Integrals Test

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Integrals Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \int_{0}^{1}\frac{d}{dx}$$ $$tan^{-1} \left( \displaystyle \frac { 1 }{ x } \right) $$

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

B
$$\dfrac{-\pi}{4}$$

C
$$\dfrac{-\pi}{2}$$

D
$$\dfrac{\pi}{2}$$

Solution

Let $$\displaystyle I=\int_{0}^{1}\frac{d}{dx}$$ $$\tan^{-1} (\displaystyle \frac{1}{x})dx$$

$$\displaystyle \frac{d}{dx}$$ $$\tan^{-1} (\displaystyle \frac{1}{x}) = \frac{1}{1+(\frac{1}{x})^{2}} \times \frac{(-1)}{x^{2}}=-\frac{1}{1+x^{2}}$$

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

$$=\displaystyle [\cot^{-1}x]_{0}^{1}+C$$

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

Evaluate: $$\int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ ({ \tan }^{ 4 }x+{ \tan }^{ 2 }x)dx }$$

A
$$1$$

B
$$1/2$$

C
$$1/3$$

D
$$1/4$$

Solution

$$\int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ ({ \tan }^{ 4 }x+{ \tan }^{ 2 }x)dx } \\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x({ \tan }^{ 2 }x+1)dx } \\ = \\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x(\sec ^{ 2 }x)dx } $$

We know that:
$$\int { [f(x)]^{ n } } f'(x)dx=\dfrac { [f(x)]^{ n+1 } }{ n+1 } +C\\ = \int _{ 0 }^{ \tfrac { \pi }{ 4 } }{ { \tan }^{ 2 }x\; \times \; \sec ^{ 2 }x\; dx } \\ = \left[\dfrac { { \tan }^{ 3 }x }{ 3 } \right]_{ 0 }^{ \tfrac { \pi }{ 4 } }\\ = \dfrac { 1 }{ 3 } [\tan ^{ 3 }(\tfrac { \pi }{ 4 }) -\tan ^{ 3 }0]\\ = \dfrac { 1 }{ 3 } [1-0]\\ = \dfrac { 1 }{ 3 }$$
Question 3
1 / -0

Evaluate: $$\displaystyle \int_{1/2}^{1}\frac{1}{\sqrt{x-x^{2}}}dx$$

A
$$\pi/8$$

B
$$\pi/4$$

C
$$\pi/2$$

D
$$\pi$$

Solution

$$ \displaystyle\int_{1/2}^{1} \dfrac{1}{\sqrt{x-x^2}}dx$$

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

we know that $$\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}dx=\sin^{-1} {\dfrac{x}{a}}$$

Now, our integration becomes:-

$$\displaystyle\left[\sin^{-1} \left(\dfrac{x-\dfrac{1}{2}}{\dfrac{1}{2}}\right)\right]_{1/2}^{1}$$

$$=\left[\sin^{-1} {1}-\sin^{-1} {0}\right]$$

$$=\dfrac{\pi}{2}$$

Hence,answer is option_(C).
Question 4
1 / -0

If $$f(x)=\begin{vmatrix}\sin x+\sin2x+\sin3x & \sin2x & \sin3x\\ 3+4\sin x & 3 & 4\sin x\\ 1+\sin x & \sin x & 1\end{vmatrix} $$, then the value of $$\displaystyle \int_{0}^{\frac{\pi}2}f(x)dx$$, is

A
$$3$$

B
$$\dfrac23$$

C
$$\dfrac13$$

D
$$0$$

Solution

Using Determinant properties:
$$C_1 \rightarrow C_1 - (C_2+C_3)$$

We get,
$$\begin{vmatrix} \sin\ x &\sin\ 2x &\sin\ 3x \\0 &3 &4\sin\ x \\0 &\sin\ x &1 \end{vmatrix}$$
$$=\sin\ x(3-4 \sin^{2}x)$$
$$=3\ \sin\ x-4\ \sin^{3}x$$
$$=\sin\ 3x$$

$$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin\ 3x\ dx=\left.\dfrac{-\cos\ 3x}{3}\right|_{0}^{\frac{\pi}{2}}=0+\dfrac{1}{3}$$
$$=\dfrac{1}{3}$$
Question 5
1 / -0

Evaluate: $$\displaystyle \int_{0}^{\pi /4}\tan^{6}xdx$$

A
$$\displaystyle \frac{13}{15}-\frac{\pi}{4}$$

B
$$\displaystyle \frac{13}{15}+\frac{\pi}{4}$$

C
$$\displaystyle \frac{\pi}{4}-\frac{2}{3}$$

D
$$\displaystyle \frac{13}{15}-4\pi$$

Solution

$$\tan^6{x}=\tan^4{x}(\sec^2{x}-1)=\tan^4{x}\sec^2{x}-\tan^4{x}$$

$$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}(\sec^2{x}-1)$$

$$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}\sec^2{x}+\tan^2{x}$$

$$\implies\tan^6{x}=\tan^4{x}\sec^2{x}-\tan^2{x}\sec^2{x}+\sec^2{x}-1$$

Now, let $$z=\tan{x}\implies dz=\sec^2{x}dx$$ And when $$ x=0, z=0$$ and when $$x=\dfrac{\pi}{4}, z=1$$

