NCERT Solutions for Class 12 Maths Chapter 8 - Exercise

Exercise 8.2

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

Ans:

Solving the given equation of circle, \[4{x^2} + 4{y^2} = 9\]and the parabola \[{x^2} = 4y\]

\[4\left( {4y} \right) + 4{y^2} = 9\]

\[4{y^2} + 16y - 9 = 0\]

\[y = \dfrac{{ - 16 \pm \sqrt {{{16}^2} - 4 \times 4 \times \left( { - 9} \right)} }}{{2 \times 4}}\]

\[y = \dfrac{1}{2}{\text{ or  }} - 4.5\]

The value of \[y\] cannot be negative as \[{x^2} = 4y\]

We obtain the point of intersection as \[B\left( {\sqrt 2 ,\dfrac{1}{2}} \right)\] and\[D\left( { - \sqrt 2 ,\dfrac{1}{2}} \right)\]

It can be observed that the required area is symmetrical about y-axis.

\[\therefore {\text{Area OBCDO  =  2 .  Area OBCO}}\]

We draw \[{\text{BM}}\] perpendicular to \[{\text{OA}}\].

Therefore, the coordinates of \[{\text{M}}\]are \[\left( {\sqrt 2 ,0} \right)\],

Therefore, \[{\text{Area OBCO  =  Area OMBCO -- Area OMBO}}\]

\[{\text{Area OMBCO}} = \]Area under the circle \[4{x^2} + 4{y^2} = 9\] between \[x = 0\]and \[x = \sqrt 2 \]

\[{\text{Area OMBCO}}\]= \[\int\limits_0^{\sqrt 2 } {\sqrt {\dfrac{{9 - 4{x^2}}}{4}} dx} \]

\[ = \int\limits_0^{\sqrt 2 } {\sqrt {\dfrac{{9 - 4{x^2}}}{4}} dx} \]

Substitute \[x = \dfrac{3}{2}\sin \theta \],\[dx = \dfrac{3}{2}\cos \theta d\theta \]

then,

\[\int {\sqrt {\dfrac{9}{4} - \dfrac{9}{4}{{\sin }^2}\theta } \dfrac{3}{2}\cos \theta d\theta } \]

\[ = \dfrac{9}{4}\int {\sqrt {1 - {{\sin }^2}\theta } \cos \theta d\theta } \]

\[ = \dfrac{9}{4}\int {{{\cos }^2}\theta d\theta } \]

\[ = \dfrac{9}{4}\int {\dfrac{{1 + \cos 2\theta }}{2}d\theta } \]

\[ = \dfrac{9}{4}\int {\dfrac{1}{2} + \dfrac{{\cos 2\theta }}{2}d\theta } \]

\[ = \dfrac{9}{8}\left[ {\theta  + \dfrac{{\sin 2\theta }}{2}} \right]\]

\[ = \dfrac{9}{8}\left[ {\theta  + \dfrac{{2\sin \theta \cos \theta }}{2}} \right]\]

\[ = \dfrac{9}{8}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{3} + \dfrac{{2x}}{3}\sqrt {1 - \dfrac{{4{x^2}}}{9}} } \right]\]

Applying the limits, then 

\[ = \dfrac{9}{8}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{3} + \dfrac{{2x}}{9}\sqrt {4{x^2} - 9} } \right]_0^{\sqrt 2 }\]

\[ = \dfrac{9}{8}\left[ {{{\sin }^{ - 1}}\dfrac{{2\sqrt 2 }}{3} + \dfrac{{2\sqrt 2 }}{9}\sqrt {4 \times 2 - 9}  - 0} \right]\]

\[{\text{Area OMBCO}}\]\[ = \dfrac{1}{4}\left[ {\sqrt 2  + \dfrac{9}{2}{{\sin }^{ - 1}}\dfrac{{2\sqrt 2 }}{3}} \right]\]…………….. (1)

\[{\text{Area OMBCO}}\]\[{\text{ = }}\]Area under between\[{x^2} = 4y\]and\[x = 0\]and\[x = \sqrt 2 \]

\[{\text{Area OMBO}}\]\[ = \int\limits_0^{\sqrt 2 } {\dfrac{{{x^2}}}{4}} \]

\[ = \dfrac{1}{4}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^{\sqrt 2 }\]

\[{\text{Area OMBO}}\]\[ = \dfrac{1}{{12}}\left[ {2\sqrt 2 } \right] = \dfrac{{\sqrt 2 }}{6}\]…………… (2)

