Application of Integrals Test

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Application of Integrals Test - 62

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Question 1
1 / -0

The whole area of the curves $$x=a\cos^3t, y=b\sin^3t$$ is given by?

A
$$\dfrac{3}{8}\pi ab$$

B
$$\dfrac{5}{8}\pi ab$$

C
$$\dfrac{1}{8}\pi ab$$

D
None of these

Solution

$$x=a{ \cos }^{ 3 }\theta $$
$$y=b{ \sin }^{ 3 }\theta $$
To get the area,
$$A=\int _{ 0 }^{ 2x }{ x } dy$$
$$=\int _{ 0 }^{ 2\pi }{ a } { \cos }^{ 3 }\theta 3b{ \sin }^{ 2 }\theta \cos\theta d\theta $$
$$=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }\theta { \cos }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 1 \right) $$
Substituting $$\theta \rightarrow \theta +\pi /2$$ in $$1$$,
$$A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta { \sin }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 2 \right) $$
Adding $$(1)$$ and $$(2)$$
$$2A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta } { \sin }^{ 2 }\theta d\theta \quad \longrightarrow \left( 3 \right) $$
$$\left\{ { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right\} $$
Multiplying $$(3)$$ by $$(4)$$
$$8A=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 4 \right) $$
Substituting $$\theta \rightarrow \theta +\pi /4$$ in $$(4)$$
$$8A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 5 \right) $$
Adding $$(4)$$ and $$(5)$$
$$16A=3ab\int _{ 0 }^{ 2\pi }{ 1 } d\theta $$
$$=6ab\pi $$
$$A=\dfrac { 3\pi ab }{ 8 } $$
Question 2
1 / -0

The area between the curves y=tan x, cot x and axis in the interval $$\left[0,\pi \right/2$$]is ?

A
log $$2$$

B
log $$3$$

C
log $$5$$

D
none of these

Solution

$$y=tanx,y=cotx$$ are symmetrical about $$x=\frac { \pi }{ 4 } $$,where they intersect

$$\frac { \pi }{ 4 } Req Area=2|∫tanxdx|$$

$$\frac { \pi }{ 4 } |2∫cotxdx|=2\times \left| \left( lnsec\frac { \pi }{ 4 } \right) -lnsec0 \right| $$

$$\frac { \pi }{ 22 } \times |ln\sqrt { 2 } -0|=2\times \frac { 1 }{ 2 } \times ln2=ln2$$
Question 3
1 / -0

The area bounded by the curves $$y = \sin \left( {x - \left[ x \right]} \right),\,y = \sin 1$$ and the x-axis is

A
$$\sin 1$$

B
$$1 - \sin 1$$

C
$$1 + \sin 1$$

D
None of these

Solution
Question 4
1 / -0

The area (in square units) bounded by the curves $$y = {\cos ^{ - 1}}\left| {\cos \,x} \right|$$ and $$y = {\left( {{{\cos }^{ - 1}}\left| {\cos \,x} \right|} \right)^2},x \in \left[ {0,\pi } \right]$$ is

A
$$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$$

B
$$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)$$

C
$$\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$$

D
$$\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)$$

Solution
Question 5
1 / -0

The area bounded by the curves $$y = sin^{-1} |sin \, x| $$ and $$y = (sin^{-1} | sin \, x|)^2 , \, 0 \le x \le 2 \pi$$ is

A
$$\left(\dfrac{\pi^3}{3} + \dfrac{4}{3} \right)$$ sq. unit

B
$$\left(\dfrac{\pi^3}{6} - \dfrac{\pi^2}{2} + \dfrac{4}{3} \right)$$ sq. unit

C
$$\left(\dfrac{\pi^2}{2} - \dfrac{4}{3} \right)$$ sq. unit

D
$$\left(\dfrac{\pi^2}{6} - \dfrac{\pi}{4} + \dfrac{4}{3} \right)$$ sq. unit

Solution

$$y=\sin ^{-1}|\sin n |$$ and $$y=(\sin ^{-1}|\sin n|^2)^2$$
Required area $$=4\int_{1}^{0}[x-(\sin^{-1}(\sin n))^2]dn+4\int_{\dfrac{\pi}{1}}^{1}[(\sin^{-1}(\sin n))^2-n]dn$$
$$=4\left(\dfrac{1}{2}\right)-4\int_{1}^{0}(\sin ^{-1}(\sin x))^2dn+4\int_{\dfrac{\pi}{2}}^{1}(\sin ^{-1}(\sin n))^2dn-\dfrac{4}{2}\left(\dfrac{\pi^2}{4}-1\right)^1$$
$$=4-\dfrac{\pi}{2}-4\int_{1}^{0}(\sin ^{-1}(\sin n))^2dn +4\int_{\dfrac{\pi}{2}}^{0}(\sin ^{-1}(\sin n))^2dn$$
$$=4-\dfrac{\pi^2}{2}-4\int_{1}^{0}n^2dn+4\int_{\dfrac{\pi}{2}}^{1}n^2dn$$
$$=4-\dfrac{\pi^2}{2}-\dfrac{4}{3}+\dfrac{4}{3}\left[\dfrac{\pi^3}{8}-1^1\right]$$
$$=4-\dfrac{\pi^2}{2}-\dfrac{8}{3}+\dfrac{\pi^3}{6}$$
$$=\boxed{\left(\dfrac{4}{3}-\dfrac{\pi^2}{2}+\dfrac{\pi^3}{6}\right)sq\, unit}$$
Question 6
1 / -0