Hence, $$\displaystyle\int_{0}^{\pi/4}\tan^6{x}dx=\int_{0}^{\pi/4}\tan^4{x}\sec^2{x}dx-\int_{0}^{\pi/4}\tan^2{x}\sec^2{x}dx+\int_{0}^{\pi/4} \sec^2{x}dx-\int_{0}^{\pi/4} 1.dx$$

$$=\displaystyle\int_{0}^{1}z^4dz-\int_{0}^{1}z^2dz+\int_{0}^{1}dz-\int_{0}^{\pi/4}1.dx$$

$$=\left[\dfrac{z^5}{5}\right]_{0}^{1}-\left[\dfrac{z^3}{3}\right]_{0}^{1}+\left[z\right]_{0}^{1}-\left[x\right]_{0}^{\pi/4}$$

$$=\dfrac{1}{5}-\dfrac{1}{3}+1-\dfrac{\pi}{4}$$

$$=\dfrac{13}{15}-\dfrac{\pi}{4}$$
Question 6
1 / -0

$$(\displaystyle \sum_{n=1}^{10}\int_{-2n-1}^{-2n}\sin^{27}xdx)+(\sum_{n=1}^{10}\int_{2n}^{2n+1}\mathrm{s}in^{27}$$ $$xdx)=$$

A
$$27^{2}$$

B
-54

C
54

D
0

Solution

$$\sum_{n=1}^{10}\int_{-2n-1}^{-2n}sin^{27}x\ dx+\sum_{n=1}^{10}\int_{2n+1}^{2n}sin^{27}x\ dx$$
$$I=\int_{-21}^{21}sin^{27}x\ dx$$
$$I=\int_{-21}^{21}sin^{27}x\ dx-\int_{-21}^{21}sin^{27}x\ dx$$
$$\to I=-I \Rightarrow I=0$$
Question 7
1 / -0

lf $$I_{\mathrm{n}}=\displaystyle \int_{0}^{\dfrac{\pi}{4}}$$ $$\tan^{n} xdx$$, then $$\displaystyle \lim_{n\rightarrow\infty}n[I_{n}+I_{n+2}]=$$

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

B
$$1$$

C
$$\infty$$

D
$$0$$

Solution

Given, $$I_{n}=\int_{0}^{\pi /4}\tan ^{n}x dx$$
and we have to find $$\lim_{n\rightarrow \infty }n[I_{n}+I_{n+2}]$$
Let us first solve the function and get the answer in terms of $$n$$ and then apply limit .
Now $$n[I_{n}+I_{n+2}]$$
$$=n[\int_{0}^{\pi /4}\tan^{n}x dx $$ $$+\int_{0}^{\pi /4}\tan^{n+2}xdx]$$
$$=n[\int_{0}^{\pi/4}(\tan^{n}x+\tan^{n+2}x)dx]$$
Since the limits of both the integrals is same it can be combined into one integral
$$=n[\int_{0}^{\pi/4}(\tan^{n}x(1+\tan^{2}x)dx]$$
Since $$1+\tan^{2}x=\sec^{2}x$$
Therefore, $$=n[\int_{0}^{\pi /4}\tan^{n}x\sec^{2}x]$$
Taking $$\tan x=t$$
$$dx=\sec^{2}xdt $$
Upper limit becomes $$\tan(\pi/4)=1$$ and lower limit becomes $$\tan(0)=0$$
Using this the integrand becomes
$$n\int_{0}^{1} t^{n}dt$$
$$=n(t^{n+1}/n+1)_{0}^{1}$$
Substituting limits, we get $$\dfrac {n}{n+1}$$
Now applying limit tending to infinity to this value
$$\lim_{n\rightarrow \infty }n[I_{n}+I_{n+2}]$$
$$=\lim_{n \to \infty }\dfrac {n}{n+1}$$
$$=1$$
Question 8
1 / -0

The value of $$\displaystyle \int_{3}^{5}\frac{x^{2}}{x^{2}-4} dx$$ is

A
$$2(1-\displaystyle \mathrm{l} \mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$$

B
$$2(1+\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$$

C
$$2(1+4\log_{\mathrm{e}}3-4\log_{\mathrm{e}}7+4\log_{\mathrm{e}}5)$$

D
$$2(1-\displaystyle \tan^{-1}(\frac{15}{7}))$$

Solution

$$\displaystyle \int_{3}^{5}\dfrac{x^{2}}{x^{2}-4}dx$$

$$\displaystyle =\int_{3}^{5}1dx+4\int_{3}^{5}\dfrac{1}{x^{2}-4}dx=2+4.\dfrac{1}
{2}log(\dfrac{x-2}{x+2})|_{3}^{5}$$

$$\displaystyle 2+2[log(3/7)-log(1/5)]$$

$$\displaystyle =2(1+log(\dfrac{15}{7}))$$
Hence, option 'B' is correct.
Question 9
1 / -0

