Application of Integrals Test

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Application of Integrals Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curves $$y=xe^{x}, y=xe^{-x}$$ and the line $$\left| x \right| =1,y=0$$ is:

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

Since $$\left| x \right| =1\quad \therefore x=\pm 1$$
$$y=x{ e }^{ \left| x \right| }=\begin{cases} { x }e^{ -x }\quad -1<x<0 \\ { xe }^{ x }\quad 0\le x<1 \end{cases}$$
Therefore $$=\left| \int _{ -1 }^{ 0 }{ { xe }^{ -x } } dx \right| +\left| \int _{ 0 }^{ 1 }{ { xe }^{ x } } dx \right| \\ =\left| \left[ { -xe }^{ -x }-{ e }^{ -x } \right] _{ -1 }^{ 0 } \right| +\left| \left[ { xe }^{ x }-{ e }^{ x } \right] _{ 0 }^{ 1 } \right| =2$$
Question 2
1 / -0

The area of the region bounded by $$y=|x-1|$$ and $$\mathrm{y}=1$$ in sq. units is:

A
1

B
1/2

C
2

D
3

Solution

$$y=|x-1|$$
area of the $$\triangle ^{le}=1/2 \:base\times\:height$$
$$=1/2(2)(1)=1sq\:unit.$$
Question 3
1 / -0

Area bounded by the curve $$\mathrm{y}=\mathrm{x}+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}$$ and its inverse function between the ordinates $$\mathrm{x}=0$$ and $$\mathrm{x}=2\pi$$ is

A
8 $$\pi$$ sqp. units

B
4 $$\pi$$ sq. units

C
8 sq. units

D
3 $$\pi$$ sq. units

Solution

Inverse function is the mirror image with respect to $$y=x$$
Then area bounded by $$x+\sin x$$ and its inverse function is $$4\int _{ 0 }^{ \pi }{ \left( x+\sin { x } -x \right) dx } =8$$
Question 4
1 / -0

Area of the region bounded by $$y=x-[x],\ y=[x]$$ and $$\mathrm{x}$$-axis in $$[$$0,2 $$]$$ is:

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

B
$$\displaystyle \frac{3}{2}$$

C
1

D
2

Solution

$$y=x-[x],\ y=[x]$$
The parallelogram Area=$$2(1/2\times1\times1)$$
$$=1\:sq\:units.$$
Question 5
1 / -0

The area bounded by the parabolas $${y}=4{x}^{2},\ y=\displaystyle \dfrac{x^{2}}{9}$$ and the line $${y}=2$$ is

A
$$\displaystyle \frac{20\sqrt{2}}{3}$$

B
$$\displaystyle \frac{10\sqrt{2}}{3}$$

C
$$\displaystyle \frac{40\sqrt{2}}{3}$$

D
$$\displaystyle \frac{5\sqrt{2}}{3}$$

Solution

The area bounded by the curves $${y}=4{x}^{2},\ y=\displaystyle \dfrac{x^{2}}{9}$$ and the line $$y=2$$ is
$$A=2\times \displaystyle \overset{2}{\underset{0}{\int}}\left (3\sqrt{y}-\dfrac{\sqrt{y}}{2} \right )dy$$
$$=2\left [\left[\dfrac{2}{3} 3 y^{3/2}\right]_{0}^{2} -\left[\dfrac{2}{3}\dfrac{y^{3/2}}{2} \right ]_{0}^{2} \right ]$$
$$=2\left [ 2(\sqrt{8})-\dfrac{\sqrt{8}}{3} \right ]$$
$$=2\left [ \sqrt{8}\left ( \dfrac{5}{3} \right ) \right ]=\dfrac{20\sqrt{2}}{3}sq\:units.$$
Question 6
1 / -0

The area between the parabolas $$y^{2}=4a(x+a)$$ and $$y^{2}=-4a(x-a)$$ in sq. units is

A
$$\displaystyle \frac{4a^{2}}{3}$$

B
$$\displaystyle \frac{8a^{2}}{3}$$

C
$$\displaystyle \frac{12a^{2}}{3}$$

D
$$\displaystyle \frac{16a^{2}}{3}$$

Solution

$$y^{2}=4a(x+a)$$ and $$y^{2}=-4a(x-a)$$
$$4\overset{0}{\underset{-a}{\int}}\sqrt{4a(x+a)}dx=4\times 2\sqrt{a}\times \dfrac{2}{3}(x+a)^{3/2}|_{-a}^{0}$$
$$=\dfrac{16}{3}\sqrt{a}(a^{3/2})$$
$$=\dfrac{16a^2}{3}sq\:units.$$
Question 7
1 / -0

The area of the portion of the circle $${ x }^{ 2 }+{ y }^{ 2 }=1$$, which lies inside the parabola $${ y }^{ 2 }=1-x$$, is

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

B
$$\displaystyle \frac { \pi }{ 2 } +\frac { 2 }{ 3 } $$

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

D
$$\displaystyle \frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$

Solution

The required area
$$\displaystyle =2\int _{ 0 }^{ 1 }{ \sqrt { 1-x } dx } +2.\left( \frac { 1 }{ 4 } .\pi .{ 1 }^{ 2 } \right) $$
$$\displaystyle =-2{ \left[ \frac { 2{ \left( 1-x \right) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 1 }+\frac { \pi }{ 2 } =\frac { \pi }{ 2 } -\frac { 4 }{ 3 } $$
Question 8
1 / -0

The area bounded by the parabolas $${ y }^{ 2 }=4a\left( x+a \right) $$ and $${ y }^{ 2 }=-4a\left( x-a \right) $$ is:

A
$$\displaystyle \frac { 16 }{ 3 } { a }^{ 2 }$$

B
$$\displaystyle \frac { 8 }{ 3 } { a }^{ 2 }$$

C
$$\displaystyle \frac { 4 }{ 3 } { a }^{ 2 }$$

D
none of these

Solution

The required area
$$\displaystyle =4\int _{ 0 }^{ a }{ \sqrt { 4a\left( a-x \right) } dx } $$
$$\displaystyle =4\times 2\sqrt { a } .{ \left[ \frac { -2{ \left( a-x \right) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ a }=\frac { 16 }{ 3 } { a }^{ 2 }$$
Question 9
1 / -0

The area bounded by the curve $$x^2=4y$$ and straight line $$x=4y-2$$ is

A
$$\frac{3}{8}$$

B
$$\frac{5}{8}$$

C
$$\frac{7}{8}$$

D
$$\frac{9}{8}$$

Solution

(x+24)=x24
Question 10
1 / -0

The area bounded by the curve $$f(x) = ce^x(c > 0)$$, the x-axis and the two ordinates x = p and x = q is proportional to

A
$$f(p) . f(q)$$

B
$$|f(p)-f(q)|$$

C
$$f(p) + f(q)$$

D
$$\sqrt{f(p)f(q)}$$

Solution

Area = $$\int _{ p }^{ q }{ { ce }^{ x } } $$
$$=c({ e }^{ q }-{ e }^{ p })$$
$$= f(q)-f(p) = |f(p) - f(q)|$$