Application of Integrals Test

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Application of Integrals Test - 52

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Question 1
1 / -0

The area bounded by the parabola $$\mathrm{y}^{2}=4\mathrm{a}(\mathrm{x}+\mathrm{a})$$ and $$\mathrm{y}^{2}=-4\mathrm{a}(\mathrm{x}-\mathrm{a})$$ is

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

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

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

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

Solution

$$\mathrm{y}^{2}=4\mathrm{a}(\mathrm{x}+\mathrm{a})$$ and $$\mathrm{y}^{2}=-4\mathrm{a}(\mathrm{x}-\mathrm{a})$$
The area bounded by the parabola in the 2nd quadrant,
$$\overset{0}{\underset{-a}{\int}}\sqrt{4a(x+a)}dx$$
Since the figure is symmetric, multiply the given area by 4.
$$4\times \: 2\sqrt{a}\overset{0}{\underset{-a}{\int}}\sqrt{(x+a)}dx$$
$$\displaystyle \frac{2}{3}\times 8\sqrt{a}(x+a)^{3/2}|_{-a}^{0}$$
$$\displaystyle \frac{2}{3}\times 8\times a^2=\displaystyle \frac{16}{3}a^2sq\:units$$.
Question 2
1 / -0

$$\sin x$$ & $$\cos x$$ meet each other at a number of points and develop symmetrical area. Area of one such region is

A
$$4\sqrt{2}$$

B
$$3\sqrt{2}$$

C
$$2\sqrt{2}$$

D
$$\sqrt{2}$$

Solution

Considering one such case:
$$\sin x$$ ans $$\cos x$$ meet each other at $$\dfrac{\pi}4$$ and $$\dfrac{5\pi}4$$.
Area enclosed by $$\sin x$$ and $$\cos x$$ between $$\dfrac{\pi}4$$ and $$\dfrac{5\pi}4$$ is twice the area enclosed between $$\dfrac{\pi}4$$ and $$\dfrac{3\pi}4$$.
$$\therefore A = \displaystyle \int_{\tfrac{\pi}4}^{\tfrac{5\pi}{4}}(\sin x - \cos x)\>dx = 2\int_{\tfrac{\pi}4}^{\tfrac{3\pi}{4}}(\sin x - \cos x)\>dx$$
$$A = 2\left(\displaystyle\int_{\tfrac{\pi}4}^{\tfrac{\pi}{2}}(\sin x - \cos x)\>dx + \int_{\tfrac{\pi}2}^{\tfrac{3\pi}{4}}(\sin x - \cos x)\>dx\right)$$
$$A = 2\sqrt2$$
Question 3
1 / -0

Let $$\displaystyle \mathrm{f}(\mathrm{x})=\min\{x+1,\ \sqrt{1-x}\}$$, then the area bounded by $$\mathrm{y}={f}({x})$$ and $${x}$$-axis is:

A
$$\dfrac76$$

B
$$\dfrac56$$

C
$$\dfrac16$$

D
$$\dfrac{11}{6}$$

Solution

$$\displaystyle \mathrm{f}(\mathrm{x})=\min\{x+1,\ \sqrt{1-x}\}$$
$$y=x+1$$ and $$y=\sqrt{1-x}$$ curves intersect at $$(0,1)$$ and
$$f(x) = x+1$$ for $$x\in (-1,0)$$
$$f(x) = \sqrt{1-x}$$ for $$x\in (0,1)$$
$$\therefore$$ Area under the curve and x-axis is
$$A = \displaystyle \int _{ -1 }^{ 0 }{ (x+1) } dx+\int _{ 0 }^{ 1 }{ \sqrt { 1-x } } dx$$
=$$\displaystyle\left[ \frac { x^{ 2 } }{ 2 } +x \right]^0_{-1} -\left[ \frac { \left( 1-x \right) ^{ 3/2 } }{ 3/2 } \right] ^{ 1 }_{ 0 }\\ =\dfrac { 1 }{ 2 } +\dfrac { 2 }{ 3 } =\dfrac { 7 }{ 6 } $$
$$\therefore$$ required area is $$ \displaystyle \frac { 7 }{ 6 }$$
Question 4
1 / -0

Area bounded by the curves $$\displaystyle \frac{y}{x}=\log x$$ and $$\displaystyle \frac{y}{2}=-x^{2}+x$$ equals:

A
7/12

B
12/7

C
7/6

D
6/7

Solution

The two curves intersect at $$x=0$$ and $$x=1$$ can be found graphically.
So area bound= $$ \int _0 ^1 (-2x^2+2x - xlogx)dx $$
$$ = (\cfrac{-2x^3}{3} + x^2) | _0 ^1 - \int xlogxdx $$
to Integrate $$xlogxdx$$, substitute $$logx =t,$$
$$dx= x dt$$
Integral becomes $$ \int e^{2t}tdt $$
Integration by parts and then putting limits gives $$= -1/4$$
So bound area $$= -2/3 + 1 +1/4$$
$$= 7/12$$
Question 5
1 / -0

