Application of Integrals Test

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Application of Integrals Test - 12

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Weekly Quiz Competition

Question 1
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The area enclosed by the curves $$y=sinx+$$cosx and $$y=|cosx-$$sinx $$|$$over the interval $$[0, \frac{\pi}{2}]$$is

A
$$4$$ $$(\sqrt{2}-1)$$

B
$$2\sqrt{2}(\sqrt{2}-1)$$

C
$$2$$ $$(\sqrt{2}+1)$$

D
$$2\sqrt{2}(\sqrt{2}+1)$$

Solution

$$\displaystyle \cos { x } > \sin { x\quad , } \forall x\epsilon \left( 0, \frac { \pi }{ 4 } \right) $$, and $$\displaystyle \cos { x } < \sin { x\quad , } \forall x\epsilon \left( \frac { \pi }{ 4 } , \frac { \pi }{ 2 } \right) $$
$$y_{1}=\sin x+\cos x $$
$$y_{2}=\left| \cos { x } -\sin { x } \right| $$
$$
\Rightarrow $$Area

$$=\int _{ 0 }^{ \pi/2 }{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) dx } $$

$$=$$ $$\int_{0}^{ \pi/4 }$$ $$(( sin x+cosx)-(cos x-sin x))dx$$ $$+\int_{ \pi/4 }^{\pi/2} ((sin x+cos x)-(sin x-cos x)
)dx$$

$$=4-2\sqrt{2}$$
Question 2
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Area enclosed between the curves $$y=8-{x}^{2}$$ and $$y={x}^{2}$$, is:

A
$$32/3$$

B
$$64/3$$

C
$$30/4$$

D
$$9$$

Solution

$$y=8-{ x }^{ 2 }\quad y={ x }^{ 2 }$$
Intersection points are
$${ x }^{ 2 }=8-{ x }^{ 2 }$$
$$2{ x }^{ 2 }=8$$
$${ x }^{ 2 }=4\quad x=-2,2$$
$$\int _{ -2 }^{ 2 }{ \left( 8-{ x }^{ 2 }-{ x }^{ 2 } \right) dx } $$
$$\int _{ -2 }^{ 2 }{ \left( 8-2{ x }^{ 2 } \right) dx } $$
$$2{ \left( 8x-\cfrac { 2{ x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }$$
$$=2\left( 8\left( 2 \right) -\cfrac { 2\left( 8 \right) }{ 3 } \right) $$
$$=2\left( 16-\cfrac { 16 }{ 3 } \right) $$
$$=2\left( \cfrac { 32 }{ 13 } \right) $$
$$=\cfrac { 64 }{ 3 } $$ sq. units
Question 3
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If area bounded by the curves $$x=at^2$$ and $$y=ax^2$$ is $$1$$, then a$$=$$ __________.

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

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

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

D
$$3$$

Solution

$$x = ay^2$$ and $$y = ax^2$$
$$\Rightarrow a^3x^3 = 1 \Rightarrow (x,y) = (\dfrac{1}{a}, \dfrac{1}{a})$$
The area bounded by the curves is computed by:
$$\displaystyle \int_{0}^{1/a} |ax^2 - \sqrt{\dfrac{x}{a}}| dx = 1 \Rightarrow \dfrac{1}{3a^2} = 1 \Rightarrow a = \dfrac{1}{\sqrt{3}}$$
So option $$C$$ is the correct answer.
Question 4
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Calculate the area of the shaded region in the figure, where $$\square ABCD$$ is a square with side 8 cm each. $$(\pi =3.14)$$

A
$$36.48 \,cm^2$$

B
$$25.40 \,cm^2$$

C
$$15 \,cm^2$$

D
$$65 \,cm^2$$

Solution

Area of square =$$ Side^2 $$
=$$ 8^2 $$
=$$ 64$$ sq cm
Area of shaded portion = Double area of segment formed by diagonal of the square
Area of two quadrants =$$\dfrac{1}{2}\pi r^2$$

