Application of Integrals Test

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Application of Integrals Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The area bounded by curve $$y=x^{2}-1$$ and tangents to it at $$(2,3)$$ and $$y-$$axis is

A
$$8/3$$

B
$$2/3$$

C
$$4/3$$

D
$$1/3$$

Solution

$$x-axis:(-1,0)$$

Area $$=\int_{-1}^{0}(x^2-1)dx$$

$$=\left [ \dfrac{x^3}{3}-x \right ]_{-1}^0$$

$$=\left [ \dfrac{-1}{3}-(-1) \right ]-[0]$$

$$=-\dfrac{1}{3}+1$$

$$=\dfrac{2}{3}$$ sq. units
Question 2
1 / -0

Area included between $${ y }=\dfrac { { x }^{ { 2 } } }{ 4{ a } } $$ and $$y = \dfrac { 8 a ^ { 3 } } { x ^ { 2 } + 4 a ^ { 2 } }$$ is

A
$$\dfrac { a ^ { 2 } } { 3 } ( 6 \pi - 4 )$$

B
$$\dfrac { a ^ { 2 } } { 3 } ( 4 \pi + 3 )$$

C
$$\dfrac { a ^ { 2 } } { 3 } ( 8 \pi + 3 )$$

D
None of these

Solution
Question 3
1 / -0

The area of the figure formed by $$a|x|+b|y|+c=0$$, is

A
$$\dfrac{c^{2}}{|ab|}$$

B
$$\dfrac{2c^{2}}{|ab|}$$

C
$$\dfrac{c^{2}}{2|ab|}$$

D
$$None\ of\ these$$

Solution
Question 4
1 / -0

The area of the region bounded by the curves $$y = \sin x$$ and $$y = \cos x ,$$ and lying between the lines $$x = \dfrac { \pi } { 4 }$$ and $$x = \dfrac { 5 \pi } { 4 } ,$$ is

A
$$2 + \sqrt { 2 }$$

B
$$2$$

C
$$2 \sqrt { 2 }$$

D
$$2 - \sqrt { 2 }$$

Solution
Question 5
1 / -0

Find the area of bounded by $$y=\sin x $$ from $$x=\dfrac{\pi}{4} $$ to $$x=\dfrac{\pi}{2}$$

A
$$\dfrac{\sqrt 2-1}{\sqrt2}$$

B
$$\dfrac12$$

C

$$\dfrac 14$$

D
None of these

Solution

The area bounded is given as $$\displaystyle \int _{\pi/4}^{\pi/2}\sin xdx\\\left.-\cos x \right|_{\pi/4}^{\pi/2}\\-\cos \dfrac {\pi}2+\cos \dfrac{\pi}4\\\dfrac{\sqrt 2-1}{\sqrt2}$$
Question 6
1 / -0

The area of the region by curves $$y=x\log x$$ and $$y=2x-2x^{2}=$$

A
$$1/12$$

B
$$3/12$$

C
$$7/12$$

D
$$None\ of\ these$$

Solution
Question 7
1 / -0

The area bounded by x-axis the curve $$y=f(x)$$ and the lines $$x =1,x=b$$ equal to $$\left( \sqrt { \left( { b }^{ 2 }+1 \right) } -\sqrt { 2 } \right) for\quad all\quad b>1,then\quad f(x)$$

A
$$\sqrt { \left( x-1 \right) } $$

B
$$\sqrt { \left( x+1 \right) } $$

C
$$\sqrt { \left( { x }^{ 2 }+1 \right) }$$

D
$$\dfrac { x }{ \sqrt { 1+{ x }^{ 2 } } } $$

Solution
Question 8
1 / -0

The area bounded by the curve $$y={ e }^{ x }$$ and the lines y = |x - 1|, x = 2 is given by :

A
$${ e }^{ 2 }+1$$

B
4$${ e }^{ 2}-1$$

C
$${ e }^{ 2 }-2$$

D
$$e - 2$$

Solution

Given,

$$y=e^x,y=|x-1|,x=2$$

$$\rightarrow e^x=|x-1|$$

$$\rightarrow e^x-(1-x)=0$$

intervals:

$$|x-1|=0\Rightarrow x=1$$

$$x=2$$

Area is given by,

$$A=\int_{0}^{1}(e^x-(1-x))dx+\int_{1}^{2}(e^x-(1-x))dx$$

$$=\int_{0}^{1}(e^x+x-1)dx+\int_{1}^{2}(e^x+x-1)dx$$

$$=\left [ e^x+\dfrac{x^2}{2}-x \right ]_0^1+\left [ e^x+\dfrac{x^2}{2}-x \right ]_1^2$$

$$=e+\dfrac{1}{2}-1+(e^2-2+2)-\left ( e-\dfrac{1}{2}+1 \right )$$

$$=e^2-2$$
Question 9
1 / -0

Area bounded by the curve $$y^2(2a-x)=x^3$$ and the line $$x=2a$$, is

A
$$3\pi a^2$$

B
$$\dfrac {3\pi a^2}2$$

C
$$\dfrac {3\pi a^2}4$$

D
$$\dfrac {\pi a^2}4$$

Solution

Given,

$$y^2(2a-x)=x^3$$

$$y=\sqrt{\dfrac{x^3}{2a-x}}$$

let, $$x=2a\sin ^2\theta $$

$$dx=4a\sin \theta \cos \theta d\theta $$

$$\Rightarrow A =\int_{0}^{2a}\sqrt{\dfrac{x^3}{2a-x}}dx$$

$$A=\int_{0}^{\tfrac{\pi }{2}}\sqrt{\dfrac{8a^3\sin ^6\theta }{2a\cos ^2\theta }}4a\sin \theta \cos \theta d\theta $$

$$=8a^2\int_{0}^{\tfrac{\pi }{2}}\sin ^4\theta d\theta $$

$$=8a^2\int_{0}^{\tfrac{\pi }{2}}\sin ^2\theta (1-\cos ^2\theta )$$

$$=8a^2\int_{0}^{\tfrac{\pi }{2}}\left ( \dfrac{1-\cos 2\theta }{2} \right )-\dfrac{1}{4}\sin ^22\theta d\theta $$

$$=8a^2\left [ \dfrac{\pi }{4}-0-\dfrac{1}{4}\left ( \dfrac{\pi }{4}-0 \right ) \right ]$$

$$=8a^2\left [\dfrac{\pi }{4}-\dfrac{\pi }{16} \right ]=\dfrac{3}{2}\pi a^2$$
Question 10
1 / -0

Area of the region bounded by $$y=\sin^{-1}{\left| \sin{x}\right|}$$ and $$y=-\cos^{-1}{\left| \cos{x}\right|}$$ in the interval $$[0,2\pi]$$ is equal to

A
$$\cfrac{\pi^2}{2}$$

B
$$\pi^2$$

C
$$\cfrac{\pi^2}{4}$$

D
$$2\pi^2$$

Solution