Integrals Test

Result

Integrals Test - 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Evaluate the following definite integral:
$$\displaystyle \int_{0}^{\pi_/{2}}\frac{1}{4+5\cos x}dx=$$

A
$$\displaystyle \frac{1}{5}\log 2$$

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

C
$$\displaystyle \frac{1}{3}\log 3$$

D
$$\displaystyle \frac{1}{3}\log 2$$

Solution

Let $$\displaystyle I=\int _{ 0 }^{\frac{\pi}{2}} { \cfrac { 1 }{ 4+5cosx } dx } $$

Substitute $$tan\left( \cfrac { x }{ 2 } \right) =t\Rightarrow \cfrac { 1 }{ 2 } { sec }^{ 2 }\left( \cfrac { x }{ 2 } \right) dx=dt$$

So, $$x \rightarrow 0 \ \Rightarrow t \rightarrow 0$$ and $$x \rightarrow \dfrac{\pi}{2} \ \Rightarrow t \rightarrow {1}$$

gives $$\cos x=\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } } $$

$$\Rightarrow$$ $$\displaystyle I=\int _{ 0 }^{ 1 } { \cfrac { 1 }{ 4+5\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } } \ \cfrac { 2dt }{ 1+{ t }^{ 2 } } } =2\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 9-{ t }^{ 2 } } dt }$$

$$\Rightarrow$$ $$\displaystyle I =\cfrac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 1-\cfrac { { t }^{ 2 } }{ 9 } } dt } $$

Substitute $$\cfrac { t }{ 3 } =u\Rightarrow dt=3du$$
So, $$t \rightarrow 0 \ \Rightarrow u \rightarrow 0$$ and $$t \rightarrow 1 \ \Rightarrow u \rightarrow \dfrac{1}{3}$$

$$\Rightarrow$$ $$\displaystyle I=\frac { 2 }{ 3 } \int _{ 0 }^{ \frac { 1 }{ 3 } }{ \cfrac { 1 }{ 1-{ u }^{ 2 } } du } =\cfrac { 2 }{ 3 } { \left[ \cfrac { 1 }{ 2 } \log { \cfrac { 1+u }{ 1-u } } \right] }_{ 0 }^{ \frac { 1 }{ 3 } }$$

$$\Rightarrow$$ $$\displaystyle I=\cfrac { 1 }{ 3 } \left( \log { 2 } -0 \right) =\cfrac { 1 }{ 3 } \log { 2 } $$
Question 2
1 / -0

$$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{2+3x}}dx_{=}$$

A
$$\displaystyle \frac{2}{3}(\sqrt{5}-\sqrt{2})$$

B
$$\displaystyle \frac{2}{3}(\sqrt{5}+\sqrt{2})$$

C
$$\displaystyle \frac{3}{5}(\sqrt{5}-\sqrt{2})$$

D
$$\displaystyle \frac{2}{3}(\sqrt{3}-\sqrt{2})$$

Solution

$$\displaystyle I=\int_{0}^{1}\frac{1}{\sqrt{2+3x}}dx$$

Put $$3x+2=t$$
$$\Rightarrow dx=\displaystyle \frac{dt}{3}$$

$$\displaystyle I=\frac {1}{3}\int_{2}^{5}\frac{1}{\sqrt{t}}dt$$

$$\Rightarrow \displaystyle I =\frac{2}{3}[\sqrt{t}]_{2}^{5}$$

$$\Rightarrow I=\dfrac{2}{3} (\sqrt{5}-\sqrt{2})$$
Question 3
1 / -0

$$\displaystyle \int_{0}^{\infty}(a^{-x}-b^{-x})dx=(\mathrm{a}>1,\mathrm{b}>1)$$

A
$$\displaystyle \dfrac{1}{\log a}-\dfrac{1}{\log b}$$

B
$$\log a-\log b$$

C
$$\log a + \log b$$

D
$$\displaystyle \dfrac{1}{\log a}+\dfrac{1}{\log b}$$

Solution

$$\displaystyle I=\int_{0}^{\infty}(a^{-x}-b^{-x})dx$$

$$\displaystyle = -\frac{1}{\log a}(a^{-x})_{0}^{\infty}+\frac{1}{\log b}(b^{-x})_{0}^{\infty}$$

