Application of Integrals Test

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Application of Integrals Test - 17

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Weekly Quiz Competition

Question 1
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The area bounded by the curve $$y^{2}=x$$ and the line $$\mathrm{x}=4$$ is:

A
$$\displaystyle \frac{32}{3}$$ sq. units

B
$$\displaystyle \frac{16}{3}$$ sq. units

C
$$\displaystyle \frac{8}{3}$$ sq. units

D
$$\displaystyle \frac{4}{3}$$ sq. units

Solution

Symmetric about x-axis
So, area is
$$2\overset{4}{\underset{0}{\int}}y\:dx$$
$$2\overset{4}{\underset{0}{\int}}\sqrt{x}\:dx$$
$$2\times\dfrac{2}{3}(2^3-0)=\dfrac{32}{3}sq\:units$$
Question 2
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The area of the region between the curve $$y=4x^{2}$$ and the line $$y=6x-2$$ is:

A
$$\displaystyle \frac{1}{9}$$ sq. units

B
$$\displaystyle \frac{1}{12}$$ sq. units

C
$$\displaystyle \frac{3}{2}$$ sq. units

D
$$\displaystyle \frac{1}{5}$$ sq. units

Solution

The curve $$y=4{ x }^{ 2 }$$ and $$y=6x-2$$ intersect at $$\displaystyle x=\frac { 1 }{ 2 } ,y=1$$ and $$y=1,y=4$$
Therefore area $$\displaystyle =\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \left( 6x-2-4{ x }^{ 2 } \right) dx } $$
$$\displaystyle ={ \left[ \frac { 6{ x }^{ 2 } }{ 2 } -2x-\frac { 4{ x }^{ 3 } }{ 3 } \right] }_{ \frac { 1 }{ 2 } }^{ 1 }=\frac { 1 }{ 12 } $$
Question 3
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The area bounded by the parabola $$y^{2}=4x$$ and the line $$y=2x-4$$:

A
9 sq. units

B
5 sq. units

C
4 sq. units

D
2 sq. units

Solution

$$y^{2}=4x$$ and the line $$y=2x-4$$
area in $$QI$$+area in QIV
$$QI=$$ $$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx- area of \triangle ^{le}$$
$$=$$ $$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx-1/2(2)(4)=\dfrac{32}{3}-4$$
$$QIV=$$ $$|\overset{1}{\underset{0}{\int}}-\sqrt{x}\:dx|+1/2\times1\times1$$
$$+\dfrac{4}{3}+1$$
$$QI$$+QIV$$= 12-3=9\:sq\:units$$
Question 4
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The area bounded by the line $$\mathrm{x}=1$$ and the curve $$\sqrt{\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}}=4$$ is

A
$$2\sqrt{3}$$

B
$$\sqrt{3}$$

C
$$3\sqrt{2}$$

D
$$4\sqrt{3}$$

Solution

Let $$\sqrt{\dfrac{y}{x}}=t$$.
Then
$$t+\dfrac{1}{t}=4$$
$$t^{2}-4t+1=0$$
$$t=\dfrac{4\pm\sqrt{12}}{2}$$

$$=2\pm\sqrt{3}$$
Hence
$$\dfrac{y}{x}=(2\pm\sqrt{3})^{2}$$
$$y=x(7-4\sqrt{3})$$ and $$y=x(7+4\sqrt{3})$$
Hence the required area is the area of the triangle formed by the lines $$y=x(7-4\sqrt{3})$$ and $$y=x(7+4\sqrt{3})$$ and $$x=1$$.
Hence we get the area
=(Area of the right angled triangle formed by $$y=(7+4\sqrt{3})$$, x-axis and x=1 )-(Area of the right angled triangle formed by $$y=(7-4\sqrt{3})$$, x-axis and x=1 )

$$=\dfrac{1}{2}[(7+4\sqrt{3})\times 1]-\dfrac{1}{2}[(7-4\sqrt{3})\times1)]$$

$$=\dfrac{1}{2}[7+4\sqrt{3}-7+4\sqrt{3}]$$

$$=4\sqrt{3}$$ units.
Question 5
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Area of the region enclosed by $$y^{2}=8x$$ and $${y}=2{x}$$ is

A
$$\dfrac{4}{3}$$

B
$$\dfrac{3}{4}$$

C
$$\dfrac{1}{4}$$

D
$$\dfrac{1}{2}$$

Solution

Given curves are $$y^2=8x$$ and $$y=2x$$
Let's find out their intersection point
$$\Rightarrow 4x^{2}=8x$$
$$\Rightarrow x^{2}-2x=0$$
$$\Rightarrow x(x-2)=0$$
$$\Rightarrow x=0,2$$
The required area is
$$A=\displaystyle \overset{2}{\underset{0}{\int}}(\sqrt{8x}-2x)dx$$
$$=\displaystyle \overset{2}{\underset{0}{\int}}\sqrt{8x}\:dx-\overset{2}{\underset{0}{\int}}2x\:dx$$
$$=\sqrt{8}\times\dfrac{2}{3}\times [x^{3/2}]_{0}^{2}-[x^2]_{0}^{2}$$
$$=\dfrac{2\sqrt{8}}{3}\sqrt{8}-4$$
$$=\dfrac{16}{3}-4=\dfrac{4}{3}\:sq\:units$$
Question 6
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The area between the curves $$y=\sqrt{x}$$ and $$y=x^{3}$$ is

