none of these

Solution

Let $$\displaystyle I={\int}_0^{\pi} \frac{dx}{1+2 \sin^2 x}$$

Since, $$\sin (\pi-x)=\sin x$$
$$\displaystyle =2\int _{0}^{\frac{\pi}{2}} \frac{dx}{1+2\sin^{2}x}$$

$$\displaystyle=2\int _{0}^{\frac{\pi}2}\frac{\sec^{2}x dx}{\sec^{2}x+2\tan^{2} x}=2\int _{0}^{\frac{\pi}2}\frac{\sec^{2}x dx}{1+3\tan^{2} x}$$

Put $$\tan x=t\Rightarrow \sec^{2}x dx=dt$$

So, $$\displaystyle I=2\int _{0}^{\infty} \frac{dt}{1+3t^{2}} = \frac{2}{\sqrt{3}} [\tan^{-1}{(\sqrt{3}t)}]_{0}^{\infty}$$

$$\displaystyle I=\frac{\pi}{\sqrt{3}}$$
Question 10
1 / -0

$$\displaystyle \int_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\sin^{2}x. \cos^{3} x dx$$ is equal to

A
$$\displaystyle \dfrac{4\pi}{30}$$

B
$$\displaystyle \dfrac{4}{15}$$

C
$$\displaystyle \dfrac{\pi}{30}$$

D
$$\displaystyle \dfrac{1}{15}$$

Solution

Let $$\displaystyle I=\int _{ \dfrac { -\pi }{ 2 } }^{ \dfrac { \pi }{ 2 } }{ \sin ^{ 2 }{ x\left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } } $$

Substitute $$\sin { x } =t\quad \Rightarrow \cos { xdx=dt } $$
$$\displaystyle I=\int _{ -1 }^{ 1 }{ \left( { t }^{ 2 }\left( 1-{ t }^{ 2 } \right) \right) } dt=\int _{ -1 }^{ 1 }{ \left( { t }^{ 2 }-{ t }^{ 4 } \right) } dt$$

$$\displaystyle ={ \left( \frac { { t }^{ 3 } }{ 3 } -\frac { { t }^{ 5 } }{ 5 } \right) }_{ -1 }^{ 1 }=\frac { 4 }{ 15 } $$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 47

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Integrals Test - 47

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SHARING IS CARING

Question 1

tanx

$$e^{x}-1$$

sinx

2sinx

Solution

Question 2

$$16$$

$$62$$

$$64$$

$$15$$

Solution

Question 3

$$\displaystyle \frac{\pi }{3}$$

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

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

$$\displaystyle \frac{\pi }{6}$$

Solution

Question 4

$$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

$$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

$$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

$$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$$

Solution

Question 5

$$\lambda I_{\mathrm{n}}$$

$$\displaystyle \frac{1}{\lambda}I_{\mathrm{n}}$$

$$\displaystyle \frac{I_{\mathrm{n}}}{\lambda^{\mathrm{n}}}$$

$$\lambda^{n}I_{n}$$

Solution

Question 6

$$0$$

$$\dfrac {\pi}{2\sqrt 3}$$

$$\dfrac {\pi}{\sqrt 3}$$

None of these

Solution

Question 7

$$\dfrac{29}{24}$$

$$\dfrac{7}{6}$$

$$\dfrac{5}{4}$$

none of these

Solution

Question 8

2

1

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

$$\frac{5}{2}$$

Question 9

$$0$$

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

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

none of these

Solution

Question 10

$$\displaystyle \dfrac{4\pi}{30}$$

$$\displaystyle \dfrac{4}{15}$$

$$\displaystyle \dfrac{\pi}{30}$$

$$\displaystyle \dfrac{1}{15}$$

Solution

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