$$\displaystyle =\frac{1}{\log a}-\frac{1}{\log b}$$
Question 4
1 / -0

$$\displaystyle \int_{0}^{a}\frac{x-a}{x+a} dx =$$

A
$$a+2a\log 2$$

B
$$a - 2a\log 2$$

C
$$2a\log-a$$

D
$$2a\log 2$$

Solution

To find : $$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot dx$$
Consider, $$\displaystyle \int \dfrac{x-a}{x+a} \cdot dx=\int 1 - 2a \int \dfrac{1}{(x+a)}dx$$
$$=x-2a \log (x+a)+c$$
Then, $$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a} \cdot dx=\left[x-2a \log (x+a)+c\right]_{0}^{a}$$
$$=(a-2a \log (1a)+c)- [0-2a \log (a)+c]$$
$$=a-2a \log 2$$
$$\displaystyle \int_{0}^{a}\dfrac{x-a}{x+a}\cdot dx= a-2a \log 2$$
Question 5
1 / -0

If $$\displaystyle \int_{0}^{60}\frac{dx}{2x+1}=\log a$$, then $$ a= $$

A
$$3$$

B
$$11$$

C
$$81$$

D
$$40$$

Solution

Let $$I=\displaystyle \int_{0} ^{60} \dfrac{dx}{2x+1}$$
$$=\left [\dfrac{\log(2x+1)}{2}\right]_{0} ^{60}$$
$$=\dfrac{1}{2}[\log (121)-\log(1)]$$
$$=\dfrac{\log (121)}{2}$$
$$=\log(\sqrt{121})$$
$$=\log 11$$
Thus $$\log 11=\log a$$
$$\Rightarrow a=11$$
Question 6
1 / -0

$$\displaystyle \int_{1}^{2}\frac{\mathrm{d}\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}=$$

A
$$\displaystyle \log_{\mathrm{e}}(\frac{2+\sqrt{5}}{\sqrt{2}+1})$$

B
$$\displaystyle \log_{\mathrm{e}}(\frac{\sqrt{2}+1}{2+\sqrt{5}})$$

C
$$\displaystyle \log_{\mathrm{e}}(\frac{2-\sqrt{5}}{\sqrt{2}-1})$$

D
0

Solution

$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+a}|+c$$
$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log |x+\sqrt{x^2+1}|+c$$
$$=\log |2+\sqrt{5}|-\log |1+\sqrt{2}|$$
$$=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$$
$$\int_{1}^{2}\dfrac{dx}{\sqrt{1+x^2}}=\log \left | \dfrac{2+\sqrt{5}}{1+\sqrt{2}} \right |$$
Question 7
1 / -0

Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{1-x}{1+x}dx$$

A
$$\log 4$$

B
$$\log (\dfrac{4}{e})$$

C
$$1$$

D
$$\log (\dfrac{e}{4})$$

Solution

$$\displaystyle \int_{0}^{1} \left ( \dfrac{1-x}{1+x} \right ) dx$$
$$\displaystyle \int \dfrac{1-x}{1+x} dx = \displaystyle \int \dfrac{2}{1+x}dx -\displaystyle \int1 \cdot dx$$
$$=2\log (1+x) - x + c$$
$$\displaystyle \int \dfrac{1-x }{1+x}dx = 2 \log (1+x)- x+c$$
$$\displaystyle \int_{0}^{1}\dfrac{1-x}{1+x}dx= \left(2 \log (1+x)-x+c\right)_{0}^{1}$$
$$=2 \log (2)-1$$
$$=\log 4- \log \ e$$
$$=\log \left(\dfrac{4}{e}\right)$$
Question 8
1 / -0

$$\int_{-1}^{1} \displaystyle \frac{d{x}}{1+x^{2}}=$$

A
0

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

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

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

Solution

$$=\int_{-1}^{1}\dfrac{dx}{1+x^2}$$
$$\int \dfrac{dx}{1+x^2}=tan^{-1}x+c$$
$$\int_{-1}^{1}\dfrac{dx}{1+x^2}=(tan^{-1}x+c)\int_{-1}^{1}$$
$$=tan^{-1}1-tan^{-1}(-1)$$
$$=\dfrac{\pi}{4}-(\dfrac{-\pi}{4})$$
$$=\dfrac{\pi}{2}$$
So,
$$\int_{-1}^{1}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$$
Question 9
1 / -0

$$\displaystyle \int_{1}^{\infty}\left( \frac { 1 }{ 1+x^{ 2 } } \right) d{ x }=$$

