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NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals Ex 8.2

NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals Ex 8.2

Page No 371:

Question 1:

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

Answer:

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

Answer:

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 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, xx = 0 and x = 3

Answer:

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).

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 = 2x +1, y = 3x + 1 and = 4.

Answer:

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)

Page No 372:

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)

Answer:

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. 

Answer:

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).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)
 
 square units

Thus, the correct answer is B.

Courtesy : CBSE