# NCERT solution class 12 chapter 2 Application of Integrals exercise 2.2 mathematics part 2

## EXERCISE 2.2

#### Question 1:

Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y

The required area is represented by the shaded area OBCDO. Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as .

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 .

Therefore, Area OBCO = Area OMBCO – Area OMBO

=∫02(9-4×2)4dx-∫02x24dx

=∫02322-x2dx-14∫02x2dx

=x2322-x2+98sin-12×302-14×3302

=24+98sin-1223-11223

=122+98sin-1223-132

=162+98sin-1223

=1226+94sin-1223Therefore, the required area OBCDO is units

#### Question 2:

Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1

The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by the shaded area as On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as A and B .

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 . Therefore, required area OBCAO = units

#### Question 3:

Find the area of the region bounded by the curves y = x+ 2, xx = 0 and x = 3

The area bounded by the curves, y = x+ 2, xx = 0, and x = 3, is represented by the shaded area OCBAO as Then, Area OCBAO = Area ODBAO – Area ODCO #### Question 4:

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

BL and CM are drawn perpendicular to x-axis.

It can be observed in the following figure that,

Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1) Equation of line segment AB is Equation of line segment BC is Equation of line segment AC is Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 4) = 4 units

#### Question 5:

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

The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and = 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 (ΔACB) = Area (OLBAO) –Area (OLCAO) #### Question 6:

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is

A. 2 (π – 2)

B. π – 2

C. 2π – 1

D. 2 (π + 2)

The smaller area enclosed by the circle, x2 + y2 = 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 (ΔOAB) Thus, the correct answer is B.

#### Question 7:

Area lying between the curve y2 = 4x and y = 2x is

A. B. C. D. The area lying between the curve, y2 = 4x and y = 2x, 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).

âˆ´ Area OBAO = Area (OCABO) â€“ Area (Î”OCA)      square units

Thus, the correct answer is B.

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