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1 | 4218-4221 | Example 5 Find the area bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
= and the ordinates x = 0
and x = ae, where, b2 = a2 (1 – e2) and e < 1 Solution The required area (Fig 8 12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae Note that the area of the region BOB′RFSB
=
20
∫ae ydx
=
2
2
0
2
ae
b
a
x dx
a
−
∫
=
2
2
2
–1
0
2
sin
2
2
ae
b x
a
x
a
x
a
a
−
+
=
2
2 2
2
–1
2
sin
2
b ae a
a e
a
e
a
−
+
=
2
–1
1
sin
ab e
e
e
−
+
EXERCISE 8 |
1 | 4219-4222 | Solution The required area (Fig 8 12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae Note that the area of the region BOB′RFSB
=
20
∫ae ydx
=
2
2
0
2
ae
b
a
x dx
a
−
∫
=
2
2
2
–1
0
2
sin
2
2
ae
b x
a
x
a
x
a
a
−
+
=
2
2 2
2
–1
2
sin
2
b ae a
a e
a
e
a
−
+
=
2
–1
1
sin
ab e
e
e
−
+
EXERCISE 8 1
1 |
1 | 4220-4223 | 12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae Note that the area of the region BOB′RFSB
=
20
∫ae ydx
=
2
2
0
2
ae
b
a
x dx
a
−
∫
=
2
2
2
–1
0
2
sin
2
2
ae
b x
a
x
a
x
a
a
−
+
=
2
2 2
2
–1
2
sin
2
b ae a
a e
a
e
a
−
+
=
2
–1
1
sin
ab e
e
e
−
+
EXERCISE 8 1
1 Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis in the first quadrant |
1 | 4221-4224 | Note that the area of the region BOB′RFSB
=
20
∫ae ydx
=
2
2
0
2
ae
b
a
x dx
a
−
∫
=
2
2
2
–1
0
2
sin
2
2
ae
b x
a
x
a
x
a
a
−
+
=
2
2 2
2
–1
2
sin
2
b ae a
a e
a
e
a
−
+
=
2
–1
1
sin
ab e
e
e
−
+
EXERCISE 8 1
1 Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis in the first quadrant 2 |
1 | 4222-4225 | 1
1 Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis in the first quadrant 2 Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant |
1 | 4223-4226 | Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis in the first quadrant 2 Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant Fig 8 |
1 | 4224-4227 | 2 Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant Fig 8 12
366
MATHEMATICS
3 |
1 | 4225-4228 | Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant Fig 8 12
366
MATHEMATICS
3 Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant |
1 | 4226-4229 | Fig 8 12
366
MATHEMATICS
3 Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant 4 |
1 | 4227-4230 | 12
366
MATHEMATICS
3 Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant 4 Find the area of the region bounded by the ellipse
2
2
1
16
9
x
+y
= |
1 | 4228-4231 | Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant 4 Find the area of the region bounded by the ellipse
2
2
1
16
9
x
+y
= 5 |
1 | 4229-4232 | 4 Find the area of the region bounded by the ellipse
2
2
1
16
9
x
+y
= 5 Find the area of the region bounded by the ellipse
2
2
1
4
9
x
+y
= |
1 | 4230-4233 | Find the area of the region bounded by the ellipse
2
2
1
16
9
x
+y
= 5 Find the area of the region bounded by the ellipse
2
2
1
4
9
x
+y
= 6 |
1 | 4231-4234 | 5 Find the area of the region bounded by the ellipse
2
2
1
4
9
x
+y
= 6 Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4 |
1 | 4232-4235 | Find the area of the region bounded by the ellipse
2
2
1
4
9
x
+y
= 6 Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4 7 |
1 | 4233-4236 | 6 Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4 7 Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
2
x=a |
1 | 4234-4237 | Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4 7 Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
2
x=a 8 |
1 | 4235-4238 | 7 Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
2
x=a 8 The area between x = y2 and x = 4 is divided into two equal parts by the line
x = a, find the value of a |
1 | 4236-4239 | Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line
2
x=a 8 The area between x = y2 and x = 4 is divided into two equal parts by the line
x = a, find the value of a 9 |
1 | 4237-4240 | 8 The area between x = y2 and x = 4 is divided into two equal parts by the line
x = a, find the value of a 9 Find the area of the region bounded by the parabola y = x2 and y = x |
