Application of Integrals Test

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Application of Integrals Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by $$3x\pm 4y\pm 6=0$$ in sq. units is

A
$$3$$

B
$$1.5$$

C
$$4.5$$

D
$$6$$

Solution

Required area $$= 4\times$$ Area of triangle $$OAB$$ $$=4\times\cfrac{1}{2}\times 2\times \cfrac{3}{2}=6 $$ sq. units
Question 2
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The area between the curve $$y^{2}=9x$$ and the line $$y=3x$$ is

A
$$\displaystyle \frac{1}{3}$$ sq. units

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

C
$$\displaystyle \frac{1}{2}$$ sq, units

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

Solution

$$y^{2}=9x$$ and the line $$y=3x$$
$$\overset{1 }{\underset{0}{\int}}(3\sqrt{x}-3x)dx$$
$$3\overset{1 }{\underset{0}{\int}}\sqrt{x}dx-3\overset{1 }{\underset{0}{\int}}x\:dx$$
$$\Rightarrow 3\times\dfrac{2}{3}x^{3/2}|_{0}^{1}-3x^2/2|_{0}^{1}$$
$$=6/3-3/2=1/2 \:sq \:units$$.
Question 3
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The area of the smaller part of the circle $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$, cut off by the line $$\displaystyle x=\frac { a }{ \sqrt { 2 } } $$, is given by:

A
$$\displaystyle \frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } +1 \right) $$

B
$$\displaystyle \frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } -1 \right) $$

C
$$\displaystyle { a }^{ 2 }\left( \frac { \pi }{ 2 } -1 \right) $$

D
None of these

Solution

Required area $$\displaystyle 2\int _{ \frac { a }{ \sqrt { 2 } } }^{ a }{ ydx } =2\int _{ \frac { a }{ \sqrt { 2 } } }^{ a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } $$
$$\displaystyle =2{ \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \frac { x }{ a } } \right] }_{ \frac { a }{ \sqrt { 2 } } }^{ a }$$
$$\displaystyle =2\left[ \frac { { a }^{ 2 } }{ 2 } .\frac { \pi }{ 2 } -\left( \frac { a }{ 2\sqrt { 2 } } .\frac { a }{ \sqrt { 2 } } +\frac { { a }^{ 2 } }{ 2 } .\frac { \pi }{ 4 } \right) \right] $$
$$\displaystyle =\frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } -1 \right) $$
Question 4
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The area of the curve $$x=a\cos^{3}t$$,$$y=b\sin^{3}t$$ in sq. units is :

A
$$\displaystyle \frac{3\pi ab}{4}$$

B
$$\displaystyle \frac{3\pi ab}{8}$$

C
$$\displaystyle \frac{\pi ab}{4}$$

D
$$\displaystyle \frac{\pi ab}{8}$$

Solution

required area $$=\left|4\int_{0}^{a}y dx\right|$$

$$=\left|4\int_{0}^{\frac {\pi}{2}}y \dfrac {dx}{dt}dt\right|$$

$$=\left|4\int_{\frac {\pi}{2}}^{0}b\sin ^3t\cdot 3a \cos ^2t(-\sin t) dt\right|$$

$$=12ab\int_{0}^{\frac {\pi}{2}}\sin ^4t\cdot \cos ^2t dt$$

$$=12ab\dfrac{(4-1)(4-3)(2-1)}{6 \times 4 \times 2}\dfrac{\pi}{2}$$ [using reduction formula]

$$=\dfrac{3ab\pi}{8}$$
Question 5
1 / -0

The area bounded by the parabola $$y^{2}=4x$$ and its latusrectum is:

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

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

C
12 sq. units

D
$$\displaystyle \frac{1}{3}$$ sq. units

Solution

Area bounded=$$2\overset{1}{\underset{0}{\int}}2\sqrt{x}\: dx$$
Area bounded= $$2\left ( 2\times \dfrac{2}{3} \right )=2(4/3)=8/3$$
Question 6
1 / -0

