Miscellaneous Exercise Application of Integrals Solution - NCERT Class 12


Chapter 8 Ex.8.ME Question 1

Find the area under the given curves and given lines:

(i) \(y = {x^2},x = 1,x = 2\) and x-axis

(ii) \(y = {x^4},x = 1,x = 5\) and x-axis

 

Solution

 

(i) \(y = {x^2},x = 1,x = 2\) and x-axis

\[\begin{align}ar\left( {ADCBA} \right) &= \int_1^2 {ydx} \\ &= \int_1^2 {{x^2}dx} \\ &= \left[ {\frac{{{x^3}}}{3}} \right]_1^2\\& = \frac{8}{3} - \frac{1}{3}\\ &= \frac{7}{3}\end{align}\]

(ii) \(y = {x^4},x = 1,x = 5\) and x-axis

\[\begin{align}ar\left( {ADCBA} \right) &= \int_1^5 {ydx} \\& = \int_1^5 {{x^4}dx} \\ &= \left[ {\frac{{{x^5}}}{5}} \right]_1^5\\ &= \frac{{{{\left( 5 \right)}^5}}}{5} - \frac{1}{5}\\ &= {\left( 5 \right)^4} - \frac{1}{5}\\ &= 625 - \frac{1}{5}\\& = 624.8\end{align}\]

Chapter 8 Ex.8.ME Question 2

Find the area between the curves \(y = x\) and \(y = {x^2}\).

 

Solution

 

Point of intersection of \(y = x\,\,{\rm{and}}\,{\rm{ }}y = {x^2}\)is A (1,1).

Draw AC perpendicular to x-axis.

\[\begin{align}ar\left( {OBAO} \right) &= ar\left( {\Delta OCA} \right) - ar\left( {OCABO} \right)\\& = \int_0^1 {xdx} - \int_0^1 {{x^2}dx} \\& = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \left[ {\frac{{{x^3}}}{3}} \right]_0^1\\ &= \frac{1}{2} - \frac{1}{3}\\& = \frac{1}{6}\end{align}\]

Chapter 8 Ex.8.ME Question 3

Find the area of the region lying in the first quadrant and bounded by \(y = 4{x^2},x = 0,y = 1\) and \(y = 4\).

 

Solution

 

\[\begin{align}ar\left( {ABCD} \right) &= \int_1^4 {xdy} \\ &= \int_1^4 {\frac{{\sqrt y }}{2}dy} \\ &= \frac{1}{2}\left[ {\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^4\\ &= \frac{1}{3}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - 1} \right]\\& = \frac{1}{3}\left[ {8 - 1} \right]\\ &= \frac{7}{3}\end{align}\]

Chapter 8 Ex.8.ME Question 4

Question 4:

Sketch the graph of \(y = \left| {x + 3} \right|\) and evaluate \(\int_{ - 6}^0 {\left| {x + 3} \right|dx} \).

 

Solution

 

X

– 6

– 5

– 4

– 3

– 2

– 1

0

Y

3

2

1

0

1

2

3

\(\left( {x + 3} \right) \le 0\) for \( - 6 \le x \le - 3\) and \(\left( {x + 3} \right) \ge 0\) for \( - 3 \le x \le 0\)

\[\begin{align}\int_{ - 6}^0 {\left| {\left( {x + 3} \right)} \right|dx}& = - \int_{ - 6}^{ - 3} {\left( {x + 3} \right)dx + \int_{ - 3}^0 {\left( {x + 3} \right)} } dx\\ &= - \left[ {\frac{{{x^2}}}{2} + 3x} \right]_{ - 6}^{ - 3} + \left[ {\frac{{{x^2}}}{2} + 3x} \right]_{ - 3}^0\\ &= - \left[ {\left( {\frac{{{{\left( { - 3} \right)}^2}}}{2} + 3\left( { - 3} \right)} \right) - \left( {\frac{{{{\left( { - 6} \right)}^2}}}{2} + 3\left( { - 6} \right)} \right)} \right]\\&\qquad + \left[ {0 - \left( {\frac{{{{\left( { - 3} \right)}^2}}}{2} + 3\left( { - 3} \right)} \right)} \right]\\ &= - \left[ { - \frac{9}{2}} \right] - \left[ { - \frac{9}{2}} \right]\\ &= 9\end{align}\]

Chapter 8 Ex.8.ME Question 5

Find the area bounded by the curve \(y = \sin x\) between \(x = 0\) and \(x = 2\pi \).