From (1) and (2)

\[{\text{Area OBCO }}\]\[ = \dfrac{{\sqrt 2 }}{2} + \dfrac{9}{8}{\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3} - \dfrac{{\sqrt 2 }}{6}\]

\[ = \dfrac{{\sqrt 2 }}{{12}} + \dfrac{9}{8}{\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3}\]

\[ = \dfrac{1}{2}\left( {\dfrac{{\sqrt 2 }}{6} + \dfrac{9}{4}{{\sin }^{ - 1}}\dfrac{{2\sqrt 2 }}{3}} \right)\]

Then the required area \[{\text{ OBCDO}}\] is\[ = 2 \times \dfrac{1}{2}\left( {\dfrac{{\sqrt 2 }}{6} + \dfrac{9}{4}{{\sin }^{ - 1}}\dfrac{{2\sqrt 2 }}{3}} \right)\]

\[ = \dfrac{{\sqrt 2 }}{6} + \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3}sq.units\]

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

Ans: The are bounded by the curves, \[{\left( {x - 1} \right)^2} + {y^2} = 1\]and\[{x^2} + {y^2} = 1\], is represented by the shaded area as

On solving the equations,\[{\left( {x - 1} \right)^2} + {y^2} = 1\] and \[{x^2} + {y^2} = 1\]

\[{\left( {x - 1} \right)^2} + 1 - {x^2} = 1\]

\[ - 2x + 1 = 0\]

\[x = \dfrac{1}{2}\]

\[{y^2} = 1 - {x^2}\]

\[ = 1 - \dfrac{1}{4} = \dfrac{3}{4}\]

we obtain the point of intersection as \[A\left( {\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right)\]and\[B\left( {\dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2}} \right)\]

It can be observed that the required area is symmetrical about x-axis. 

\[\therefore {\text{ Area OBCAO  = }}2 \times {\text{Area OCAO}}\]

We join \[{\text{AB}}\] which intersects \[{\text{OC}}\]at \[{\text{M}}\], such that \[{\text{AM}}\] is perpendicular to \[{\text{OC}}\].

The coordinates of M are\[\left( {\dfrac{1}{2},0} \right)\].

\[ \Rightarrow {\text{ Area OCAO =  Area OMAO}} + {\text{Area MCAM}}\]

\[ = \int\limits_0^{\dfrac{1}{2}} {\sqrt {1 - {{\left( {x - 1} \right)}^2}} dx + } \int\limits_{\dfrac{1}{2}}^1 {\sqrt {1 - {x^2}} dx} \]

\[ = \left[ {\dfrac{{x - 1}}{2}\sqrt {1 - {{\left( {x - 1} \right)}^2}}  + \dfrac{1}{2}{{\sin }^{ - 1}}\left( {x - 1} \right)} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{x}{2}\sqrt {1 - {x^2}}  + \dfrac{1}{2}{{\sin }^{ - 1}}x} \right]_{\dfrac{1}{2}}^1\]

\[ = \left[ {\dfrac{{ - 1}}{4}\sqrt {1 - {{\left( { - \dfrac{1}{2}} \right)}^2}}  + \dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{1}{2} - 1} \right) - \dfrac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right]_0^{\dfrac{1}{2}} + \left[ {\dfrac{1}{2}{{\sin }^{ - 1}}\left( 1 \right) - \dfrac{1}{4}\sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}}  - \dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right]_{\dfrac{1}{2}}^1\]

\[ = \left[ { - \dfrac{{\sqrt 3 }}{8} + \dfrac{1}{2}\left( { - \dfrac{\pi }{6}} \right) - \dfrac{1}{2}\left( { - \dfrac{\pi }{2}} \right)} \right] + \left[ {\dfrac{1}{2}\left( {\dfrac{\pi }{2}} \right) - \dfrac{{\sqrt 3 }}{8} - \dfrac{1}{2}\left( {\dfrac{\pi }{6}} \right)} \right]\]

\[ =  - \dfrac{{\sqrt 3 }}{4} - \dfrac{\pi }{{12}} + \dfrac{\pi }{4} + \dfrac{\pi }{4} - \dfrac{\pi }{{12}}\]

\[ =  - \dfrac{{\sqrt 3 }}{4} + \dfrac{{2\pi }}{6}\]