1 | 4238-4241 | The area between x = y2 and x = 4 is divided into two equal parts by the line
x = a, find the value of a 9 Find the area of the region bounded by the parabola y = x2 and y = x 10 |
1 | 4239-4242 | 9 Find the area of the region bounded by the parabola y = x2 and y = x 10 Find the area bounded by the curve x2 = 4y and the line x = 4y – 2 |
1 | 4240-4243 | Find the area of the region bounded by the parabola y = x2 and y = x 10 Find the area bounded by the curve x2 = 4y and the line x = 4y – 2 11 |
1 | 4241-4244 | 10 Find the area bounded by the curve x2 = 4y and the line x = 4y – 2 11 Find the area of the region bounded by the curve y2 = 4x and the line x = 3 |
1 | 4242-4245 | Find the area bounded by the curve x2 = 4y and the line x = 4y – 2 11 Find the area of the region bounded by the curve y2 = 4x and the line x = 3 Choose the correct answer in the following Exercises 12 and 13 |
1 | 4243-4246 | 11 Find the area of the region bounded by the curve y2 = 4x and the line x = 3 Choose the correct answer in the following Exercises 12 and 13 12 |
1 | 4244-4247 | Find the area of the region bounded by the curve y2 = 4x and the line x = 3 Choose the correct answer in the following Exercises 12 and 13 12 Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines
x = 0 and x = 2 is
(A) π
(B) 2
π
(C) 3
π
(D) 4
π
13 |
1 | 4245-4248 | Choose the correct answer in the following Exercises 12 and 13 12 Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines
x = 0 and x = 2 is
(A) π
(B) 2
π
(C) 3
π
(D) 4
π
13 Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
(A) 2
(B)
49
(C)
39
(D)
9
2
8 |
1 | 4246-4249 | 12 Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines
x = 0 and x = 2 is
(A) π
(B) 2
π
(C) 3
π
(D) 4
π
13 Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
(A) 2
(B)
49
(C)
39
(D)
9
2
8 3 Area between Two Curves
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by
cutting the region into a large number of small strips of elementary area and then
adding up these elementary areas |
1 | 4247-4250 | Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines
x = 0 and x = 2 is
(A) π
(B) 2
π
(C) 3
π
(D) 4
π
13 Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
(A) 2
(B)
49
(C)
39
(D)
9
2
8 3 Area between Two Curves
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by
cutting the region into a large number of small strips of elementary area and then
adding up these elementary areas Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig 8 |
1 | 4248-4251 | Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is
(A) 2
(B)
49
(C)
39
(D)
9
2
8 3 Area between Two Curves
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by
cutting the region into a large number of small strips of elementary area and then
adding up these elementary areas Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig 8 13 |
1 | 4249-4252 | 3 Area between Two Curves
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by
cutting the region into a large number of small strips of elementary area and then
adding up these elementary areas Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig 8 13 Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves |
1 | 4250-4253 | Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig 8 13 Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves For setting up a formula for the integral, it is convenient to take elementary area in
the form of vertical strips |
1 | 4251-4254 | 13 Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves For setting up a formula for the integral, it is convenient to take elementary area in
the form of vertical strips As indicated in the Fig 8 |
1 | 4252-4255 | Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves For setting up a formula for the integral, it is convenient to take elementary area in
the form of vertical strips As indicated in the Fig 8 13, elementary strip has height
APPLICATION OF INTEGRALS 367
y
=f x
(��)
X
Y
y
g x
=����(��)
x
=a
