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Find the area of the region bounded by the curves y = + 2, y = x, x = 0 and x = 3
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)
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.
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
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
=∫2√0(9−4x2)4‾‾‾‾‾‾‾√dx−∫2√0x24dx=∫02(9-4x2)4dx-∫02x24dx
=∫2√0(32)2−x2‾‾‾‾‾‾‾‾‾‾√dx−14∫2√0x2dx=∫02322-x2dx-14∫02x2dx
=(x2(32)2−x2‾‾‾‾‾‾‾‾‾‾√+98sin−12x3)2√0−14(x33)2√0=x2322-x2+98sin-12x302-14x3302
=2√4+98sin−122√3−112(2‾√)3=24+98sin-1223-11223
=122√+98sin−122√3−132√=122+98sin-1223-132
=162√+98sin−122√3=162+98sin-1223
=12[2√6+94sin−122√3]=1226+94sin-1223
Therefore, the required area OBCDO is 
units
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
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
.
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
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3
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
Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
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
Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.
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)
Area lying between the curve y2 = 4x and y = 2x is
A. 
B. 
C. 
D. 
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.
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