Therefore, required area \[{\text{OBCAO}}\]\[ = 2 \times \left( {\dfrac{{2\pi }}{6} - \dfrac{{\sqrt 3 }}{4}} \right)\]

\[ = \left( {\dfrac{{2\pi }}{3} - \dfrac{{\sqrt 3 }}{2}} \right)sq.units\]

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

Ans: The area bounded by the curves,\[y = {x^2} + 2\],\[y = x\],\[x = 0\] and \[x = 3\], is represented by the area \[{\text{OCBAO}}\]as

Then, \[{\text{Area OCBAO =  Area ODBAO -- Area ODCO}}\]

\[ = \int\limits_0^3 {\left( {{x^2} + 2} \right)dx}  - \int\limits_0^3 {xdx} \]

\[ = \left[ {\dfrac{{{x^3}}}{3} + 2x} \right]_0^3 - \left[ {\dfrac{{{x^2}}}{2}} \right]_0^3\]

\[ = \left[ {9 + 6} \right] - \left[ {\dfrac{9}{2}} \right]\]

\[ = 15 - \dfrac{9}{2}\]

Therefore, required area\[ = \dfrac{{21}}{2}sq.units\]

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

Ans: BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

\[{\text{Area}}\left( {\Delta {\text{ACB}}} \right) = {\text{Area}}\left( {{\text{ALBA}}} \right) + {\text{Area}}\left( {{\text{BLMCB}}} \right){\text{ - Area}}\left( {{\text{AMCA}}} \right)\]………

Equation of line segment \[{\text{AB}}\]is

\[{\text{y - 0 = }}\dfrac{{3 - 0}}{{1 + 1}}\left( {x + 1} \right)\]

\[y = \dfrac{3}{2}\left( {x + 1} \right)\]

\[\therefore {\text{Area}}\left( {{\text{ALBA}}} \right) = \int\limits_{ - 1}^1 {\dfrac{3}{2}\left( {x + 1} \right)dx}  = \dfrac{3}{2}\left[ {\dfrac{{{x^2}}}{2} + x} \right]_{ - 1}^1\]

\[ = \dfrac{3}{2}\left[ {\dfrac{1}{2} + 1 - \dfrac{1}{2} + 1} \right] = 3{\text{ }}sq.units\]

Equation of line segment \[{\text{BC}}\] is

\[{\text{y - 3 = }}\dfrac{{2 - 3}}{{3 - 1}}\left( {x - 1} \right)\]

\[y = \dfrac{1}{2}\left( { - x + 7} \right)\]

\[\therefore {\text{Area}}\left( {{\text{BLMCB}}} \right) = \int\limits_1^3 {\dfrac{1}{2}\left( { - x + 7} \right)dx}  = \dfrac{1}{2}\left[ { - \dfrac{{{x^2}}}{2} + 7x} \right]_1^3\]

\[ = \dfrac{1}{2}\left[ { - \dfrac{9}{2} + 21 + \dfrac{1}{2} - 7} \right] = 5{\text{ }}sq.units\]

Equation of line segment \[{\text{AC}}\]is

\[{\text{y - 0 = }}\dfrac{{2 - 0}}{{3 + 1}}\left( {x + 1} \right)\]

\[y = \dfrac{1}{2}\left( {x + 1} \right)\]

\[\therefore {\text{Area}}\left( {{\text{AMCA}}} \right) = \int\limits_{ - 1}^3 {\dfrac{1}{2}\left( {x + 1} \right)dx}  = \dfrac{1}{2}\left[ {\dfrac{{{x^2}}}{2} + x} \right]_{ - 1}^3\]

\[ = \dfrac{1}{2}\left[ {\dfrac{9}{2} + 3 - \dfrac{1}{2} + 1} \right] = 4{\text{ }}sq.units\]

Therefore, From equation,we obtain

\[{\text{Area}}\left( {\Delta {\text{ACB}}} \right) = 3 + 5 - 4 = 4{\text{ }}sq.units\]

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

Ans:The equations of sides of the triangle are \[y = 2x + 1\],\[y = 3x + 1\]and \[x = 4\].