x
c
=
y
g x
=����(��)
y
=f x
(��)
x
b
=
A
B
R
C
D
Q
O
P
X′
Y′
f(x) – g(x) and width dx so that the elementary area
Fig 8 |
1 | 4253-4256 | For setting up a formula for the integral, it is convenient to take elementary area in
the form of vertical strips As indicated in the Fig 8 13, elementary strip has height
APPLICATION OF INTEGRALS 367
y
=f x
(��)
X
Y
y
g x
=����(��)
x
=a
x
c
=
y
g x
=����(��)
y
=f x
(��)
x
b
=
A
B
R
C
D
Q
O
P
X′
Y′
f(x) – g(x) and width dx so that the elementary area
Fig 8 13
Fig 8 |
1 | 4254-4257 | As indicated in the Fig 8 13, elementary strip has height
APPLICATION OF INTEGRALS 367
y
=f x
(��)
X
Y
y
g x
=����(��)
x
=a
x
c
=
y
g x
=����(��)
y
=f x
(��)
x
b
=
A
B
R
C
D
Q
O
P
X′
Y′
f(x) – g(x) and width dx so that the elementary area
Fig 8 13
Fig 8 14
dA = [f (x) – g(x)] dx, and the total area A can be taken as
A =
[ ( )
( )]
b
a f x
−g x dx
∫
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
=
( )
( )
b
b
a
a
f x dx
g x dx
−
∫
∫
=
[
]
( )
( )
,
b
a f x
g x
dx
−
∫
where f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8 |
1 | 4255-4258 | 13, elementary strip has height
APPLICATION OF INTEGRALS 367
y
=f x
(��)
X
Y
y
g x
=����(��)
x
=a
x
c
=
y
g x
=����(��)
y
=f x
(��)
x
b
=
A
B
R
C
D
Q
O
P
X′
Y′
f(x) – g(x) and width dx so that the elementary area
Fig 8 13
Fig 8 14
dA = [f (x) – g(x)] dx, and the total area A can be taken as
A =
[ ( )
( )]
b
a f x
−g x dx
∫
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
=
( )
( )
b
b
a
a
f x dx
g x dx
−
∫
∫
=
[
]
( )
( )
,
b
a f x
g x
dx
−
∫
where f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8 14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
=
[
]
[
]
( )
( )
( )
( )
c
b
a
c
f x
g x
dx
g x
f x dx
−
+
−
∫
∫
368
MATHEMATICS
Y
O
P�(4,�4)
C�(4,�0)
Y�
X�
X
Q�(8,�0)
Fig 8 |
1 | 4256-4259 | 13
Fig 8 14
dA = [f (x) – g(x)] dx, and the total area A can be taken as
A =
[ ( )
( )]
b
a f x
−g x dx
∫
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
=
( )
( )
b
b
a
a
f x dx
g x dx
−
∫
∫
=
[
]
( )
( )
,
b
a f x
g x
dx
−
∫
where f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8 14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
=
[
]
[
]
( )
( )
( )
( )
c
b
a
c
f x
g x
dx
g x
f x dx
−
+
−
∫
∫
368
MATHEMATICS
Y
O
P�(4,�4)
C�(4,�0)
Y�
X�
X
Q�(8,�0)
Fig 8 16
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x |
1 | 4257-4260 | 14
dA = [f (x) – g(x)] dx, and the total area A can be taken as
A =
[ ( )
( )]
b
a f x
−g x dx
∫
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
=
( )
( )
b
b
a
a
f x dx
g x dx
−
∫
∫
=
[
]
( )
( )
,
b
a f x
g x
dx
−
∫
where f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8 14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
=
[
]
[
]
( )
( )
( )
( )
c
b
a
c
f x
g x
dx
g x
f x dx
−
+
−
∫
∫
368
MATHEMATICS
Y
O
P�(4,�4)
C�(4,�0)
Y�
X�
X
Q�(8,�0)
Fig 8 16
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8 |
1 | 4258-4261 | 14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
=
[
]
[
]
( )
( )
( )
( )
c
b
a
c
f x
g x
dx
g x
f x dx
−
+
−
∫
∫
368
MATHEMATICS
Y
O
P�(4,�4)
C�(4,�0)
Y�
X�
X
Q�(8,�0)
Fig 8 16
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8 15 |
1 | 4259-4262 | 16
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8 15 Here, we can set y 2 = x or y =
x = f(x) and y = x2
= g(x), where, f (x) ≥ g (x) in [0, 1] |
1 | 4260-4263 | Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8 15 Here, we can set y 2 = x or y =
x = f(x) and y = x2
= g(x), where, f (x) ≥ g (x) in [0, 1] Therefore, the required area of the shaded region
=
[
]
1
0
( )
( )
f x
g x
dx
−
∫
=
1
2
0
x
x
dx
−
∫
1
3
3
2
0
32
x3
x
=
−
=
2
1
1
3
3
3
−
=
Example 7 Find the area lying above x-axis and included between the circle
x2 + y2 = 8x and inside of the parabola y2 = 4x |
1 | 4261-4264 | 15 Here, we can set y 2 = x or y =
x = f(x) and y = x2
= g(x), where, f (x) ≥ g (x) in [0, 1] Therefore, the required area of the shaded region
=
[
]
1
0
( )
( )
f x
g x
dx
−
∫
=
1
2
0
x
x
dx
−
∫
1
3
3
2
0
32
x3