 

Solution

 

Area bounded by the curve = Area OABO + Area BCDB

\[\begin{align}ar\left( {OABO} \right) + ar\left( {BCDB} \right) &= \int_0^\pi {\sin xdx} + \left| {\int_\pi ^{2\pi } {\sin xdx} } \right|\\& = \left[ { - \cos x} \right]_0^\pi + \left| {\left[ { - \cos x} \right]_\pi ^{2\pi }} \right|\\ &= \left[ { - \cos \pi + \cos 0} \right] + \left| { - \cos 2\pi + \cos \pi } \right|\\ &= 1 + 1 + \left| {\left( { - 1 - 1} \right)} \right|\\ &= 2 + \left| { - 2} \right|\\& = 2 + 2\\ &= 4\end{align}\]

Chapter 8 Ex.8.ME Question 6

Find the area enclosed between the parabola \({y^2} = 4ax\) and the line \(y = mx\).

 

Solution

 

Points of intersection of curves are \(\left( {0,0} \right)\) and \(\left( {\frac{{4a}}{{{m^2}}},\frac{{4a}}{m}} \right)\)

Draw AC perpendicular to x-axis.

\[\begin{align}ar\left( {OABO} \right) &= ar\left( {OCABO} \right) - ar\left( {\Delta OCA} \right)\\ &= \int_0^{\frac{{4a}}{{{m^2}}}} {2\sqrt {ax} dx - \int_0^{\frac{{4a}}{{{m^2}}}} {mxdx} } {\rm{ }}\\ &= 2\sqrt a \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^{\frac{{4a}}{{{m^2}}}} - m\left[ {\frac{{{x^2}}}{2}} \right]_0^{\frac{{4a}}{{{m^2}}}}\\ &= \frac{4}{3}\sqrt a {\left( {\frac{{4a}}{{{m^2}}}} \right)^{\frac{3}{2}}} - \frac{m}{2}\left[ {{{\left( {\frac{{4a}}{{{m^2}}}} \right)}^2}} \right]\\& = \frac{{32{a^2}}}{{3{m^3}}} - \frac{m}{2}\left( {\frac{{16{a^2}}}{{{m^4}}}} \right)\\ &= \frac{{32{a^2}}}{{3{m^3}}} - \frac{{8{a^2}}}{{{m^3}}}\\& = \frac{{8{a^2}}}{{3{m^3}}}\end{align}\]

Chapter 8 Ex.8.ME Question 7

Find the area enclosed by the parabola \(4y = 3{x^2}\) and the line \(2y = 3x + 12\).

 

Solution

 

Points of intersection of curves are \(A\left( { - 2,3} \right)\) and \(B\left( {4,12} \right)\).

Draw AC and BD perpendicular to x-axis.

\[\begin{align}ar\left( {OBAO} \right) &= ar\left( {CDBA} \right) - ar\left( {ODBO + OACO} \right)\\ &= \int_{ - 2}^4 {\frac{1}{2}\left( {3x + 12} \right)dx - \int_{ - 2}^4 {\frac{{3{x^2}}}{4}} } dx\\ &= \frac{1}{2}\left[ {\frac{{3{x^2}}}{2} + 12x} \right]_{ - 2}^4 - \frac{3}{4}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 2}^4\\ &= \frac{1}{2}\left[ {24 + 48 - 6 + 24} \right] - \frac{1}{4}\left[ {64 + 8} \right]\\ &= \frac{1}{2}\left[ {90} \right] - \frac{1}{4}\left[ {72} \right]\\& = 45 - 18\\& = 27\end{align}\]

Chapter 8 Ex.8.ME Question 8

Find the area of the smaller region bounded by the ellipse \(\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\) and the line \(\frac{x}{3} + \frac{y}{2} = 1\).