=$$\dfrac{1}{2} \times 3.14 \times 8^2$$

=$$100.48$$ sq cm
Area of square - Area of two triangles formed by radii and the cord
So,
Area of shaded portion =$$ 100.48-64$$
=$$ 36.48 sq$$ cm
Question 5
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The area included between the parabolas
$$y=\dfrac { { x }^{ 2 } }{ 4a }$$ and $$y=\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } }$$ is

A
$${ a }^{ 2 }\left( 2\pi +\dfrac { 2 }{ 3 } \right)$$

B
$${ a }^{ 2 }\left( 2\pi -\dfrac { 8 }{ 3 } \right)$$

C
$${ a }^{ 2 }\left( \pi +\dfrac { 4 }{ 3 } \right)$$

D
$${ a }^{ 2 }\left( \pi -\dfrac { 4 }{ 3 } \right)$$

Solution

Firstly we have to find the point of intersection.

$$\dfrac { { x }^{ 2 } }{ 4a } =\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } $$

On solving the equation We get $$x^{2}=4a^{2}$$ and $$x^{2}= -8a^{2}$$ (which is not possible)

We have $$x=2a$$ and $$x=-2a$$

Now area between these 2 curves is

$$\displaystyle \int _{ -2a }^{ 2a }{ \left(\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ 4a } \right)}dx$$

$$\displaystyle =\int _{ -2a }^{ 2a }{ \dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } dx } -\int _{ -2a }^{ 2a }{ \dfrac { { x }^{ 2 } }{ 4a } } dx$$

$$=\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { x }{ 2a } } -\dfrac { { x }^{ 3 } }{ 12a } $$

After putting the limits we get

$$\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { 2a }{ 2a } } -\dfrac { { 8a }^{ 3 } }{ 12a } -(\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { -2a }{ 2a } } -\dfrac { { -8a }^{ 3 } }{ 12a } )$$

$$=\pi { a }^{ 2 }-\dfrac { 4 }{ 3 } { a }^{ 2 }$$
Question 6
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Find the area of the region bounded by the curve $$y^2=4x$$ and the line $$x=3$$.

A
$$4\sqrt{3}$$

B
$$8\sqrt{3}$$

C
$$6$$

D
$$2\sqrt{3}$$

Solution

Required area = area of OACO = 2 ( Area of OAB)
$$\displaystyle =2\int_{0}^{3} 2\sqrt{x} dx$$

$$=\dfrac{8}{3} \times (3)^{3/2}=8\sqrt{3}$$ units.
Question 7
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The area enclosed between the $${y}^{2}=x$$ and $$y=|x|$$ is

A
$$\dfrac {1}{3}$$

B
$$\dfrac {2}{3}$$

C
$$1$$

D
$$\dfrac {1}{6}$$

Solution

$$y = x - \left( 1 \right)\,\,and\,{y^2} = x - \left( 2 \right)$$
$$solving\,1\,and\,2$$
$${y^2} = y$$
$${y^2} - y = 0$$
$$y\left( {y - 1} \right) = 0$$
so, $$y = 0,1$$
thus points of intersection are $$(0,0)$$ ,$$(1,1)$$
Area enclosed=$$\int_0^1 {\left( {{y_1} - {y_2}} \right)dx} $$
$$ = \int_0^1 {\left( {\sqrt x - x} \right)dx} $$
$$ = {\left[ {\dfrac{2}{3}{x^{3/2}}\dfrac{{ - {x^2}}}{2}} \right]_0}^1$$
$$ = \dfrac{2}{3} - \dfrac{1}{2} = \dfrac{1}{6}$$
Question 8
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The value of $$a$$ for which the area between the curves $${y^2} = 4ax$$ and $${x^2} = 4ay$$ is $$1\,sq.\,unit$$, is-

