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

### Integrals Exercise 8.2 Solutions

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

**Solution**

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4*x*^{2} + 4*y*^{2} = 9, and parabola, *x*^{2} = 4*y* , we obtain the point of intersection as B(√2, 1/2) and D(-√2, 1/2).

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

Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are (√2, 0).

Therefore, Area OBCO = Area OMBCO - Area OMBO

Therefore, the required area OBCDO is

**2. Find the area bounded by curves ( x – 1)^{2} + y^{2} = 1 and x^{2} + y^{ 2} = 1**

**Solution**

The area bounded by the curves, (x - 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1, is represented by the shaded area as

On solving the equations, (x - 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1, we obtain the point of intersections as A (1/2, √3/2) and B(1/2, √3/2)

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

∴ Area OBCAO = 2 × Area OCAO

We join AB, which intersects OC at M, such that AM is perpendicular to OC.

The coordinates of M are (1/2, 0)

⇒ Area OCAO = Area OMAO + Area MCAM

Therefore, required area OBCAO = 2 × [2Ï€/6 - √3/4) = (2Ï€/3 - √3/2) units

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

**Solution**

The area bounded by the curves, y = *x*^{2 }+ 2, *y *= *x*, *x* = 0 and x = 3, is represented b the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO - Area ODCO

= (9 + 6) - (9/2)

= 15 - (9/2)

= 21/2 units

**4. Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).**

**Solution**

BL and CM are drawn perpendicular to x - axis.

It can be observed in the following figure that,

Area (Î” ACB) = Area (ALBA) + Area (BLMCB) - Area (AMCA) **...(1)**

Equation of line segment AB is

y - 0 = [(3 - 0)/(1 + 1)] (x + 1)

y = 3/2 (x + 1)

Equation of line segment BC is

y - 3 = [(2 - 3)/(3 - 1)](x - 1)

y = 1/2 (-x + 7)

Equation of line segment AC is

y - 0 = [(2 -0)/(3 + 1)] (x + 1)

y = 1/2 (x + 1)

Therefore, from equation (1), we obtain

Area (Î”ABC) = (3 + 5 - 4) = 4 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.**

**Solution**

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 A(0, 1), B(4, 13), and C(4, 9)

It can be observed that,

Area (∆ACB) = Area (OLBAO) - Area (OLCAO)

= (24 + 4) - (16 + 4)

= 28 - 20

= 8 units

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

A. 2 (Ï€ – 2)

B. Ï€ – 2

C. 2Ï€ – 1

D. 2 (Ï€ + 2)

**Solution**

The smaller area enclosed by the circle, *x*^{2} + *y*^{2} = 4 and the line *x* + *y* = 2 , is represented by the shaded area ACBA as

It can be observed that,

Area ACBA = Area OACBO - Area (Î” OAB)

= (Ï€ - 2) units

Thus, the correct answer is B.

**7. Area lying between the curve y^{2} = 4x and y = 2x is**

A. 2/3

B. 1/3

C. 1/4

D. 3/4

**Solution**

The area lying between the curve, y^{2} = 4*x* and *y* = 2*x* , is represented by the shaded area OBAO as

The points of intersection of these curves are O(0, 0) and A(1, 2)

We draw AC perpendicular to x - axis such that the coordinates of C are (1, 0).

Area OBAO = Area (Î”OCA) - Area (OCABO)

= |1 - 4/3|

= |-1/3|

= 1/3 units

Thus, the correct answer is B.