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

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

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

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

Solution

$$\int_{1}^{\infty }\left [ \dfrac{1}{1+x^2} \right ] dx$$
$$tan^{-1}x \int_{1}^{\infty }$$
$$\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$$
Question 10
1 / -0

Evaluate: $$\displaystyle \int_{0}^{\pi /8} \cos^{3}4x\ dx $$

A
$$1/6$$

B
$$1/5$$

C
$$-1/3$$

D
$$1/8$$

Solution

Given that $$\int_{0}^{\tfrac{\pi}{8}}\cos^3(4x)dx$$
We can write $$\cos^3(4x)= \cos^2(4x)\cos(4x) =[1-\sin^2(4x)]\cos(4x)$$
$$\int_{0}^{\tfrac{\pi}{8}}[1-\sin^2(4x)]\cos(4x)$$ ............1
Now let $$\sin(4x)= t$$ ........2
when $$x=0, t = 0$$; When $$x=\cfrac{\pi}{8}, t = 1$$
On differentiating equation 2 we get,
$$\cfrac{\cos(4x)}{4}dx=dt$$ .........3
Substituting the values of equation 2 and 3 in equation 1.
$$\int_{0}^{1}\cfrac{[1-t^2]}{4}dt$$
$$=\int_{0}^1 \cfrac{1}{4}dt - \int_{0}^1 \cfrac{t^2}{4}dt$$
$$=[\cfrac{t}{4}]_{0}^{1} - [\cfrac{t^3}{12}]_{0}^{1}$$
$$=\cfrac{1}{4} - \cfrac{1}{12}$$
$$=\cfrac{3-1}{12}$$
$$=\cfrac{2}{12}$$
$$=\cfrac{1}{6}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Integrals Test - 19

Result

Integrals Test - 19

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{5}\log 2$$

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

$$\displaystyle \frac{1}{3}\log 3$$

$$\displaystyle \frac{1}{3}\log 2$$

Solution

Question 2

$$\displaystyle \frac{2}{3}(\sqrt{5}-\sqrt{2})$$

$$\displaystyle \frac{2}{3}(\sqrt{5}+\sqrt{2})$$

$$\displaystyle \frac{3}{5}(\sqrt{5}-\sqrt{2})$$

$$\displaystyle \frac{2}{3}(\sqrt{3}-\sqrt{2})$$

Solution

Question 3

$$\displaystyle \dfrac{1}{\log a}-\dfrac{1}{\log b}$$

$$\log a-\log b$$

$$\log a + \log b$$

$$\displaystyle \dfrac{1}{\log a}+\dfrac{1}{\log b}$$

Solution

Question 4

$$a+2a\log 2$$

$$a - 2a\log 2$$

$$2a\log-a$$

$$2a\log 2$$

Solution

Question 5

$$3$$

$$11$$

$$81$$

$$40$$

Solution

Question 6

$$\displaystyle \log_{\mathrm{e}}(\frac{2+\sqrt{5}}{\sqrt{2}+1})$$

$$\displaystyle \log_{\mathrm{e}}(\frac{\sqrt{2}+1}{2+\sqrt{5}})$$

$$\displaystyle \log_{\mathrm{e}}(\frac{2-\sqrt{5}}{\sqrt{2}-1})$$

0

Solution

Question 7

$$\log 4$$

$$\log (\dfrac{4}{e})$$

$$1$$

$$\log (\dfrac{e}{4})$$

Solution

Question 8

0

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

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

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

Solution

Question 9

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

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

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

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

Solution

Question 10

$$1/6$$

$$1/5$$

$$-1/3$$

$$1/8$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.