Integrals Test

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Integrals Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

Value of the definite integral $$\displaystyle \int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ \left( 1-x+{ x }^{ 2 } \right) } dx }$$ is:

A
$$\pi -\log { 2 }$$

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

C
$$\pi +\log { 2 }$$

D
$$\dfrac{\pi}{2} +\log { 2 }$$
Question 2
1 / -0

If $$\displaystyle I=\int _{ { -\pi }/{ 6 } }^{ { \pi }/{ 6 } }{ \dfrac { \pi +4{ x }^{ 5 } }{ 1-\sin { \left( \left| x \right| +\dfrac { \pi }{ 6 } \right) } } } dx$$, then I equals to

A
$$4\pi$$

B
$$2\pi +\dfrac { 1 }{ \sqrt { 3 } }$$

C
$$2\pi -\dfrac { 1 }{ \sqrt { 3 } }$$

D
$$4\pi +\sqrt { 3 } -\dfrac { 1 }{ \sqrt { 3 } }$$

Solution
Question 3
1 / -0

$$\displaystyle \int^{\pi/2}_{-\pi/2}\sqrt {\cos x-\cos^{3}x}dx=$$

A
$$1$$

B
$$4/3$$

C
$$-1/3$$

D
$$0$$

Solution
Question 4
1 / -0

$$For x>0$$,let $$f\left( x \right) =\int _{ 1 }^{ 2 }{ \dfrac { \log { t } }{ 1+t } } dt$$,then$$f\left( x \right) +f\left( \dfrac { 1 }{ x } \right)$$is equal to:

A
$$\dfrac { 1 }{ 4 } \left( \log { x } \right) ^{ 2 }$$

B
$$\dfrac { 1 }{ 2 } \left( \log { x } \right) ^{ 2 }$$

C
$$\log { x }$$

D
$$\dfrac { 1 }{ 4 } \log { { x }^{ 2 } }$$

Solution
Question 5
1 / -0

If, $$\int^{\frac{\pi}{2}} _0 \frac{sin^2x}{(1 + cosx^2)} dx = 0$$

A
$$\frac{4 - \pi}{2}$$

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

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

D
$$\frac{4 + \pi}{2}$$

Solution
Question 6
1 / -0

$${I}_{n}=\int_{1}^{e}{\left(logx\right)^{n}dx}$$ and $${I}_{n}=A+{BI}_{n-1}$$ then
A=........., B=............

A
$$e,-n$$

B
$$1/e,n$$

C
$$-e,n$$

D
$$-e-n$$

Solution
Question 7
1 / -0

If, $$\displaystyle\int^{\displaystyle\frac{\pi}{2}}_0\frac{\sin^2x}{(1+\cos x)^2}dx=$$.

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

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

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

D
$$\displaystyle\frac{4+\pi}{2}$$

Solution
Question 8
1 / -0

What is $$\int _{ 0 }^{ 1 }{ x{ \left( 1-x \right) }^{ 9 }dx } $$ equal to?

A
$$\dfrac{1}{110}$$

B
$$\dfrac{1}{132}$$

C
$$\dfrac{1}{148}$$

D
$$\dfrac{1}{240}$$

Solution

We need to find $$\int_{0}^{1}x(1-x)^9dx$$

$$\int_{0}^{1}x(1-x)^9dx=\int_{0}^{1}(1-x)x^9dx$$

$$=\int_{0}^{1}(x^9-x^{10})dx$$

$$=\left[\dfrac{x^{10}}{10}-\dfrac{x^{11}}{11}\right]_{0}^{1}$$

$$=\dfrac{1}{10}-\dfrac{1}{11}$$

$$=\dfrac{11-10}{110}$$

$$=\dfrac{1}{110}$$

$$\therefore \int_{0}^{1}x(1-x)^9dx=\dfrac{1}{110}$$
Question 9
1 / -0

If $$(-1, 2)$$ and $$(2, 4)$$ are two points on the curve $$y=f(x)$$ and if $$g(x)$$ is the gradient of the curve at point (x, y), then the value of the integral $$\displaystyle\int^{2}_{-1}g(x)dx$$, is?

A
$$2$$

B
$$-2$$

C
$$0$$

D
$$1$$

Solution
Question 10
1 / -0

The integral $$\int _{ 0 }^{ \pi }{ \sqrt { 1+4\sin { ^{ 2 }\frac { x }{ 2 } -4 } \sin { \frac { x }{ 2 } } } dx }$$ is equal to

A
$$\pi-4$$

B
$$\frac{2\pi}{3}-4-4\sqrt{3}$$

C
$$4sqrt{3}-4$$

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