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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Question 1

π4\dfrac{\pi}{4}4π​

−π4\dfrac{-\pi}{4}4−π​

−π2\dfrac{-\pi}{2}2−π​

π2\dfrac{\pi}{2}2π​

Solution

Question 2

111

1/21/21/2

1/31/31/3

1/41/41/4

Solution

Question 3

π/8\pi/8π/8

π/4\pi/4π/4

π/2\pi/2π/2

π\piπ

Solution

Question 4

333

23\dfrac2332​

13\dfrac1331​

000

Solution

Question 5

1315−π4\displaystyle \frac{13}{15}-\frac{\pi}{4}1513​−4π​

1315+π4\displaystyle \frac{13}{15}+\frac{\pi}{4}1513​+4π​

π4−23\displaystyle \frac{\pi}{4}-\frac{2}{3}4π​−32​

1315−4π\displaystyle \frac{13}{15}-4\pi1513​−4π

Solution

Question 6

27227^{2}272

-54

54

0

Solution

Question 7

12\displaystyle \dfrac{1}{2}21​

111

∞\infty∞

000

Solution

Question 8

2(1−loge(157))2(1-\displaystyle \mathrm{l} \mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))2(1−loge​(715​))

2(1+loge(157))2(1+\displaystyle \mathrm{l}\mathrm{o}\mathrm{g}_{\mathrm{e}}(\frac{15}{7}))2(1+loge​(715​))

2(1+4log⁡e3−4log⁡e7+4log⁡e5)2(1+4\log_{\mathrm{e}}3-4\log_{\mathrm{e}}7+4\log_{\mathrm{e}}5)2(1+4loge​3−4loge​7+4loge​5)

2(1−tan⁡−1(157))2(1-\displaystyle \tan^{-1}(\frac{15}{7}))2(1−tan−1(715​))

Solution

Question 9

log⁡65\log \dfrac{6}{5}log56​

6log⁡656 \log \dfrac{6}{5}6log56​

17log⁡65\dfrac{1}{7} \log \dfrac{6}{5}71​log56​

15 log⁡65\dfrac{1}{5} \log \dfrac{6}{5}51​ log56​

Solution

Question 10

34\displaystyle \frac{3}{4}43​

−34-\displaystyle \frac{3}{4}−43​

43\displaystyle \frac{4}{3}34​

000

Solution

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$\dfrac{\pi}{4}$

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

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

$\dfrac{\pi}{2}$

$1$

$1/2$

$1/3$

$1/4$

$\pi/8$

$\pi/4$

$\pi/2$

$\pi$

$3$

$\dfrac23$

$\dfrac13$

$0$

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

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

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

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

$27^{2}$

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

$1$

$\infty$

$0$

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

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

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

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

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

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

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

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

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

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

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

$0$