The area bounded by the curves $$\mathrm{y}=2^{\mathrm{x}}$$,$$\mathrm{y}=2\mathrm{x}-\mathrm{x}^{2}$$ between the lines $$\mathrm{x}=0,\ \mathrm{x}=2$$ is

A
$$\displaystyle \frac{3}{\log 2}-\frac{4}{3}$$ sq. units

B
$$\displaystyle \frac{3}{\log 2}+\frac{4}{3}$$ sq.units

C
$$3-4\log 2$$ sq. units

D
$$\displaystyle \frac{4}{3}-\frac{3}{\log 2}$$ sq.units

Solution

$$\mathrm{y}=2^{\mathrm{x}}$$ , $$\mathrm{y}=2\mathrm{x}-\mathrm{x}^{2}$$
$$\overset{2}{\underset{0}{\int}}2^{\mathrm{x}}dx-\overset{2}{\underset{0}{\int}}(2x-x^2)dx$$
$$\dfrac{2^{\mathrm{x}}}{\log 2}\left | _{0}^{2} -\left ( x^2-\dfrac{x^3}{3} \right )\right |_{0}^{2}$$
$$\dfrac{3}{\log 2}-\left [ 4-\dfrac{8}{3} \right ]$$
$$=\left ( \dfrac{3}{\log 2}-4/3 \right )sq\:units.$$
Question 6
1 / -0

The area bounded by two branches of the curve $$(y-x)^{2}=x^{3} \& x=1$$ equals

A
$$3/5$$

B
$$5/4$$

C
$$6/5$$

D
$$4/5$$

Solution

$$ (y-x)^2 = x^3 $$
$$ y = x^{3/2} +x $$ and $$ y = -x^{3/2} +x $$
Area 1 $$= _0 ^1\int ( x^{3/2} +x) dx = \cfrac{2x^{5/2}}{5} + \cfrac{x^2}{2} |_0 ^1 = 2/5 + 1/2$$
Area 2 $$= _0 ^1\int ( -x^{3/2} +x) dx = \cfrac{-2x^{5/2}}{5} + \cfrac{x^2}{2} |_0 ^1 = -2/5 +1/2$$
Total Area $$=$$ Area 1 - Area 2$$ =$$ $$\cfrac{4}{5}$$
Question 7
1 / -0

Area bounded by $$x^{2}=4ay$$ and $$y=\displaystyle \frac{8a^{3}}{x^{2}+4a^{2}}$$ is:

A
$$\displaystyle \frac{a^{2}}{3}(6\pi-4)$$

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

C
$$\displaystyle \frac{a^{2}}{3}(6\pi+4)$$

D
$$0$$

Solution

points of intersection
$$x^{2}=4ay$$ , $$y=\displaystyle \frac{8a^{3}}{x^{2}+4a^{2}}$$
$$y(4ay+4a^{2})=8a^3$$
$$\Rightarrow 4ay^2+4a^{2}y-8a^3=0$$
$$y^2+ay-2a^2=0$$
$$y=-2a;y=a$$
there is only 1 point of intersection for $$y=a$$
So, they don't enclose area
So, zero
Question 8
1 / -0

Area bounded by the curves satisfying the conditions $$\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{36}\leq 1\leq\frac{x}{5}+\frac{y}{6}$$ is given by

A
$$15(\displaystyle \dfrac{\pi}{2}+1)$$ sq.units

B
$$\dfrac{15}{4}(\dfrac{\pi}{2}-1)$$ sq.units

C
$$30(\pi-1)$$ sq.unit

D
$$\displaystyle \dfrac{15}{2}(\pi-2)$$ sq.unit

Solution

Area bound b/w the ellipse and the line
Area of quarter ellipse $$= \cfrac{\pi ab}{4} = \cfrac{\pi \times 5 \times 6}{4} $$
Area under the line$$= \cfrac{1}{2} \times 5 \times 6 = \cfrac{30}{2} $$
Area b/w the two curves $$= \cfrac{15 \pi}{2} -\cfrac{30}{2} =\dfrac{15}{2}(\pi-2)$$
Question 9
1 / -0

The area of the region bounded by the curve y $$\displaystyle =\frac{16-x^{2}}{4}$$ and $$\displaystyle y=sec^{-1}[-sin^{2}x],$$ where [.] stands for the greatest integer function is:

A
$$(4-\pi )^{3/2}$$

B
$$\dfrac{8}{3}(4-\pi )^{3/2}$$

C
$$\dfrac{4}{3}(4-\pi )^{3/2}$$

D
$$\dfrac{8}{3}(4-\pi )$$

Solution

$$[-sin^{2}x]=0\, or\, -1.\, But\, sec^{-1}0$$ is not defined .
$$\therefore [-sin^{2}x]=-1.$$
Hence the required area $$=$$ area between the parabola $$x^{2}=-4(y-4)$$ and the straight line
$$y=sec^{-1}=\pi .$$
Hence the required area$$=$$ 2area MAB
$$2\int_{\pi }^{4}x\, dy=2\int_{\pi }^{4}\sqrt{16-4y}\, dy =4\int_{\pi }^{4}\sqrt{4-y}\, dy$$
$$=-4\left \{ \dfrac{(4-y)^{3/2}}{\dfrac{3}{2}} \right \}_{\pi }^{4}=\dfrac{8}{3}(4-\pi )^{3/2}$$
Question 10
1 / -0

