## EXERCISE 2.2

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#### Question 1:

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

#### Answer:

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.

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} + *y*^{2} = 1 and *x*^{2} + *y*^{ 2} = 1

#### Answer:

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 intersection as Aand 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 }+ 2, *y *= *x*, *x* = 0 and *x* = 3

#### Answer:

The area bounded by the curves, *y* = *x*^{2 }+ 2, *y *= *x*, *x* = 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).

#### Answer:

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* = 2*x* +1, *y* = 3*x* + 1 and *x *= 4.

#### Answer:

The equations of sides of the triangle are *y* = 2*x* +1, *y* = 3*x* + 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 (ΔACB) = Area (OLBAO) –Area (OLCAO)

#### Page No 372:

#### Question 6:

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

**A.** 2 (π – 2)

**B.** π – 2

**C.** 2π – 1

**D. **2 (π + 2)

#### Answer:

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 (ΔOAB)

Thus, the correct answer is B.

#### Question 7:

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

**A.**

**B.**

**C.**

**D. **

#### Answer:

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).

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

square units

Thus, the correct answer is B.