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

A
$$\frac{{32}}{3}$$

B
$$\frac{{16}}{3}$$

C
$$\frac{8}{3}$$

D
$$0$$

Solution

$$x^2=4y$$ $$y^2=4x$$
$$x=2\sqrt{y}$$ $$=4\times 2\sqrt{y}$$
$$y^2=8\sqrt{y}$$
$$\Rightarrow y^4-8y=0$$
then $$y=0, 4$$
$$x^2=4y$$
we get $$x=4, 0$$
So, the points of intersection are $$(0, 0)$$ & $$(4, 4)$$
Area $$=\displaystyle\int^4_0\displaystyle\int^{x^2/4}_{2\sqrt{x}}dydx=\displaystyle\int^4_0[y]^{x^2/4}_{2\sqrt{x}}dx$$
$$=\displaystyle\int^4_0\left[\dfrac{x^2}{4}-2\sqrt{x}\right]dx$$
$$=\left[\dfrac{x^3}{12}-\dfrac{4}{3}\cdot x^{3/2}\right]^4_0$$
$$=\left[\dfrac{4^3}{12}-\dfrac{4}{3}(4)^{3/2}\right]$$
$$=\dfrac{16}{3}$$ sq. units.
Question 7
1 / -0

The area bounded by $$y=2-\left| 2-x \right|$$ and $$y=\frac { 3 }{ \left| x \right| }$$ is :

A
$$\frac { 4+3\ell n3 }{ 2 } $$

B
$$\frac { 4-3\ell n3 }{ 2 } $$

C
$$\frac { 3 }{ 2 } +\ell n3$$

D
$$\frac { 1 }{ 2 } +\ell n3$$

Solution

$$y=2-\left|2-x\right|$$ $$y=\dfrac{3}{\left|x\right|}$$
$$1$$ can be rewriiten as $$y=x if x<+2$$
On solving these two equation, we get
$$x=\sqrt{3}$$ and $$x=3$$
$$Area=\displaystyle \int_{\sqrt{3}}^{3}{\left({y}_{1}-{y}_{2}\right)}dx$$
$$=\displaystyle \int_{\sqrt{3}}^{3}{\left(2-\left|2-x\right|\right)}-\dfrac{3}{\left|x\right|}dx$$
$$=\displaystyle \int_{\sqrt{3}}^{2}{x-\dfrac{3}{\left|x\right|}+\int_{2}^{3}{4-x-\dfrac{3}{\left|x\right|}}}dx$$
$${\left[\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{\sqrt{3}}^{2}+{\left[4x-\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{2}^{3}$$
$$=\dfrac{1}{2}+3\log_{e}{\dfrac{\sqrt{3}}{2}}+12-\dfrac{9}{2}-3\log_{e}{3}-8+2+3\log_{e}{2}$$
$$=\dfrac{1}{2}+3\ln{\dfrac{\sqrt{3}}{2}}+\dfrac{3}{2}3\ln{\dfrac{2}{3}}$$
$$2+3\ln{\dfrac{\sqrt{3}}{2}}=2-3\ln{\sqrt{3}}=\dfrac{4-3\ln{3}}{2}$$
Question 8
1 / -0

Area of the region defined by $$1\ \le |x|+|y|$$ and $$x^{2}-2x+1 \le 1-y^{2}$$ is $$k \pi$$ then $$k=.....sq$$ units

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

B
$$\dfrac {7}{6}$$

C
$$\dfrac {128}{5}$$

D
$$\dfrac {10}{3}$$

Solution
Question 9
1 / -0

The maximum area bounded by the curves $${y^2} = 4ax,\,\,\,\,y = ax\,\,a$$ and $$y = \frac{x}{a}\,\,,1 \le a \le 2$$ is

A
$$ 44 sq. units $$

B
$$ 74 sq. units $$

C
$$84 sq.units $$

D
$$ 114 sq. units $$

Solution
Question 10
1 / -0

Consider two curves $${C_1}\,:\,y = \frac{1}{x}\,and\,{C_2}:\,y = \,\ell nx$$ on the $$xy$$ plane. Let $${D_1}$$ denotes the region surrounded by $${C_1},{C_2}$$ and the lines $$x=1$$ and $${D_2}$$ denotes the region the region surrounded by $${C_1},{D_2}$$ and the line $$x=a$$. If a $${D_1}={D_2}$$ then the value of $$'a'$$ -

A
$$\frac{e}{2}$$

B
$$e$$

C
$$ 1$$

D
$$2\left( {e - 1} \right)$$

Solution