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
unitsQuestion 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 A
and B
.
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
unitsQuestion 3:
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
Answer:
The area bounded by the curves, y = x2 + 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 = 2x +1, y = 3x + 1 and x = 4.
Answer:
The equations of sides of the triangle are y = 2x +1, y = 3x + 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 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.
