## NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals Ex 8.1

#### Page No 365:

#### Question 1:

Find the area of the region bounded by the curve

*y*^{2}=*x*and the lines*x*= 1,*x*= 4 and the*x*-axis.#### Answer:

The area of the region bounded by the curve,

*y*^{2}=*x*, the lines,*x*= 1 and*x*= 4, and the*x*-axis is the area ABCD.#### Question 2:

Find the area of the region bounded by

*y*^{2}= 9*x*,*x*= 2,*x*= 4 and the*x*-axis in the first quadrant.#### Answer:

The area of the region bounded by the curve,

*y*^{2}= 9*x*,*x*= 2, and*x*= 4, and the*x*-axis is the area ABCD.#### Page No 366:

#### Question 3:

Find the area of the region bounded by

*x*^{2}= 4*y*,*y*= 2,*y*= 4 and the*y*-axis in the first quadrant.#### Answer:

The area of the region bounded by the curve,

*x*^{2}= 4*y*,*y*= 2, and*y*= 4, and the*y*-axis is the area ABCD.#### Question 4:

Find the area of the region bounded by the ellipse

#### Answer:

The given equation of the ellipse, , can be represented as

It can be observed that the ellipse is symmetrical about

*x*-axis and*y*-axis.
∴ Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

#### Question 5:

Find the area of the region bounded by the ellipse

#### Answer:

The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about

*x*-axis and*y*-axis.
∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse =

#### Question 6:

Find the area of the region in the first quadrant enclosed by

*x*-axis, line and the circle#### Answer:

The area of the region bounded by the circle, , and the

*x*-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of OAC

Area of ABC

Therefore, required area enclosed =

32 + π3 – 32 = π3 square units

#### Question 7:

Find the area of the smaller part of the circle

*x*^{2}+*y*^{2}=*a*^{2}cut off by the line#### Answer:

The area of the smaller part of the circle,

*x*^{2}+*y*^{2}=*a*^{2}, cut off by the line, , is the area ABCDA.
It can be observed that the area ABCD is symmetrical about

*x*-axis.
∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle,

*x*^{2}+*y*^{2}=*a*^{2}, cut off by the line, , is units.#### Question 8:

The area between

*x*=*y*^{2}and*x*= 4 is divided into two equal parts by the line*x*=*a*, find the value of*a*.#### Answer:

The line,

*x*=*a*, divides the area bounded by the parabola and*x*= 4 into two equal parts.
âˆ´ Area OAD = Area ABCD

It can be observed that the given area is symmetrical about

*x*-axis.
â‡’ Area OED = Area EFCD

From (1) and (2), we obtain

Therefore, the value of

*a*is .#### Question 9:

Find the area of the region bounded by the parabola

*y*=*x*^{2}and#### Answer:

The area bounded by the parabola,

*x*^{2}=*y*,and the line,, can be represented as
The given area is symmetrical about

*y*-axis.
âˆ´ Area OACO = Area ODBO

The point of intersection of parabola,

*x*^{2}=*y*, and line,*y*=*x*, is A (1, 1).
Area of OACO = Area Î”OAM â€“ Area OMACO

Area of Î”OAM

Area of OMACO

â‡’ Area of OACO = Area of Î”OAM â€“ Area of OMACO

Therefore, required area = units

#### Question 10:

Find the area bounded by the curve

*x*^{2}= 4*y*and the line*x*= 4*y*– 2#### Answer:

The area bounded by the curve,

*x*^{2}= 4*y*, and line,*x*= 4*y*– 2, is represented by the shaded area OBAO.
Let A and B be the points of intersection of the line and parabola.

Coordinates of point .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to

*x*-axis.
It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =

#### Question 11:

Find the area of the region bounded by the curve

*y*^{2}= 4*x*and the line*x*= 3#### Answer:

The region bounded by the parabola,

*y*^{2}= 4*x*, and the line,*x*= 3, is the area OACO.
The area OACO is symmetrical about

*x*-axis.
∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

#### Question 12:

Area lying in the first quadrant and bounded by the circle

*x*^{2}+*y*^{2}= 4 and the lines*x*= 0 and*x*= 2 is**A.**π

**B.**

**C.**

**D.**

#### Answer:

The area bounded by the circle and the lines,

*x*= 0 and*x*= 2, in the first quadrant is represented as
Thus, the correct answer is A.

#### Question 13:

Area of the region bounded by the curve

*y*^{2}= 4*x*,*y*-axis and the line*y*= 3 is**A.**2

**B.**

**C.**

**D.**

#### Answer:

The area bounded by the curve,

*y*^{2}= 4*x*,*y*-axis, and*y*= 3 is represented as
Thus, the correct answer is B.

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