Evaluate the integral
$$\displaystyle \int_{1}^{\sqrt[7]{2}}\frac{1}{x(2x^{7}+1)} dx$$

A
$$\log \dfrac{6}{5}$$

B
$$6 \log \dfrac{6}{5}$$

C
$$\dfrac{1}{7} \log \dfrac{6}{5}$$

D
$$\dfrac{1}{5} \log \dfrac{6}{5}$$

Solution

$$\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x(2x^{7}+1)}dx=\displaystyle \int_{1}^{\sqrt[7]{2}}\dfrac{1}{x^{8}(2+1/x^{7})dx}$$
$$1/x^{7}= 1\Rightarrow\dfrac{-7}{x^{8}}dx=dt$$
$$\Rightarrow \dfrac{dx}{x^{8}}-\dfrac{dx}{7}$$
$$\Rightarrow \dfrac{1}{-7}\displaystyle \int_{1}^{1/2}\dfrac{dt}{(2+t)}=\dfrac{1}{7}\displaystyle \int_{1/2}^{1}\left [ \log(2+t) \right ]_{1/2}^{1}$$
$$=\dfrac{1}{7} \left [ \log(3)-\log\dfrac{5}{2} \right ]$$
$$=\dfrac{1}{7} \log \dfrac{6}{5}$$
Question 10
1 / -0

Evalaute: $$\displaystyle \int_{-\pi/2}^{\pi/2}\sqrt{\cos x-\cos^{3}x}\ dx$$

A
$$\displaystyle \frac{3}{4}$$

B
$$-\displaystyle \frac{3}{4}$$

C
$$\displaystyle \frac{4}{3}$$

D
$$0$$

Solution

$$\displaystyle\int_{-\pi/2}^{\pi/2}\sqrt{\cos{x}-\cos^3{x}}dx$$

$$=\displaystyle\int_{-\pi/2}^{\pi/2}\sqrt{\cos{x}}|\sin{x}|dx$$

Now, the given function is an even function:-

$$=2\displaystyle\int_{0}^{\pi/2}\sqrt{\cos{x}}\sin{x}dx$$ (since,$$ \sin{x}$$ is positive in interval $$ (0,\dfrac{\pi}{2})$$)

Let, $$z=\cos{x}\implies dz=-\sin{x}dx$$
And when$$ x=0, z=1$$ and when $$ x=\dfrac{\pi}{2}, z=0$$

Hence, integration becomes:-

$$2\displaystyle\int_{1}^{0} -\sqrt{z}dz=\int_{0}^{1} \sqrt{z}dz$$

$$=2\times\dfrac{2}{3}\displaystyle\left[z^{3/2}\right]_{0}^{1}$$

$$=\dfrac{4}{3}$$

Hence, answer is option-(C).

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Integrals Test - 24

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Integrals Test - 24

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SHARING IS CARING

Question 1

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

$$\dfrac{-\pi}{4}$$

$$\dfrac{-\pi}{2}$$

$$\dfrac{\pi}{2}$$

Solution

Question 2

$$1$$

$$1/2$$

$$1/3$$

$$1/4$$

Solution

Question 3

$$\pi/8$$

$$\pi/4$$

$$\pi/2$$

$$\pi$$

Solution

Question 4

$$3$$

$$\dfrac23$$

$$\dfrac13$$

$$0$$

Solution

Question 5

$$\displaystyle \frac{13}{15}-\frac{\pi}{4}$$

$$\displaystyle \frac{13}{15}+\frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{4}-\frac{2}{3}$$

$$\displaystyle \frac{13}{15}-4\pi$$

Solution

Question 6

$$27^{2}$$

-54

54

0

Solution

Question 7

$$\displaystyle \dfrac{1}{2}$$

$$1$$

$$\infty$$

$$0$$

Solution

Question 8

$$2(1-\displaystyle \mathrm{l} \mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$$

$$2(1+\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))$$

$$2(1+4\log_{\mathrm{e}}3-4\log_{\mathrm{e}}7+4\log_{\mathrm{e}}5)$$

$$2(1-\displaystyle \tan^{-1}(\frac{15}{7}))$$

Solution

Question 9

$$\log \dfrac{6}{5}$$

$$6 \log \dfrac{6}{5}$$

$$\dfrac{1}{7} \log \dfrac{6}{5}$$

$$\dfrac{1}{5} \log \dfrac{6}{5}$$

Solution

Question 10

$$\displaystyle \frac{3}{4}$$

$$-\displaystyle \frac{3}{4}$$

$$\displaystyle \frac{4}{3}$$

$$0$$

Solution

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