NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.2

Go back to  'Application of Integrals'

Chapter 8 Ex.8.2 Question 1

Find the area of the circle \(4{x^2} + 4{y^2} = 9\) which is interior to the parabola \({x^2} = 4y\).

Solution

                                                                

Solving \(4{x^2} + 4{y^2} = 9\) and \({x^2} = 4y\), point of intersection \(B\left( {\sqrt 2 ,\frac{1}{2}} \right)\) and \(D\left( { - \sqrt 2 ,\frac{1}{2}} \right)\).

Required area is symmetrical about y-axis.

\(ar\left( {OBCDO} \right) = 2 \times ar\left( {OBCO} \right)\)

Draw BM perpendicular to OA

Coordinates of M are \(\left( {\sqrt 2 ,0} \right)\)

\[\begin{align}ar\left( {OBCO} \right)&= ar\left( {OMBCO} \right)--ar\left( {OMBO} \right)\\ &= \int_0^{\sqrt 2 } {\sqrt {\frac{{\left( {9 - 4{x^2}} \right)}}{4}} dx - \int_0^{\sqrt 2 } {\frac{{{x^2}}}{4}} } dx\\ &= \frac{1}{2}\int_0^{\sqrt 2 } {\sqrt {9 - 4{x^2}} } dx - \frac{1}{4}\int_0^{\sqrt 2 } {{x^2}dx} \\ &= \frac{1}{4}\left[ {x\sqrt {9 - 4{x^2}} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2x}}{3}} \right]_0^{\sqrt 2 } - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_0^{\sqrt 2 }\\& = \frac{1}{4}\left[ {\sqrt 2 \sqrt {9 - 8} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right] - \frac{1}{{12}}{\left( {\sqrt 2 } \right)^3}\\ &= \frac{{\sqrt 2 }}{4} + \frac{9}{8}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3} - \frac{{\sqrt 2 }}{6}\\ &= \frac{{\sqrt 2 }}{{12}} + \frac{9}{8}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\\ &= \frac{1}{4}\left( {\frac{{\sqrt 2 }}{3} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right)\end{align}\]

Required area OBCDO \( = \left( {2 \times \frac{1}{4}\left[ {\frac{{\sqrt 2 }}{3} + \frac{9}{2}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right]} \right) = \left[ {\frac{{\sqrt 2 }}{6} + \frac{9}{4}{{\sin }^{ - 1}}\frac{{2\sqrt 2 }}{3}} \right]\) units.

Chapter 8 Ex.8.2 Question 2

Find the area bounded by curves \({\left( {x - 1} \right)^2} + {y^2} = 1\) and \({x^2} + {y^2} = 1\).

Solution

                                                             

Solving \({\left( {x - 1} \right)^2} + {y^2} = 1\) and \({x^2} + {y^2} = 1\), point of intersection \(A\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)\) and \(B\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)\)

Required area is symmetrical about x-axis.

\(ar\left( {OBCAO} \right) = 2 \times ar\left( {OCAO} \right)\)

Join AB, intersects OC at M

AM is perpendicular to OC

Coordinates of \(M\left( {\frac{1}{2},0} \right)\)

\[\begin{align}ar\left( {OCAO} \right) &= ar\left( {OMAO} \right) + ar\left( {MCAM} \right)\\& = \left[ {\int_0^{\frac{1}{2}} {\sqrt {1 - {{\left( {x - 1} \right)}^2}} dx} + \int_{\frac{1}{2}}^1 {\sqrt {1 - {x^2}} dx} } \right]\\ &= \left[ {\frac{{x - 1}}{2}\sqrt {1 - {{\left( {x - 1} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {x - 1} \right)} \right]_0^{\frac{1}{2}} + \left[ {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x} \right]_{\frac{1}{2}}^1\\ &= \left[ { - \frac{1}{4}\sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2} - 1} \right) - \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right] \\& \qquad+ \left[ { + \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right) - \frac{1}{4}\sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} - \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{8} + \frac{1}{2}\left( { - \frac{\pi }{6}} \right) - \frac{1}{2}\left( { - \frac{\pi }{2}} \right)} \right] + \left[ {\frac{1}{2}\left( {\frac{\pi }{2}} \right) - \frac{{\sqrt 3 }}{8} - \frac{1}{2}\left( {\frac{\pi }{6}} \right)} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{4} - \frac{\pi }{{12}} + \frac{\pi }{4} + \frac{\pi }{4} - \frac{\pi }{{12}}} \right]\\ &= \left[ { - \frac{{\sqrt 3 }}{4} - \frac{\pi }{6} + \frac{\pi }{2}} \right]\\& = \left[ {\frac{{2\pi }}{6} - \frac{{\sqrt 3 }}{4}} \right]\end{align}\]

Required Area OBCAO is \(\left( {2 \times \left[ {\frac{{2\pi }}{6} - \frac{{\sqrt 3 }}{4}} \right]} \right) = \left[ {\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}} \right]\) units.