On solving these equations, we obtain the vertices of triangle as \[{\text{A}}\left( {0,1} \right)\],\[{\text{B}}\left( {4,13} \right)\]and \[{\text{C}}\left( {4,9} \right)\].

It can be observed that, 

\[{\text{Area }}\left( {\Delta {\text{ACB}}} \right) = {\text{Area }}\left( {{\text{OLBAO}}} \right) - {\text{Area}}\left( {{\text{OLCAO}}} \right)\]

\[ = \int\limits_0^4 {\left( {3x + 1} \right)dx}  - \int\limits_0^4 {\left( {2x + 1} \right)dx} \]

\[ = \left[ {\dfrac{{3{x^2}}}{2} + x} \right]_0^4 - \left[ {\dfrac{{2{x^2}}}{2} + x} \right]_0^4\]

\[ = \left( {24 + 4} \right) - \left( {16 + 4} \right)\]

\[ = 28 - 20\]

Therefore, required area \[ = 8{\text{ }}sq.units\]

6. Smaller area enclosed by the circle \[{x^2} + {y^2} = 4\]and the line \[x + y = 2\]is

(A)  \[2\left( {\pi  - 2} \right)\](B)\[\pi  - 2\]   (C) \[2\pi  - 1\]  (D)\[2\left( {\pi  + 2} \right)\]

Ans: The smaller area enclosed by the circle\[{x^2} + {y^2} = 4\] and the line\[x + y = 2\], is represented by the area \[{\text{ACBA}}\] as

It can be observed that

\[{\text{Area }}\left( {{\text{ACBA}}} \right) = {\text{Area }}\left( {{\text{OACBO}}} \right) - {\text{Area}}\left( {\Delta {\text{OAB}}} \right)\]

\[ = \int\limits_0^2 {\sqrt {4 - {x^2}} dx}  - \int\limits_0^2 {\left( {2 - x} \right)dx} \]

\[ = \left[ {\dfrac{x}{2}\sqrt {4 - {x^2}}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{x}{2}} \right]_0^2 - \left[ {2x - \dfrac{{{x^2}}}{2}} \right]_0^2\]

\[ = \left[ {2\dfrac{\pi }{2}} \right] - \left[ {4 - 2} \right]\]

\[ = \left( {\pi  - 2} \right){\text{ }}sq.units\]

Thus, the correct answer is \[{\text{B}}\].

7. Area lying between the curves \[{y^2} = 4x\]and\[y = 2x\] is

(A)\[\dfrac{2}{3}\]  (B)  \[\dfrac{1}{3}\] (C) \[\dfrac{1}{4}\]  (D)\[\dfrac{3}{4}\]

Ans:The area lying between the curves\[{y^2} = 4x\]and \[y = 2x\], is represented by the shaded area \[{\text{OABO}}\] as

The points of intersection of these curves are \[{\text{O}}\left( {0,0} \right)\] and \[{\text{A}}\left( {1,2} \right)\]

We draw \[{\text{AC}}\] perpendicular to \[x{\text{ - axis}}\] such that the coordinates of \[{\text{C}}\] are \[\left( {1,0} \right)\]

\[{\text{Area }}\left( {{\text{OBAO}}} \right) = {\text{Area }}\left( {\Delta {\text{OCA}}} \right) - {\text{Area}}\left( {{\text{OCABO}}} \right)\]

\[ = \int\limits_0^1 {2xdx}  - \int\limits_0^1 {2\sqrt x dx} \]

\[ = 2\left[ {\dfrac{{{x^2}}}{2}} \right]_0^1 - 2\left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_0^1\]

\[ = \left| {1 - \dfrac{4}{3}} \right|\]

\[ = \left| { - \dfrac{1}{3}} \right|\]

\[ = \dfrac{1}{3}{\text{ }}\]sq.units

Thus, the correct answer is \[{\text{B}}\].

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 8 All Exercises

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2

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