Application of Integrals Test

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Application of Integrals Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Area of the region $$\{(x,y)/x^{2}+y^{2}\leq 1\leq x+y\}$$ is:

A
$$\displaystyle \frac{\pi}{4}+\frac{1}{2}$$

B
$$\displaystyle \frac{\pi}{4}-\frac{1}{2}$$

C
$$\displaystyle \frac{\pi}{4}+\frac{3}{4}$$

D
$$\pi+1$$

Solution

Required area $$=$$ Area of quarter in first quadrant $$-$$ Area of $$\triangle AOB$$
$$=\cfrac{1}{4}\pi(1)^2-\cfrac{1}{2}.1.1 = \cfrac{\pi}{4}-\cfrac{1}{2}$$
Question 2
1 / -0

The area bounded by the curves $$y=\cos x,y=\cos 2x$$ between the ordinates $$x=0,x=\displaystyle \frac{\pi}{3}$$ are in the ratio

A
$$2\sqrt{3}:4-\sqrt{3}$$

B
$$2: 1$$

C
$$2\sqrt{3}:4+\sqrt{3}$$

D
$$1: 3$$

Solution

$$y=\cos x,y=\cos 2x$$
$$\overset{\pi /3}{\underset{0}{\int}}(\cos x)dx$$ area of $$\cos x$$ and area of $$\cos 2x$$
$$\overset{\pi /4}{\underset{0}{\int}}(\cos 2x)dx-\overset{\pi /3}{\underset{\pi /4}{\int}}(\cos 2x)dx=\dfrac{4-\sqrt{3}}{4}$$
$$\overset{\pi /3}{\underset{0}{\int}}(\cos x)dx=\dfrac{\sqrt{3}}{2} sq\:units$$
S0, ratio is $$\dfrac{\sqrt{3}}{2}:\dfrac{4-\sqrt{3}}{4}=2\sqrt{3}:4-\sqrt{3}$$
Question 3
1 / -0

The area bounded by $$y^{2}=4ax$$ and $$y=mx$$ is $$\displaystyle \frac{a^{2}}{3}$$ sq. units then $$\mathrm{m}$$

A
1

B
2

C
3

D
4

Solution

The two curves $${ y }^{ 2 }=4ax$$ and $$y=mx$$ intersect at $$\displaystyle \left( \dfrac { 4a }{ { m }^{ 2 } } ,\dfrac { 4a }{ m } \right) $$
and the area enclosed by the two curves are given by

$$\displaystyle \int _{ 0 }^{ \dfrac { 4a }{ { m }^{ 2 } } }{ \left( \sqrt { 4ax } -mx \right) dx } $$

$$\displaystyle \therefore \int _{ 0 }^{ \dfrac { 4a }{ { m }^{ 2 } } }{ \left( \sqrt { 4ax } -mx \right) dx } =\dfrac { { a }^{ 2 } }{ 3 } $$

$$\displaystyle \Rightarrow \dfrac { 8 }{ 3 } .\dfrac { { a }^{ 2 } }{ { m }^{ 3 } } =\dfrac { { a }^{ 2 } }{ 3 } \Rightarrow { m }^{ 3 }=8\Rightarrow m=2$$
Question 4
1 / -0

Area of the segment cut off from the parabola $$x^{2}=8y$$ by the line $$x-2y+8=0$$ is:

A
$$12$$

B
$$24$$

C
$$48$$

D
$$36$$

Solution

$$x^{2}=8y$$ and $$x-2y+8=0$$
$$x^2=4(x+8)$$
$$x^2-4x-32=0$$
$$x=-4;x=8$$
area is
$$\overset{8 }{\underset{-4}{\int}}\left [ \left ( \dfrac{x+8}{2} \right )-\frac{x^2}{8} \right ]dx$$
$$\overset{8 }{\underset{-4}{\int}}(x/2+4-x^2/8)dx=\dfrac{x^2}{4}|_{-4}^{8}+4x|_{-4}^{8}-\dfrac{x^3}{24}|_{-4}^{8}$$
$$12+48-24$$
$$=36\: sq\: units$$
Question 5
1 / -0