 

Solution

 

\[\begin{align}ar\left( {BCAB} \right)&= ar\left( {OBCAO} \right) - ar\left( {OBAO} \right)\\ &= \int_0^3 {2\sqrt {1 - \frac{{{x^2}}}{9}} dx} - \int_0^3 {2\left( {1 - \frac{x}{3}} \right)dx} \\ &= \frac{2}{3}\left[ {\int_0^3 {\sqrt {9 - {x^2}} dx} } \right] - \frac{2}{3}\int_0^3 {\left( {3 - x} \right)} dx\\ &= \frac{2}{3}\left[ {\frac{x}{2}\sqrt {9 - {x^2}} + \frac{9}{2}{{\sin }^{ - 1}}\frac{x}{3}} \right]_0^3 - \frac{2}{3}\left[ {3x - \frac{{{x^2}}}{2}} \right]_0^3\\& = \frac{2}{3}\left[ {\frac{9}{2}\left( {\frac{\pi }{2}} \right)} \right] - \frac{2}{3}\left[ {9 - \frac{9}{2}} \right]\\ &= \frac{2}{3}\left[ {\frac{{9\pi }}{4} - \frac{9}{2}} \right]\\ &= \frac{2}{3} \times \frac{9}{4}\left( {\pi - 2} \right)\\& = \frac{3}{2}\left( {\pi - 2} \right)\end{align}\]

Chapter 8 Ex.8.ME Question 9

Find the area of the smaller region bounded by the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) and the line \(\frac{x}{a} + \frac{y}{b} = 1\).

 

Solution

 

\[\begin{align}ar\left( {CBA} \right) &= ar\left( {OBCAO} \right) - ar\left( {OBAO} \right)\\ &= \int_0^a b \sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} dx - \int_0^a b \left( {1 - \frac{x}{a}} \right)dx\\& = \frac{b}{a}\left[ {\left\{ {\frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right\}_0^a - \left\{ {ax - \frac{{{x^2}}}{2}} \right\}_0^a} \right]\\& = \frac{b}{a}\left[ {\left\{ {\frac{{{a^2}}}{2}\left( {\frac{\pi }{2}} \right)} \right\} - \left\{ {{a^2} - \frac{{{a^2}}}{2}} \right\}} \right]\\ &= \frac{b}{a}\left[ {\frac{{{a^2}\pi }}{4} - \frac{{{a^2}}}{2}} \right]\\& = \frac{{b{a^2}}}{{2a}}\left[ {\frac{\pi }{2} - 1} \right]\\& = \frac{{ab}}{2}\left[ {\frac{\pi }{2} - 1} \right]\\ &= \frac{{ab}}{4}\left( {\pi - 2} \right)\end{align}\]

Chapter 8 Ex.8.ME Question 10

Find the area of the region enclosed by the parabola \({x^2} = y\), the line \(y = x + 2\) and x-axis.

 

Solution

 

Point of intersection of \({x^2} = y\) and \(y = x + 2\), is \(A\left( { - 1,1} \right)\) and \(C\left( {2,4} \right)\).

Now required Area = Area of trapezium ALMB- Area of ALODBM

\[\begin{align}ar\left( {trap.ALMB} \right) - ar\left( {ALODBM} \right) &= \int_{ - 1}^2 {\left( {x + 2} \right)dx} - \int_{ - 1}^2 {{x^2}dx} \\ &= \left[ {\frac{{{x^2}}}{2} + 2x} \right]_{ - 1}^2 - \left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^2\\ &= \left[ {2 + 4 - \frac{1}{2} + 2} \right] - \left[ {\frac{8}{3} + \frac{1}{3}} \right]\\ &= \frac{{15}}{2} - 3\\ &= \frac{9}{2}\end{align}\]

Chapter 8 Ex.8.ME Question 11

Using the method of integration find the area bounded by the curve \(\left| x \right| + \left| y \right| = 1\)

[Hint: The required region is bounded by lines \(x + y = 1,x - y = 1, - x + y = 1\) and \( - x - y = 1\)]

 

Solution

 

Curve intersects axis at points \(A\left( {0,1} \right),B\left( {1,0} \right),C\left( {0, - 1} \right)\) and \(D\left( { - 1,0} \right)\).

