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Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as
The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to x-axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x-axis
(ii) y = x4, x = 1, x = 5 and x –axis
The required area is represented by the shaded area ADCBA as
The required area is represented by the shaded area ADCBA as
Find the area between the curves y = x and y = x2
The required area is represented by the shaded area OBAO as
The points of intersection of the curves, y = x and y = x2, is A (1, 1).
We draw AC perpendicular to x-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4
The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as
Area of ABCDA = ∫41 x dy =∫41 y√2 dy [as, y = 4x2] =12∫41y√ dy =12×23[y3/2]41 =13[(4)3/2 − (1)3/2] =13(8 − 1) =13×7 =73 square units
Sketch the graph of 
and evaluate
The given equation is
The corresponding values of x and y are given in the following table.
| x | – 6 | – 5 | – 4 | – 3 | – 2 | – 1 | 0 |
| y | 3 | 2 | 1 | 0 | 1 | 2 | 3 |
On plotting these points, we obtain the graph of as follows.
It is known that, 
Find the area bounded by the curve y = sin x between x = 0 and x = 2π
The graph of y = sin x can be drawn as
∴ Required area = Area OABO + Area BCDB
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and 
.
We draw AC perpendicular to x-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Find the area of the smaller region bounded by the ellipse 
and the line 
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Find the area of the smaller region bounded by the ellipse 
and the line 
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Using the method of integration find the area bounded by the curve 
[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]
The area bounded by the curve, , is represented by the shaded region ADCB as
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about x-axis and y-axis.
∴ Area ADCB = 4 × Area OBAO
Find the area bounded by curves 
The area bounded by the curves, , is represented by the shaded region as
It can be observed that the required area is symmetrical about y-axis.
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B (4, 5) and C (6, 3)
The vertices of ΔABC are A (2, 0), B (4, 5), and C (6, 3).
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment CA is
Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
The given equations of lines are
2x + y = 4 … (1)
3x – 2y = 6 … (2)
And, x – 3y + 5 = 0 … (3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Find the area of the region 
The area bounded by the curves, , is represented as
The points of intersection of both the curves are
.
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about x-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Therefore, the required area is 
units
Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is
A. – 9
B. 
C. 
D. 
Required Area = ??∫ 0−2ydx??+∫10ydx∫-2 0ydx+∫01ydx
=??∫ 0−2x3dx??+∫10x3dx=???[x44]0−2???+[x44]10=??[0−164]??+[14−0]=??−4??+14=4+14=174 sq. units=∫-2 0x3dx+∫01x3dx=x44-20+x4401=0-164+14-0=-4+14=4+14=174 sq. units
Thus, the correct answer is D.
The area bounded by the curve
, x-axis and the ordinates x = –1 and x = 1 is given by
[Hint: y = x2 if x > 0 and y = –x2 if x < 0]
A. 0
B. 
C. 
D. 
Thus, the correct answer is C.
The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A. 
B. 
C. 
D.
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)
Area bounded by the circle and parabola
=2[area(OADO)+area(ADBA)]=2[∫206x‾‾√ dx+∫4216−x2‾‾‾‾‾‾‾√ dx]=2∫206x‾‾√ dx+2∫4216−x2‾‾‾‾‾‾‾√ dx=26‾√∫20x√ dx+2∫4216−x2‾‾‾‾‾‾‾√ dx=26‾√×23[x32]20+2[x216−x2‾‾‾‾‾‾‾√+162sin−1(x4)]42 =46√3(22‾√−0)+2[{0+8sin−1(1)}−{23‾√+8sin−1(12)}]=163√3+2[8×π2−23‾√−8×π6]=163√3+2(4π−23‾√−4π3)=163√3+8π−43‾√−8π3=163√+24π−43√−8π3=16π+123√3=43[4π+3‾√] square units=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23x3202+2x216-x2+162sin-1x424 =46322-0+20+8sin-11-23+8sin-112=1633+28×π2-23-8×π6=1633+24π-23-4π3=1633+8π-43-8π3=163+24π-43-8π3=16π+1233=434π+3 square units
Area of circle = π (r)2
= π (4)2
= 16π square units
∴ Required area=16π−43(4π+3‾√)=16π−16π3−43√3=32π3−43√3=43[8π−3‾√] square units∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units
Thus, the correct answer is C.
The area bounded by the y-axis, y = cos x and y = sin x when 
A. 
B. 
C. 
D. 
The given equations are
y = cos x … (1)
And, y = sin x … (2)
Required area = Area (ABLA) + area (OBLO)
Integrating by parts, we obtain
Thus, the correct answer is B.
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