The area bounded by $$y=3x$$ and $$y=x^{2}$$ is (in square units)

A
$$10$$

B
$$5$$

C
$$4.5$$

D
$$9$$

Solution

$$y=3x$$ and $$y=x^{2} in \:sq\: units$$
$$3x=x^2$$
$$x=3 \:and\: x=0$$ are the points of intersection
$$\overset{3 }{\underset{0}{\int}}(3x-x^2)dx$$
$$=\dfrac{3x^2}2|_{0}^{3}-\dfrac{x^3}{3}|_{0}^{3}$$

$$=3\dfrac{9}{2}-\dfrac{27}{3}$$

$$=\dfrac{27}{2}-9=\dfrac{9}{2}=4.5\: sq\: units$$
Question 6
1 / -0

The area bounded by the two curves $$y=\sin x,\ y=\cos x$$ and the $$\mathrm{X}$$-axis in the first quadrant $$\left[0,\displaystyle \frac{\pi}{2}\right]$$ is

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

B
$$2+\sqrt{2}$$ sq,. units

C
$$2(\sqrt{2}-1)$$ sq. units

D
$$4$$ sq. units

Solution

The area bounded by the curves $$y=\sin x,\ y=\cos x$$ and $$\mathrm{X}$$-axis in $$\left[0,\displaystyle \frac{\pi}{2}\right]$$ is given by
$$A=\overset{\pi /4}{\underset{0}{\displaystyle \int}}\sin x\:dx+\overset{\pi /2}{\underset{\pi /4}{\displaystyle \int}}\cos x\:dx$$

$$=-[\cos x]_{0}^{\pi /4}+[\sin x]_{\pi /4}^{\pi /2}$$

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

$$=(2-\sqrt{2})sq\:units$$
Question 7
1 / -0

Area bounded by $$y=\sqrt{a^{2}-x^{2}},\ x+y=0$$ and $$\mathrm{y}$$-axis in sq. units is:

A
$$a^{2}(\displaystyle \frac{\pi}{2})$$

B
$$a^{2}(\displaystyle \frac{\pi}{4})$$

C
$$a^{2}(\displaystyle \frac{\pi}{8})$$

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

Solution

$$y=\sqrt{a^{2}-x^{2}},\ x+y=0$$
its 1/8 of the segment of the circle So, the area is $$\frac{\pi a^2}{8}$$
Question 8
1 / -0

Area ofthe region bounded by $$y=|x|$$ and $$\mathrm{y}=2$$ is

A
$$4$$ sq units

B
$$2$$ sq. units

C
$$1$$ sq. units

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

Solution

$$y=|x|$$ and $$\mathrm{y}=2$$
area of $$\triangle^{le}$$ OAB
$$=1/2\times(OC)\times(AB)$$
$$=1/2(2)(4)$$
$$=4 \:sq\:units$$
Question 9
1 / -0

The area of a region bounded by $$\mathrm{X}$$-axis and the curves defined by $$y=\tan x , 0\displaystyle \leq x\leq\frac{\pi}{4}$$ and $$y=\displaystyle \cot x,\frac{\pi}{4}\leq x\leq\frac{\pi}{2}$$ is:

A
$$log3$$ sq. unlts

B
$$log5$$ sq. unlts

C
$$log 1$$ sq. unit

D
$$log 2$$ sq. units
Question 10
1 / -0

The area bounded by $$y=\cos x,\ y=x+1$$ and $$y=0$$ in the second quadrant is

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

B
2 sq. units

C
1 sq. unit

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

Solution

$$y=\cos x,\ y=x+1$$ and $$y=0$$
Area of shaded region is required is (Area of curve OAB- area of $$\triangle^{le}$$ OCB)
$$\overset{0}{\underset{-\pi /2}{\int}}\cos x\:dx-\overset{0}{\underset{-1}{\int}}(x+1)dx$$
$$\sin x|_{-\pi /2}^{0}-\left ( \frac{x^2}{2}+x \right )|_{-1}^{0}$$
$$1-1/2=1/2 sq\:units$$