Curve is symmetrical about x-axis and y-axis.

\[\begin{align}ar\left( {ADCB} \right) &= 4 \times ar\left( {OBAO} \right)\\ &= 4\int_0^1 {\left( {1 - x} \right)dx} \\& = 4\left( {x - \frac{{{x^2}}}{2}} \right)_0^1\\ &= 4\left[ {1 - \frac{1}{2}} \right]\\& = 2\end{align}\]

Chapter 8 Ex.8.ME Question 12

Find the area bounded by curves \(\left\{ {\left( {x,y} \right):y \ge {x^2}\,{\rm{and }}y = \left| x \right|} \right\}\)

 

Solution

 

Required area is symmetrical about y-axis.

Required area \(= 2\)[Area (OCAO) – Area (OCADO)]

\[\begin{align}2\left[ {ar\left( {OCAO} \right)-ar\left( {OCADO} \right)} \right] &= 2\left[ {\int_0^1 {xdx} - \int_0^1 {{x^2}dx} } \right]\\ &= 2\left[ {\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \left[ {\frac{{{x^3}}}{3}} \right]_0^1} \right]\\ &= 2\left[ {\frac{1}{2} - \frac{1}{3}} \right]\\ &= 2\left[ {\frac{1}{6}} \right]\\ &= \frac{1}{3}\end{align}\]

Chapter 8 Ex.8.ME Question 13

Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are \(A\left( {2,0} \right),B\left( {4,5} \right)\) and \(C\left( {6,3} \right)\).

 

Solution

 

Equation of AB is

\[\begin{align}y - 0 &= \frac{{5 - 0}}{{4 - 2}}\left( {x - 2} \right)\\2y &= 5x - 10\\y &= \frac{5}{2}\left( {x - 2} \right)\end{align}\]

Equation of BC is

\[\begin{align}y - 5 &= \frac{{3 - 5}}{{6 - 4}}\left( {x - 4} \right)\\2y - 10& = - 2x + 8\\2y& = - 2x + 18\\y &= - x + 9\end{align}\]

Equation of CA is

\[\begin{align}y - 3& = \frac{{0 - 3}}{{2 - 6}}\left( {x - 6} \right)\\ - 4y + 12 &= - 3x + 18\\4y &= 3x - 6\\y& = \frac{3}{4}\left( {x - 2} \right)\end{align}\]

\[\begin{align}ar\left( {\Delta ABC} \right)& = ar\left( {ABLA} \right) + ar\left( {BLMCB} \right) - ar\left( {ACMA} \right)\\ &= \int_2^4 {\frac{5}{2}\left( {x - 2} \right)dx} + \int_4^6 {\left( { - x + 9} \right)dx - \int_2^6 {\frac{3}{4}} } \left( {x - 2} \right)dx\\ &= \frac{5}{2}\left[ {\frac{{{x^2}}}{2} - 2x} \right]_2^4 + \left[ {\frac{{ - {x^2}}}{2} + 9x} \right]_4^6 - \frac{3}{4}\left[ {\frac{{{x^2}}}{2} - 2x} \right]_2^6\\ &= \frac{5}{2}\left[ {8 - 8 - 2 + 4} \right] + \left[ { - 18 + 54 + 8 - 36} \right] - \frac{3}{4}\left[ {18 - 12 - 2 + 4} \right]\\& = 5 + 8 - \frac{3}{4}\left( 8 \right)\\ &= 13 - 6\\ &= 7\,units.\end{align}\]

Chapter 8 Ex.8.ME Question 14

Using the method of integration find the area of the region bounded by lines:

\(2x + y = 4,3x - 2y = 6\) and \(x - 3y + 5 = 0\).

