Application of Integrals Test

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Application of Integrals Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The are included between the curves $$y^2 = 4ax \,$$ and $$\, x^2 = 4 ay$$ is ____ sq units.

A
$$\dfrac{16a^2}{3}$$

B
$$\dfrac{8a^2}{3}$$

C
$$\dfrac{4a^2}{3}$$

D
$$\dfrac{5a^2}{3}$$

Solution

We have, $$y^2 = 4ax$$ --------------------------- (1)

$$x^2 = 4ay$$ ---------------------------- (2)

(1) and (2) intersects hence

$$x = \dfrac{y^2}{4a}$$ (a > 0)

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

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

$$\Rightarrow y^4 – 64a^3y = 0 $$

$$\Rightarrow y[y^3 – (4a)^{3}] = 0 $$

$$\Rightarrow y = 0, 4a $$

When y = 0, x = 0 and when y = 4a, x = 4a.

The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).

Area$$=\int_{b}^{a}\left [ y_1-y_2 \right ]dx$$

$$=\int_{0}^{4a}\left [ \sqrt{4ax}-\dfrac{x^2}{4a} \right ]dx$$

$$\left [ \dfrac{4\sqrt{a}}{3}(x)^{\dfrac{3}{2}} - \dfrac{x^3}{12a}\right ]_0^{4a}$$

$$=\dfrac{32a^2}{3}-\dfrac{16a^2}{3}$$

$$=\dfrac{16a^2}{3}$$
Question 2
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The area of the region enclosed by the curves $$y=x$$, $$x=e$$, $$y=\dfrac{1}{x}$$ and the positive $$x-axis$$ is .

A
$$ \dfrac{3}{2}$$ square units

B
$$ \dfrac{5}{2}$$ square units

C
$$ \dfrac{1}{2}$$ square units

D
$$1$$ square units

Solution

$$A=\int _{ 0 }^{ 1 }{ x } dx+\int _{ 0 }^{ e }{ \cfrac { 1 }{ x } } dx\\ A={ [\cfrac { { x }^{ 2 } }{ 2 } ] }_{ 0 }^{ 1 }+{ [\ln { x } ] }_{ 1 }^{ e }\\ A=\cfrac { 1 }{ 2 } +\ln { e } -\ln { 1 } \\ A=\cfrac { 1 }{ 2 } +1-0\\ A=\cfrac { 3 }{ 2 } $$
Question 3
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The area bounded by the curve $$y=cos ax$$ in one are of the curve is where $$a=4n+1,n\in integer$$

A
$$2a$$

B
$$1/a$$

C
$$2/a$$

D
$$2{a^2}$$

Solution

Area $$ = \left| {\int\limits_0^{\dfrac{\pi }{2}} {\cos \,ax\,dx} } \right|$$

$$ = {\left[ {\dfrac{{\sin \,\,ax\,\,}}{a}} \right]_0}^{\dfrac{\pi }{2}}$$

$$ = \dfrac{1}{a} - 0$$

$$ = \dfrac{1}{a}$$
Question 4
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Area enclosed between the curves $$\left| y \right| = 1 - {x^2}$$ and $${x^2} + {y^2} = 1$$ is

A
$$\dfrac{{3\pi - 14}}{3}$$ sq.units

B
$$\dfrac{{\pi - 8}}{3}$$ sq.units

C
$$\dfrac{{2\pi - 8}}{3}$$ sq.units

D
None of these

Solution

We have,

$${{x}^{2}}+{{y}^{2}}=1$$

$${{y}_{1}}=\sqrt{1-{{x}^{2}}}$$ ……. (1)

$$\left| y \right|=1-{{x}^{2}}$$

$$y=1-{{x}^{2}}$$ ……. (2)

So, area of first quadrant

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

$$ =\int_{0}^{1}{\left( \sqrt{1-{{x}^{2}}}-\left( 1-{{x}^{2}} \right) \right)}dx $$

$$ =\int_{0}^{1}{\left( \sqrt{1-{{x}^{2}}} \right)}dx-\int_{0}^{1}{\left( 1-{{x}^{2}} \right)}dx $$

We know that

$$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{1}{2}a\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$$

Therefore,

$$ =\left[ \dfrac{1}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}\left( x \right) \right]_{0}^{1}-\left( x-\dfrac{{{x}^{3}}}{3} \right)_{0}^{1} $$

$$ =\left[ \left( \dfrac{1}{2}\sqrt{1-1}+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( \dfrac{1}{2}\sqrt{1-0}+\dfrac{1}{2}{{\sin }^{-1}}\left( 0 \right) \right) \right]-\left[ \left( 1-\dfrac{1}{3} \right)-0 \right] $$

$$ =\left[ \left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( \sin \dfrac{\pi }{2} \right) \right)-\left( \dfrac{1}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( \sin 0 \right) \right) \right]-\dfrac{2}{3} $$

$$ =\dfrac{\pi }{4}-\dfrac{1}{2}-\dfrac{2}{3} $$

$$ =\dfrac{\pi }{4}-\left( \dfrac{3+4}{6} \right) $$

$$ =\dfrac{\pi }{4}-\dfrac{7}{6} $$

$$ =\dfrac{3\pi -14}{12} $$

Therefore,

Area of all quadrants

$$ =4\times \left( \dfrac{3\pi-14}{12} \right) $$

$$ =\dfrac{3\pi-14}{3} $$

Hence, this is the answer.
Question 5
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The area bounded by the curves $$y=xe, y=-xe$$ and the line $$x=1$$ is-