A
$$\sqrt 3 $$

B
$$4$$

C
$$4 \sqrt 3$$

D
$$\dfrac {\sqrt 3}{4}$$

Solution

$$\begin{array}{l}{y^2} = 4ax\\y = \sqrt {4ax} \\{x^2} = 4ay\\y = \dfrac{{{x^2}}}{{4a}}\\area = \int_0^{4a} {\sqrt {4ax} dx} - \int_0^{4a} {\dfrac{{{x^2}}}{{4a}}dx} \\\left. {2\sqrt a \times \dfrac{2}{3}{{\left( x \right)}^{3/2}}} \right]_0^{4a} - \left. {\dfrac{{{x^3}}}{{3\left( {4a} \right)}}} \right]_0^{4a}\\ = \dfrac{{32{a^2}}}{3} - \dfrac{{16{a^2}}}{3}\\ = \dfrac{{16{a^2}}}{3} = 1\\a = \dfrac{{\sqrt 3 }}{4}\end{array}$$
Question 9
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The area bounded by curves $$3x^2 + 5y = 32 $$ and $$y = \left|x-2\right|$$ is

A
25

B
17/2

C
33/2

D
33

Solution

$$\begin{array}{l} 3{ x^{ 2 } }+5y=32 \\ Draw\, a\, parabola\, .\, symmetrical\, about\, negative\, x-axis \\ Also, \\ y=\left| { x-2 } \right| \\ \Rightarrow y=\left( \begin{array}{l} x-2,\, \, \, \, \, x\ge 2 \\ 2-x,\, \, \, \, \, x<2 \end{array} \right) \\ y=x-2,\, \, x\ge 2\, \, represents\, a\, line,\, cutting\, the\, x-axis\, at\, \, A\left( { 2,0 } \right) \, \, and\, the\, parabola\, C\left( { 3,1 } \right) \\ And,\, y=2-x,\, x<2\, \, represents\, a\, line,\, cutting\, the\, parabola\, at\, \left( { -2,4 } \right) \\ Shaded\, area\, AEYCA \\ =\int _{ -2 }^{ 2 }{ \left[ { \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) -\left( { 2-x } \right) } \right] }dx+\int _{ 2 }^{ 3 }{ \left[ { \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) -\left( { x-2 } \right) } \right] }dx \\ =\int _{ -2 }^{ 3 }{ \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) dx- }\int _{ -2 }^{ 2 }{ \left( { 2-x } \right) dx+ }\int _{ 2 }^{ 3 }{ \left( { x-2 } \right) dx } \\ =\frac { 1 }{ 5 } \left[ { 32-x } \right] _{ -2 }^{ 3 }-\left[ { 2x-\frac { { { x^{ 2 } } } }{ 2 } } \right] _{ -2 }^{ 2 }-\left[ { \frac { { { x^{ 2 } } } }{ 2 } -3x } \right] _{ 2 }^{ 3 } \\ =\frac { 1 }{ 5 } \left[ { 32\times 3-27+64-8 } \right] -\left( { 4-2+4+2 } \right) -\left( { \frac { 9 }{ 2 } -6-2+4 } \right) \\ =\frac { 1 }{ 5 } \left[ { 96-35+64 } \right] -\left( 8 \right) -\left( { \frac { 9 }{ 2 } -4 } \right) \\ =\frac { { 125 } }{ 5 } -4-\frac { 9 }{ 2 } \\ =25-4-\frac { 9 }{ 2 } \\ =\frac { { 42-9 } }{ 2 } \\ =\frac { { 33 } }{ 2 } \, \, sq.units. \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$$
Question 10
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The area bounded by the $$x-$$axis, the curve $$y=f\left(x\right)$$ and the lines $$x=1$$ and $$x=b$$ is equal to $$\left(\sqrt{{b}^{2}+1}-\sqrt{2}\right)$$ for all $$b>1$$, then $$f\left(x\right)$$ is

A
$$\sqrt{x-1}$$

B
$$\sqrt{x+1}$$

C
$$\sqrt{{x}^{2}+1}$$

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

Solution

$$\int_{1}^{b}{f\left(x\right)dx}=\sqrt{\left({b}^{2}+1\right)}-\sqrt{2}$$
Differentiating both sides w.r.t $$b$$ , we get
$$f\left(b\right)=\dfrac{b}{\sqrt{\left({b}^{2}+1\right)}}$$
Hence $$f\left(x\right)=\dfrac{x}{\sqrt{\left({x}^{2}+1\right)}}$$