 

Solution

 

AL and CM are perpendicular on x-axis.

\[\begin{align}ar\left( {\Delta ABC} \right)& = ar\left( {ALMCA} \right) - ar\left( {ALB} \right) - ar\left( {CMB} \right)\\ &= \int_1^4 {\left( {\frac{{x + 5}}{3}dx} \right)} - \int_1^2 {\left( {4 - 2x} \right)} dx - \int_2^4 {\left( {\frac{{3x - 6}}{2}} \right)} dx\\& = \frac{1}{3}\left[ {\frac{{{x^2}}}{2} + 5x} \right]_1^4 - \left[ {4x - {x^2}} \right]_1^2 - \frac{1}{2}\left[ {\frac{{3{x^2}}}{2} - 6x} \right]_2^4\\& = \frac{1}{3}\left[ {8 + 20 - \frac{1}{2} - 5} \right] - \left[ {8 - 4 - 4 + 1} \right] - \frac{1}{2}\left[ {24 - 24 - 6 + 12} \right]\\ &= \left( {\frac{1}{3} \times \frac{{45}}{2}} \right) - \left( 1 \right) - \frac{1}{2}\left( 6 \right)\\ &= \frac{{15}}{2} - 1 - 3\\& = \frac{{15}}{2} - 4\\ &= \frac{{15 - 8}}{2}\\ &= \frac{7}{2}\end{align}\]

Chapter 8 Ex.8.ME Question 15

Find the area of the region \(\left\{ {\left( {x,y} \right):{y^2} \le 4x,4{x^2} + 4{y^2} \le 9} \right\}\)

 

Solution

 

Points of intersection of curves are \(\left( {\frac{1}{2},\sqrt 2 } \right)\) and \(\left( {\frac{1}{2}, - \sqrt 2 } \right)\).

Required area is OABCO.

Area OABCO is symmetrical about x-axis.

\[\begin{align}{\text{Area }}OABCO &= {\rm{ }}2 \times Area{\rm{ }}OBC\\ar\left( {OBCO} \right) & = ar\left( {OMC} \right) + ar\left( {MBC} \right)\\ &= \int_0^{\frac{1}{2}} {2\sqrt x dx} + \int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{2}\sqrt {9 - 4{x^2}} dx} \\ &= \int_0^{\frac{1}{2}} {2\sqrt x dx} + \int_{\frac{1}{2}}^{\frac{3}{2}} {\frac{1}{2}\sqrt {{{\left( 3 \right)}^2} - {{\left( {2x} \right)}^2}} dx}\end{align}\]

\[\begin{align}{\rm{ put}}2x &= t \Rightarrow dx = \frac{{dt}}{2}\\{\rm{ When}}x = \frac{3}{2},t& = 3\,\text{and}\,{\rm{when}} \,x = \frac{1}{2},t = 1\\ar\left( {OBCO} \right) &= \int_0^{\frac{1}{2}} {2\sqrt x } dx + \frac{1}{4}\int_1^3 {\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} } dt\\& = 2\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^{\frac{1}{2}} + \frac{1}{4}\left[ {\frac{t}{2}\sqrt {9 - {t^2}} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{t}{3}} \right)} \right]_1^3\\& = 2\left[ {\frac{2}{3}{{\left( {\frac{1}{2}} \right)}^{\frac{3}{2}}}} \right] + \frac{1}{4}\left[ {\left\{ {\frac{3}{2}\sqrt {9 - {{\left( 3 \right)}^2}} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{3}{3}} \right)} \right\} - \left\{ {\frac{1}{2}\sqrt {9 - {{\left( 1 \right)}^2}} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right\}} \right]\\ &= \frac{2}{{3\sqrt 2 }} + \frac{1}{4}\left[ {\left\{ {0 + \frac{9}{2}{{\sin }^{ - 1}}\left( 1 \right)} \right\} - \left\{ {\frac{1}{2}\sqrt 8 + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right\}} \right]\\& = \frac{{\sqrt 2 }}{3} + \frac{1}{4}\left[ {\frac{{9\pi }}{4} - \sqrt 2 - \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right]\\ &= \frac{{\sqrt 2 }}{3} + \frac{{9\pi }}{{16}} - \sqrt 2 - \frac{9}{8}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\\& = \frac{{9\pi }}{{16}} - \frac{9}{8}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right) + \frac{{\sqrt 2 }}{{12}}\end{align}\]