Chapter 8 Ex.8.2 Question 3

Find the area of the region bounded by the curves \(y = {x^2} + 2,y = x,x = 0\) and \(x = 3\)

Solution

                                                                    

\[\begin{align}ar\left( {OCBAO{\rm{ }}} \right)& = ar\left( {ODBAO} \right)--ar\left( {ODCO} \right)\\ &= \int_0^3 {\left( {{x^2} + 2} \right)dx} - \int_0^3 {xdx} \\& = \left[ {\frac{{{x^3}}}{3} + 2x} \right]_0^3 - \left[ {\frac{{{x^2}}}{2}} \right]_0^3\\& = \left[ {9 + 6} \right] - \left[ {\frac{9}{2}} \right]\\ &= 15 - \frac{9}{2}\\ &= \frac{{21}}{2}\end{align}\]

Chapter 8 Ex.8.2 Question 4

Using integration finds the area of the region bounded by the triangle whose vertices are \(\left( {-1,0} \right),\left( {1,3} \right)\) and \(\left( {3,2} \right)\).

Solution

                                                                 

BL and CM are perpendicular to x-axis.

\(ar\left( {\Delta ACB} \right) = ar\left( {ALBA} \right) + ar\left( {BLMCB} \right) - ar\left( {AMCA} \right)\)

Equation of AB is

\[\begin{align}y - 0 &= \frac{{3 - 0}}{{1 + 1}}\left( {x + 1} \right)\\y &= \frac{3}{2}\left( {x + 1} \right)\end{align}\]

\[\begin{align}ar\left( {ALBA} \right) &= \int_{ - 1}^1 {\frac{3}{2}\left( {x + 1} \right)} dx\\& = \frac{3}{2}\left[ {\frac{{{x^2}}}{2} + x} \right]_{ - 1}^1\\ &= \frac{3}{2}\left[ {\frac{1}{2} + 1 - \frac{1}{2} + 1} \right]\\& = 3\end{align}\]

Equation of BC is

\[\begin{align}y - 3 &= \frac{{2 - 3}}{{3 - 1}}\left( {x - 1} \right)\\y& = \frac{1}{2}\left( { - x + 7} \right)\end{align}\]

\[\begin{align}ar\left( {BLMCB} \right) &= \int_1^3 {\frac{1}{2}\left( { - x + 7} \right)} dx\\ &= \frac{1}{2}\left[ { - \frac{{{x^2}}}{2} + 7x} \right]_1^3\\ &= \frac{1}{2}\left[ { - \frac{9}{2} + 21 + \frac{1}{2} - 7} \right]\\ &= 5\end{align}\]

Equation of AC is

\[\begin{align}y - 0 &= \frac{{2 - 0}}{{3 + 1}}\left( {x + 1} \right)\\y& = \frac{1}{2}\left( {x + 1} \right)\end{align}\]

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

Therefore, \(ar\left( {\Delta ABC} \right) = \left( {3 + 5 - 4} \right) = 4 \,\rm{units}\)

Chapter 8 Ex.8.2 Question 5

Using integration find the area of the triangular region whose sides have the equations \(y = 2x + 1,y = 3x + 1\) and \(x = 4\).

Solution

Vertices of triangle are \(A\left( {0,1} \right),B\left( {4,13} \right)\) and \(C\left( {4,9} \right)\).

                                                                   

\[\begin{align}ar\left( {\Delta ACB} \right)& = ar\left( {OLBAO} \right) - ar\left( {OLCAO} \right)\\ &= \int_0^4 {\left( {3x + 1} \right)dx} - \int_0^4 {\left( {2x + 1} \right)dx} \\ &= \left[ {\frac{{3{x^2}}}{2} + x} \right]_0^4 - \left[ {\frac{{2{x^2}}}{2} + x} \right]_0^4\\& = \left( {24 + 4} \right) - \left( {16 + 4} \right)\\ &= 28 - 20\\ &= 8\end{align}\]

Chapter 8 Ex.8.2 Question 6

Smaller area enclosed by the circle \({x^2} + {y^2} = 4\) and the line \(x + y = 2\) is

(A) \({\rm{2}}\left( {\pi - 2} \right)\)

(B) \(\pi - 2\)

(C) \(2\pi - 1\)

(D) \({\rm{2}}\left( {\pi + 2} \right)\)

Solution

                                                                     

\[\begin{align}ar\left( {ACBA} \right)& = ar\left( {OACBO} \right) - ar\left( {\Delta OAB} \right)\\ &= \int_0^2 {\sqrt {4 - {x^2}} dx} - \int_0^2 {\left( {2 - x} \right)dx} \\& = \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\frac{x}{2}} \right]_0^2 - \left[ {2x - \frac{{{x^2}}}{2}} \right]_0^2\\& = \left[ {2 \times \frac{\pi }{2}} \right] - \left[ {4 - 2} \right]\\ &= \left( {\pi - 2} \right)\end{align}\]

Correct answer is B.

Chapter 8 Ex.8.2 Question 7

Area lying between the curve \({y^2} = 4x\) and \(y = 2x\) is

(A) \(\frac{{\rm{2}}}{3}\)

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

(C) \(\frac{1}{4}\)

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

Solution

                                                           

Points of intersection of curve \({y^2} = 4x\) and \(y = 2x\) are \({\rm{O}}\left( {0,0} \right)\) and\(A\left( {1,2} \right)\).

Draw AC perpendicular to x-axis.

Coordinates of C are (1,0)

\[\begin{align}ar\left( {OBAO} \right) &= ar\left( {\Delta OCA} \right) - ar\left( {OCABO} \right)\\& = \int_0^1 {2xdx} - \int_0^1 {2\sqrt x } dx\\ &= 2\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - 2\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1\\& = \left| {1 - \frac{4}{3}} \right|\\& = \left| { - \frac{1}{3}} \right|\\ &= \frac{1}{3}\end{align}\]

Correct answer is B.

  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0