Chapter 8 Application of Integrals Ex 8.2 Solutions
Question - 1 : - Find the area of thecircle 4x2 + 4y2 = 9 which isinterior to the parabola x2 = 4y
Answer - 1 : -
The required area isrepresented by the shaded area OBCDO.
Solving the givenequation of circle, 4x2 + 4y2 =9, and parabola, x2 = 4y, we obtain thepoint of intersection as
.
It can be observed thatthe required area is symmetrical about y-axis.
∴ Area OBCDO = 2 × AreaOBCO
We draw BM perpendicularto OA.
Therefore, thecoordinates of M are 
.
Therefore, Area OBCO = AreaOMBCO – Area OMBO
Therefore, the requiredarea OBCDO is 
units
Question - 2 : - Find the area bounded bycurves (x – 1)2 + y2 =1 and x2 + y 2 = 1
Answer - 2 : -
The area bounded by thecurves, (x – 1)2 + y2 =1 and x2 + y 2 =1, is represented by the shaded area as
On solving theequations, (x – 1)2 + y2 =1 and x2 + y 2 =1, we obtain the point of intersection as A
and B
.
It can be observed thatthe required area is symmetrical about x-axis.
∴ Area OBCAO = 2 × AreaOCAO
We join AB, whichintersects OC at M, such that AM is perpendicular to OC.
The coordinates of Mare 
.
Therefore, required areaOBCAO = 
units
Question - 3 : - Find the area of theregion bounded by the curves y = x2 +2, y = x, x = 0 and x =3
Answer - 3 : -
The area bounded by thecurves, y = x2 + 2, y = x, x =0, and x = 3, is represented by the shaded area OCBAO as
Then, Area OCBAO = AreaODBAO – Area ODCO
Question - 4 : - Using integration findsthe area of the region bounded by the triangle whose vertices are (–1, 0), (1,3) and (3, 2).
Answer - 4 : -
BL and CM are drawnperpendicular to x-axis.
It can be observed inthe following figure that,
Area (ΔACB) = Area(ALBA) + Area (BLMCB) – Area (AMCA) … (1)
Equation of line segmentAB is
Equation of line segmentBC is
Equation of line segmentAC is
Therefore, from equation(1), we obtain
Area (ΔABC) = (3 + 5 –4) = 4 units
Question - 5 : - Using integration findthe area of the triangular region whose sides have the equations y =2x +1, y = 3x + 1 and x =4.
Answer - 5 : -
The equations of sidesof the triangle are y = 2x +1, y =3x + 1, and x = 4.
On solving theseequations, 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)
Question - 6 : - Smaller area enclosed bythe circle x2 + y2 = 4and the line x + y = 2 is
A. 2 (π – 2)
B. π – 2
C. 2π – 1
D. 2 (π + 2)
Answer - 6 : -
The smaller areaenclosed by the circle, x2 + y2 =4, and the line, x + y = 2, is represented bythe shaded area ACBA as
Thus, the correct answeris B.
It can be observed that,
Area ACBA = Area OACBO –Area (ΔOAB)
Question - 7 : - Area lying between thecurve y2 = 4x and y =2x is
A. 
B. 
C. 
D. 
Answer - 7 : -
The area lying betweenthe curve, y2 = 4x and y =2x, is represented by the shaded area OBAO as
The points ofintersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicularto x-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area(OCABO) – Area (ΔOCA)
square units
Thus, the correct answeris B.