\[\begin{align}ar\left( {OABCO} \right) &= 2 \times ar\left( {OBC} \right)\\ &= 2 \times \frac{{9\pi }}{{16}} - \frac{9}{8}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right) + \frac{{\sqrt 2 }}{{12}}\\ &= \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right) + \frac{{\sqrt 2 }}{6}\\ &= \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right) + \frac{1}{{3\sqrt 2 }}\end{align}\]

Chapter 8 Ex.8.ME Question 16

Area bounded by the curve \(y = {x^3}\), the x-axis and the coordinates \(x = - 2\) and \(x = 1\) is

(A) \( - 9\)

(B) \(-\frac{{15}}{4}\)

(C) \(\frac{{15}}{4}\)

(D) \(\frac{{17}}{4}\)

 

Solution

 

\[\begin{align}{\text{Required area}} &= {\rm{ }}\int_{ - 2}^0 {ydx} + \int_0^1 {ydx} \\ &= \int_{ - 2}^0 {{x^3}dx} + \int_0^1 {{x^3}dx} \\ &= \left[ {\frac{{{x^4}}}{4}} \right]_{ - 2}^0 + \left[ {\frac{{{x^4}}}{4}} \right]_0^1\\& = \left[ {\frac{{{{\left( { - 2} \right)}^4}}}{4} + \frac{1}{4}} \right]\\ &= \left( {4 + \frac{1}{4}} \right) = \frac{{17}}{4}\end{align}\]

Correct answer is D

Chapter 8 Ex.8.ME Question 17

The area bounded by the curve \(y = x\left| x \right|\), x-axis and the coordinates \(x = - 1\) and \(x = 1\) is given by

[Hint: \(y = {x^2}\) if \(x > 0\) and \(y = - {x^2}\) if \(x < 0\)]

(A) \(0\)

(B) \(\frac{1}{3}\)

(C) \(\frac{2}{3}\)

(D) \(\frac{4}{3}\)

 

Solution

 

\[\begin{align}{\text{ Required area }}{\rm{ }}& = \int_{ - 1}^1 {ydx} \\ &= \int_{ - 1}^1 {x\left| x \right|dx} \\ &= \int_{ - 1}^0 {{x^2}dx} + \int_0^1 {{x^2}dx} \\ &= \left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^0 + \left[ {\frac{{{x^3}}}{3}} \right]_0^1\\& = - \left( { - \frac{1}{3}} \right) + \frac{1}{3}\\ &= \frac{2}{3}\end{align}\]

Correct answer is C.

Chapter 8 Ex.8.ME Question 18

The area of the circle \({x^2} + {y^2} = 16\)exterior to the parabola \({y^2} = 6x\).

(A) \(\frac{4}{3}\left( {4\pi - \sqrt 3 } \right)\)

(B) \(\frac{4}{3}\left( {4\pi + \sqrt 3 } \right)\)

(C) \(\frac{4}{3}\left( {8\pi - \sqrt 3 } \right)\)

(D) \(\frac{4}{3}\left( 8\pi +\sqrt{3} \right)~\)

 

Solution

 

