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

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

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Question 1:

Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.

Answer:

The area of the region bounded by the curve, y² = 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 y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, y² = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.

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Question 3:

Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Answer:

The area of the region bounded by the curve, x² = 4y, 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 x² + y² = a² cut off by the line 

Answer:

The area of the smaller part of the circle, x² + y² = a², 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, x² + y² = a², cut off by the line, , is  units.

Question 8:

The area between x = y² 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.

∴ Area OAD = Area ABCD

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

⇒ 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 = x² and 

Answer:

The area bounded by the parabola, x² = y,and the line,, can be represented as

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x² = y, and line, y = x, is A (1, 1).

Area of OACO = Area ΔOAM – Area OMACO

Area of ΔOAM   

Area of OMACO 

⇒ Area of OACO = Area of ΔOAM – Area of OMACO

Therefore, required area = units

Question 10:

Find the area bounded by the curve x² = 4y and the line x = 4y – 2

Answer:

The area bounded by the curve, x² = 4y, and line, x = 4y – 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 y² = 4x and the line x = 3

Answer:

The region bounded by the parabola, y² = 4x, 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 x² + y² = 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 y² = 4x, y-axis and the line y = 3 is

A. 2

B. 

C. 

D. 

Answer:

The area bounded by the curve, y² = 4x, y-axis, and y = 3 is represented as

Thus, the correct answer is B.