A
$$\dfrac{e}{2}$$

B
$$e$$

C
$$\dfrac{1}{e}$$

D
$$ \dfrac{3}{e}$$

Solution

Consider the given curves.

$$y=xe$$

$$y=-xe$$

$$x=1$$

Therefore, area bounded by these curves

$$ =\int_{0}^{1}{\left( xe-\left( -xe \right) \right)}dx $$

$$ =2e\int_{0}^{1}{x}dx $$

$$ =2e\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1} $$

$$ =e\left[ {{x}^{2}} \right]_{0}^{1} $$

$$ =e\left[ {{1}^{2}}-0 \right] $$

$$ =e $$

Hence, this is the answer.
Question 6
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The area bounded by the curves $$y=|x|-1$$ and $$y=-|x|+1$$ is?

A
$$1$$

B
$$2$$

C
$$2\sqrt{2}$$

D
$$4$$

Solution

Given, $$\begin{cases} x-1 \quad x\ge 0 \\ -x-1\quad x<0 \end{cases}$$

$$y=-|x|+1=\begin{cases} -x+1, \quad x\ge 0 \\ x+1\quad x< 0 \end{cases}$$

Plotting on graph

$$OA=OB=OC=OD=1\ unit$$
$$AD=AB=BC=CD=\sqrt{2}\ units$$

Area of bounded region $$=(\sqrt{2})^{2}=2$$

$$\therefore$$ Option $$B$$ is correct
Question 7
1 / -0

Area bounded by curve $$y = k \sin \,x$$ between $$x = \pi$$ and $$x = 2\pi$$, is

A
$$2k$$ sq. unit

B
$$0$$

C
$$\dfrac{k^2}{2}$$ sq. unit

D
None of these

Solution

$$y=k\sin x, x=[\pi,2\pi]$$
Area required = Area BCD
$$=\int_{2\pi}^{\pi}k\sin dx$$
$$=k\int_{2\pi}^{\pi}\sin xdx$$
$$k[=\cos x]^{2\pi}_\pi$$
$$=-k[\cos 2\pi-\cos \pi]$$
$$=-k[1-(-1)]$$
$$-2k$$
since area cannot be negative
$$\therefore Area =2ksq.unit$$
Question 8
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Suppose that $$F(\alpha)$$ denotes the area of the region bounded by $$x=0$$, $$x=2$$, $$y^2=4x$$ and $$y=|\alpha x-1|+|\alpha x-2|+\alpha x$$, where $$\alpha \in \{0, 1\}$$. Then the value of $$F(\alpha)+\dfrac{8\sqrt{2}}{3}$$, when $$\alpha =0$$, is

A
$$4$$

B
$$5$$

C
$$6$$

D
$$9$$

Solution
Question 9
1 / -0

The area bounded by the curves $$x^2=4ay$$ and $$y^2=4ax$$ is,

A
$$0$$

B
$$\dfrac {16a^2}{3}$$

C
$$\dfrac {8a^2}{3}$$

D
$$\dfrac {4a^2}{3}$$

Solution
Question 10
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The area of the region bounded by the curves $$y=x^2 $$ and $$y = \dfrac {2}{1+x^2} $$ is :

A
$$ \pi - \dfrac {2}{3} $$

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

C
$$ \dfrac {\pi}{3} $$

D
$$ \dfrac { 2 \pi}{3} $$

Solution

$$y=x^2$$ and $$y=\dfrac {2}{1+x^2}$$
meeting point
$$x^2 \left(1+x^2\right)=2$$
$$x=1$$ and $$x=-1$$ are roots $$\Rightarrow \left(x^2-1\right)$$ is factor
$$\qquad \quad\left( { x }^{ 2 }+2 \right) \\ \left. { x }^{ 2 }-1 \right) \overline { { x }^{ 4 }+{ x }^{ 2 }-2 } \\ \quad \quad \quad \ { x }^{ 4 }-{ x }^{ 2 }\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 2{ x }^{ 2 }-2 \qquad \qquad \qquad x^2+2=0\\ \qquad \quad 2{ x }^{ 2 }-2 \qquad \qquad \qquad x^2=-2\ not\ positive\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 0\quad $$

$$\displaystyle \therefore area=\int _{ -1 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx } $$
$$\displaystyle =-\left[ \int _{ -1 }^{ 1 }{ \left( { x }^{ 2 }-\dfrac { 2 }{ \left( 1+{ x }^{ 2 } \right) } \right) dx } \right] $$
$$\displaystyle =-\left[ \int _{ -1 }^{ 1 }{ { x }^{ 2 }dx } -\int _{ -1 }^{ 1 }{ \dfrac { 2 }{ 1+{ x }^{ 2 } } dx } \right] $$
$$\displaystyle =-\left[ { \left. \dfrac { { x }^{ 2 } }{ 3 } \right] }_{ -1 }^{ 1 }-{ \left. 2\tan ^{ -1 }{ x } \right] }_{ -1 }^{ 1 } \right] $$
$$\displaystyle =-\left[ \left[ \dfrac { 1 }{ 3 } -\left( -1/3 \right) \right] -2\left[ \pi /4-\left( -\pi /4 \right) \right] \right] $$
$$\displaystyle =-\left[ \dfrac { 2 }{ 3 } -2\left[ \pi /4+\pi /4 \right] \right] $$
$$\displaystyle =\boxed { \pi -\dfrac { 2 }{ 3 } } $$