Required area = 2[ Area (OADO) + Area (ADBA)]

\[\begin{align}2\left[ {ar\left( {OADO} \right) + ar\left( {ADBA} \right)} \right] &= 2\left[ {\int_0^2 {\sqrt {6x} dx} + \int_2^4 {\sqrt {16 - {x^2}} dx} } \right]\\ &= 2\int_0^2 {\sqrt {6x} dx} + 2\int_2^4 {\sqrt {16 - {x^2}} } dx\\ &= 2\sqrt 6 \int_0^2 {\sqrt x dx} + 2\int_2^4 {\sqrt {16 - {x^2}} } dx\\& = 2\sqrt 6 \times \frac{2}{3}\left[ {{x^{\frac{3}{2}}}} \right]_0^2 + 2\left[ {\frac{x}{2}\sqrt {16 - {x^2}} + \frac{{16}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_2^4\\& = \frac{{4\sqrt 6 }}{3}\left( {2\sqrt 2 - 0} \right) + 2\left[ {\left\{ {0 + 8{{\sin }^{ - 1}}\left( 1 \right)} \right\} - \left\{ {2\sqrt 3 + 8{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right\}} \right]\\& = \frac{{16\sqrt 3 }}{3} + 2\left[ {8 \times \frac{\pi }{2} - 2\sqrt 3 - 8 \times \frac{\pi }{6}} \right]\\ &= \frac{{16\sqrt 3 }}{3} + 8\pi - 4\sqrt 3 - \frac{{8\pi }}{3}\\&= \frac{{16\sqrt 3 + 24\pi - 12\sqrt 3 - 8\pi }}{3}\\ &= \frac{{4\sqrt 3 + 16\pi }}{3}\\& = \frac{4}{3}\left[ {4\pi + \sqrt 3 } \right]{\rm{ }}units\end{align}\]

\[\begin{align}{\text{ Area of circle}}&= \pi {\left( r \right)^2}\\ &= \pi {\left( 4 \right)^2}\\& = 16\pi \\{\rm{Required Area}} &= 16\pi - \frac{4}{3}\left[ {4\pi + \sqrt 3 } \right]\\ &= \frac{4}{3}\left[ {4 \times 3\pi - 4\pi - \sqrt 3 } \right]\\ & = \frac{4}{3}\left( {8\pi - \sqrt 3 } \right)\end{align}\]

Correct answer is C.

Chapter 8 Ex.8.ME Question 19

The area bounded by the y-axis, \(y = \cos x{\rm{ }}and{\rm{ }}y = \sin x\) when \(0 \le x \le \frac{\pi }{2}\).

(A) \(2\left( {\sqrt 2 - 1} \right)\)

(B) \(\sqrt 2 - 1\)

(C) \(\sqrt 2 + 1\)

(D) \(\sqrt 2 \)

 

Solution

 

Required area \(=\) Area (ABLA) \(+\) Area (OBLO)

\[\begin{align}ar\left( {ABLA} \right) + ar\left( {OBLO} \right) &= \int_{\frac{1}{{\sqrt 2 }}}^1 {xdy} + \int_0^{\frac{1}{{\sqrt 2 }}} {xdy} \\& = \int_{\frac{1}{{\sqrt 2 }}}^1 {{{\cos }^{ - 1}}ydy} + \int_0^{\frac{1}{{\sqrt 2 }}} {{{\sin }^{ - 1}}xdy} \\ &= \left[ {y{{\cos }^{ - 1}}y - \sqrt {1 - {y^2}} } \right]_{\frac{1}{{\sqrt 2 }}}^1 + \left[ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right]_0^{\frac{1}{{\sqrt 2 }}}\\ &= \left[ {{{\cos }^{ - 1}}\left( 1 \right) - \frac{1}{{\sqrt 2 }}{{\cos }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + \sqrt {1 - \frac{1}{2}} } \right] \\&\quad+ \left[ {\frac{1}{{\sqrt 2 }}{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + \sqrt {1 - \frac{1}{2} - 1} } \right]\\ &= \frac{{ - \pi }}{{4\sqrt 2 }} + \frac{1}{{\sqrt 2 }} + \frac{\pi }{{4\sqrt 2 }} + \frac{1}{{\sqrt 2 }} - 1\\ &= \frac{2}{{\sqrt 2 }} - 1\\ &= \sqrt 2 - 1\end{align}\]

Correct answer is B.

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