Area of the region $$R=\{[(x,y)/x^{2}\leq y\leq x]\}$$ is

A
$$1/6$$

B
$$2/3$$

C
$$4/3$$

D
$$2$$

Solution

Given curve $$x^{2}=y$$ and line $$y=x$$
Point of intersection of line and curve is (0,0) and (1,1)

Area of shaded region $$= \int_{0}^{1} x{dx}-\int_{0}^{1} x^{2}{dx}$$

$$\displaystyle A=[\displaystyle\frac{x^{2}}{2}]_{0}^{1} -[\displaystyle\frac{x^{3}}{3}]_{0}^{1} $$

$$A=\displaystyle \frac{1}{6}$$ sq.units
Question 7
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Area of the region bounded by $$x=|y+4|$$ and $$\mathrm{y}$$ axis is sq. units

A
4

B
8

C
16

D
32

Solution

$$x=|y+4|$$
the area of the shaded region
$$8\times \dfrac {1}{2}\times 4=16 sq. units.$$
Question 8
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The area of the region between the curves $$y=x^{2}$$ and $$y=x^{3}$$ is

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

B
$$\displaystyle \frac{1}{3}$$ sq. units

C
$$\displaystyle \frac{1}{4}$$ sq. units

D
$$\displaystyle \frac{1}{2}$$ sq. units

Solution

$$y=x^{2}$$ and $$y=x^{3}$$
$$\overset{1 }{\underset{0}{\int}}(x^2-x^3)dx$$

$$=\frac{x^3}{3}|_{0}^{1}-\frac{x^4}{4}|_{0}^{1}$$

$$=(1/3-0)-(1/4-0)$$

$$=1/3-1/4=1/12 sq \:units$$
Question 9
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The area enclosed between $$y=\sin 2x,y=\sqrt{3}\sin x$$ between $$x=0$$ and $$x=\displaystyle \frac{\pi}{6}$$ is

A
$$\displaystyle \frac{7}{4}-\sqrt{3}$$ sq. units

B
$$\displaystyle \frac{7}{4}+\sqrt{3}$$ sq. units

C
$$\displaystyle \frac{7\sqrt{3}}{4}$$ sq, units

D
$$7-\displaystyle \frac{\sqrt{3}}{4}$$ sq. units

Solution

$$y=\sin 2x,y=\sqrt{3}\sin x$$ $$x=0$$ and $$x=\displaystyle \frac{\pi}{6}$$
$$\sin 2x=\sqrt{3}\sin x$$
$$2\cos x=\sqrt{3}$$
$$\cos x=\sqrt{3/2}$$
$$x=\displaystyle \frac{\pi}{6}$$
$$\overset{\pi /6}{\underset{0}{\int}}(\sin 2x-\sqrt{3}\sin x)dx$$
$$\sin^2x|_{0}^{\pi /6}+\sqrt{3}\cos x|_{0}^{\pi /6}$$
$$1/4+\sqrt{3}\left ( \frac{\sqrt{3}}{2}-1 \right )$$
$$1/4+3/2-\sqrt{3}$$
$$7/4-\sqrt{3}$$
Question 10
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$$AOB$$ is the positive quadrant of the ellipse $$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ where $$\mathrm{O}\mathrm{A}={a},\ {O}\mathrm{B}={b}$$. Then area between the arc $$\mathrm{A}\mathrm{B}$$ and chord $$\mathrm{A}\mathrm{B}$$ of the ellipse is

A
$$\pi\ ab$$

B
$$(\pi-2)ab$$

C
$$\displaystyle \frac{ab(\pi+2)}{2}$$

D
$$\displaystyle \frac{ab(\pi-2)}{4}$$

Solution

The area between the chord $$AB$$ and arc $$AB$$ of ellipse will be
Area of sector $$AOB -$$ Area of $$\triangle$$ AOB
$$=\dfrac{\pi ab}{4}-\dfrac12\times a\times b$$

$$=\dfrac{\pi ab}{4}-\dfrac{2ab}{4}=\dfrac14ab (\pi-2)\ sq\: units$$