The area of the smaller region in which the curve $$y=\left [ \frac{x^{3}}{100}+\frac{x}{50} \right ],$$ where[.] denotes the greatest integer function, divides the circle $$\left ( x-2 \right )^{2}+\left ( y+1 \right )^{2}=4,$$ is equal to

A
$$\frac{2\pi-3\sqrt{3}}{3}sq. units$$

B
$$\frac{3\sqrt{3}-\pi}{3}sq. units$$

C
$$\frac{4\pi-3\sqrt{3}}{3}sq. units$$

D
$$\frac{5\pi-3\sqrt{3}}{3}sq. units$$

E
$$\frac{4\pi-3\sqrt{3}}{6}sq. units$$

Solution

Circle has (2, 1) as its center and radius of this circle is 2.
Thus, if P(x, y) be any point on it. then x $$\epsilon $$ [0,4]

$$Letg\left ( x \right )=\frac{x^{3}}{100}+\frac{x}{50}$$

$$\Rightarrow g'\left ( x \right )=\frac{3x^{2}}{100}+\frac{1}{50}> 0 \forall x \epsilon \left ( 0,4 \right )$$

Thus g(x) is increasing in [0, 4].
g(0) $$=$$ 0, g(4)$$ =$$ $$\frac{18}{25}.$$

Hence g(x) $$\epsilon \left [ 0,\frac{18}{25} \right ]\forall x \epsilon \left [ 0,4 \right ]$$

$$\Rightarrow \left [ g\left ( x \right ) \right ]=0 \forall x \epsilon \left [ 0,4 \right ].$$

Thus $$y=\left [ \frac{x^{3}}{100}+\frac{x}{50} \right ],$$ simply represents the x-axis.
CA $$=$$ CB $$=$$ 2, CD $$=$$ 1

$$\Rightarrow cos\theta =\frac{CD}{CA}=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3}\Rightarrow \angle ACB=\frac{2\pi}{3}.$$

$$\Delta _{ACB}=\frac{1}{2}.2^{2}sin\frac{2\pi}{3}=\sqrt{3}sq. units$$

Area of sector ACB $$=\frac{1}{2}.2^{2}.\frac{2\pi}{3}=\frac{4\pi}{3}sq. units$$

Thus area of smaller segment is $$=\frac{4\pi}{3}-\sqrt{3}=\frac{4\pi-3\sqrt{3}}{3}$$ sq. units

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Re-Attempt Weekly Quiz Competition

Application of Integrals Test - 52

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Application of Integrals Test - 52

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$4\sqrt{2}$$

$$3\sqrt{2}$$

$$2\sqrt{2}$$

$$\sqrt{2}$$

Solution

Question 3

$$\dfrac76$$

$$\dfrac56$$

$$\dfrac16$$

$$\dfrac{11}{6}$$

Solution

Question 4

7/12

12/7

7/6

6/7

Solution

Question 5

$$\displaystyle \frac{3}{\log 2}-\frac{4}{3}$$ sq. units

$$\displaystyle \frac{3}{\log 2}+\frac{4}{3}$$ sq.units

$$3-4\log 2$$ sq. units

$$\displaystyle \frac{4}{3}-\frac{3}{\log 2}$$ sq.units

Solution

Question 6

$$3/5$$

$$5/4$$

$$6/5$$

$$4/5$$

Solution

Question 7

$$\displaystyle \frac{a^{2}}{3}(6\pi-4)$$

$$\displaystyle \frac{\pi a^{2}}{3}$$

$$\displaystyle \frac{a^{2}}{3}(6\pi+4)$$

$$0$$

Solution

Question 8

$$15(\displaystyle \dfrac{\pi}{2}+1)$$ sq.units

$$\dfrac{15}{4}(\dfrac{\pi}{2}-1)$$ sq.units

$$30(\pi-1)$$ sq.unit

$$\displaystyle \dfrac{15}{2}(\pi-2)$$ sq.unit

Solution

Question 9

$$(4-\pi )^{3/2}$$

$$\dfrac{8}{3}(4-\pi )^{3/2}$$

$$\dfrac{4}{3}(4-\pi )^{3/2}$$

$$\dfrac{8}{3}(4-\pi )$$

Solution

Question 10

$$\frac{2\pi-3\sqrt{3}}{3}sq. units$$

$$\frac{3\sqrt{3}-\pi}{3}sq. units$$

$$\frac{4\pi-3\sqrt{3}}{3}sq. units$$

$$\frac{5\pi-3\sqrt{3}}{3}sq. units$$

$$\frac{4\pi-3\sqrt{3}}{6}sq. units$$

Solution

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