A
$$\displaystyle \frac{1}{12}$$ sq. units

B
$$\displaystyle \frac{5}{12}$$ sq. units

C
$$\displaystyle \frac{3}{5}$$ sq. units

D
$$\displaystyle \frac{4}{5}$$ sq. units

Solution

$$y=\sqrt{x}$$ and $$y=x^{3}$$
The point of intersection of these curves will be
$$\sqrt{x}=x^{3}\Rightarrow x=0,1$$
Then, the area is given by
$$A=\displaystyle \overset {1}{ \underset { 0}{ \int } } (\sqrt{x}-x^3)dx$$
$$=\left[\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{x^{4}}{4}\right]_{0}^{1}$$

$$=\dfrac{2}{3}(1)-\dfrac{1}{4}$$

$$=\dfrac{8-3}{12}$$

$$=\dfrac{5}{12}sq\:units.$$
Question 7
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The area bounded by the parabola $$x^{2}=4ay,\ \mathrm{x}$$-axis and the straight line $$\mathrm{y}=2\mathrm{a}$$ is:

A
$$16\sqrt{2}a^{2}$$ sq. units

B
$$\displaystyle \frac{16\sqrt{2}}{3}a^{2}$$ sq. units

C
$$\displaystyle \frac{32\sqrt{2}}{3}a^{2}$$ sq. units

D
$$\displaystyle \frac{32\sqrt{2}}{5}a^{2}$$ sq. units

Solution

So, area of the shaded region
$$-\overset{2\sqrt{2}a}{\underset{0}{\int}}x^2/4a\:dx+\overset{2\sqrt{2}a}{\underset{0}{\int}}(2a)\:dx$$
$$2a(2\sqrt{2a})-\dfrac{}{4a}\dfrac{x^3}{3}|_{0}^{2\sqrt{2a}}$$
$$=4\sqrt{2}a^2-\dfrac{(2^{3/2}a)^3}{4a\times 3}$$
$$4\sqrt{2}a^2-\dfrac{2\times2\sqrt{2}a^2}{3}$$
$$=\dfrac{16\sqrt{2}a^2}{3}sq\:units.$$
Question 8
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The area bounded by $$\mathrm{y}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\ \mathrm{y}=\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}$$ between any two successive intersections is:

A
$$2$$

B
$$\sqrt{2}$$

C
$$2\sqrt{2}$$

D
4

Solution

$$\sqrt{2}\overset {5\pi/4}{ \underset { \pi/4}{ \int } } \sin (x-\pi /4)dx$$
$$=-\sqrt{2}\cos (x-\pi /4)|_{\pi /4}^{5\pi /4}$$
$$=-\sqrt{2}[\cos \pi-\cos 0]$$
$$=2\sqrt{2}sq\:units.$$
Question 9
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The area bounded by the curves $$y=\sin x,y=$$ cosx and the $$\mathrm{y}$$-axis and the first point of intersection is:

A
$$\sqrt{2}$$ sq,.units

B
$$\sqrt{2}-1$$ sq. units

C
$$2+\sqrt{2}$$ sq. units

D
0 sq, units

Solution

$$\overset {\pi/4}{ \underset { 0}{ \int } } (\cos x-\sin x)\:dx$$
$$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx-\overset {\pi/4}{ \underset { 0}{ \int } } \sin x\:dx$$
$$\sin x|_{0}^{\pi /4}+\cos x|_{0}^{\pi /4}$$
$$=1/\sqrt{2}+1/\sqrt{2}-1$$
$$=\sqrt{2}-1 sq\:units.$$
Question 10
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Assertion(A): The area bounded by $$y^{2}=4x$$ and $$x^{2}=4y$$ is $$\displaystyle \frac{16}{3}$$ sq. units.

Reason(R): The area bounded by $$y^{2}=4ax$$ and $$x^{2}=4ay$$ is $$\displaystyle \frac{16a^2}{3}$$ sq. units

A
Both A and R are true and R is the correct explanation of A.

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

Given $$y^2=4ax$$ and $$x^2=4ay$$
Since both curves intersect,
$$\therefore x=\dfrac{y^2}{4a}$$

$$\Rightarrow (\dfrac{y^2}{4a})^2=4ay $$

$$\Rightarrow y^4=64a^3y$$

$$\Rightarrow y[y^3-(4a)^3]=0$$

$$\Rightarrow y=0,4a.$$
When $$y=0,x=0$$ and when $$ y=4a,x=4a,$$
Area $$A=\int_{0}^{4a}\left(\sqrt(4ax)-\dfrac{x^2}{4a}\right)dx $$
on solving we will get
$$A=\dfrac{16}{3a^2}$$

Now if we put $$ a=1 $$,
we will get $$A=\dfrac{16}{3}$$