x
=
−
=
2
1
1
3
3
3
−
=
Example 7 Find the area lying above x-axis and included between the circle
x2 + y2 = 8x and inside of the parabola y2 = 4x Solution The given equation of the circle x2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16 |
1 | 4262-4265 | Here, we can set y 2 = x or y =
x = f(x) and y = x2
= g(x), where, f (x) ≥ g (x) in [0, 1] Therefore, the required area of the shaded region
=
[
]
1
0
( )
( )
f x
g x
dx
−
∫
=
1
2
0
x
x
dx
−
∫
1
3
3
2
0
32
x3
x
=
−
=
2
1
1
3
3
3
−
=
Example 7 Find the area lying above x-axis and included between the circle
x2 + y2 = 8x and inside of the parabola y2 = 4x Solution The given equation of the circle x2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16 Thus, the centre of the
circle is (4, 0) and radius is 4 |
1 | 4263-4266 | Therefore, the required area of the shaded region
=
[
]
1
0
( )
( )
f x
g x
dx
−
∫
=
1
2
0
x
x
dx
−
∫
1
3
3
2
0
32
x3
x
=
−
=
2
1
1
3
3
3
−
=
Example 7 Find the area lying above x-axis and included between the circle
x2 + y2 = 8x and inside of the parabola y2 = 4x Solution The given equation of the circle x2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16 Thus, the centre of the
circle is (4, 0) and radius is 4 Its intersection
with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x (x – 4) = 0
or
x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0,0) and P(4,4) above the
x-axis |
1 | 4264-4267 | Solution The given equation of the circle x2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16 Thus, the centre of the
circle is (4, 0) and radius is 4 Its intersection
with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x (x – 4) = 0
or
x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0,0) and P(4,4) above the
x-axis From the Fig 8 |
1 | 4265-4268 | Thus, the centre of the
circle is (4, 0) and radius is 4 Its intersection
with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x (x – 4) = 0
or
x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0,0) and P(4,4) above the
x-axis From the Fig 8 16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
=
4
8
0
4
ydx
ydx
+
∫
∫
=
4
8
2
2
0
4
2
4
(
4)
x dx
x
dx
+
−
−
∫
∫
(Why |
1 | 4266-4269 | Its intersection
with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x (x – 4) = 0
or
x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0,0) and P(4,4) above the
x-axis From the Fig 8 16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
=
4
8
0
4
ydx
ydx
+
∫
∫
=
4
8
2
2
0
4
2
4
(
4)
x dx
x
dx
+
−
−
∫
∫
(Why )
Fig 8 |
1 | 4267-4270 | From the Fig 8 16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
=
4
8
0
4
ydx
ydx
+
∫
∫
=
4
8
2
2
0
4
2
4
(
4)
x dx
x
dx
+
−
−
∫
∫
(Why )
Fig 8 15
APPLICATION OF INTEGRALS 369
=
4
3
4
2
2
2
0
0
2
2
4
, where,
4
3
x
t dt
x
t
×
+
−
−
=
∫
(Why |
1 | 4268-4271 | 16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
=
4
8
0
4
ydx
ydx
+
∫
∫
=
4
8
2
2
0
4
2
4
(
4)
x dx
x
dx
+
−
−
∫
∫
(Why )
Fig 8 15
APPLICATION OF INTEGRALS 369
=
4
3
4
2
2
2
0
0
2
2
4
, where,
4
3
x
t dt
x
t
×
+
−
−
=
∫
(Why )
=
4
2
2
2
–1
0
32
1
4
4
sin
3
2
2
4
t
t
t
+
−
+
×
×
=
2
–1
32
4
1
0
4
sin 1
3
2
2
+
×
+
×
×
32
32
0
8
4
3
2
3
π
=
+
+
×
=
+ π
= 4(8 3 )
3
+ π
Example 8 In Fig 8 |
1 | 4269-4272 | )
Fig 8 15
APPLICATION OF INTEGRALS 369
=
4
3
4
2
2
2
0
0
2
2
4
, where,
4
3
x
t dt
x
t
×
+
−
−
=
∫
(Why )
=
4
2
2
2
–1
0
32
1
4
4
sin
3
2
2
4
t
t
t
+
−
+
×
×
=
2
–1
32
4
1
0
4
sin 1
3
2
2
+
×
+
×
×
32
32
0
8
4
3
2
3
π
=
+
+
×
=
+ π
= 4(8 3 )
3
+ π
Example 8 In Fig 8 17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6 |
1 | 4270-4273 | 15
APPLICATION OF INTEGRALS 369
=
4
3
4
2
2
2
0
0
2
2
4
, where,
4
3
x
t dt
x
t
×
+
−
−
=
∫
(Why )
=
4
2
2
2
–1
0
32
1
4
4
sin
3
2
2
4
t
t
t
+
−
+
×
×
=
2
–1
32
4
1
0
4
sin 1
3
2
2
+
×
+
×
×
32
32
0
8
4
3
2
3
π
=
+
+
×
=
+ π
= 4(8 3 )
3
+ π
Example 8 In Fig 8 17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6 Find the area between the arc AB and the
chord AB |
1 | 4271-4274 | )
=
4
2
2
2
–1
0
32
1
4
4
sin
3
2
2
4
t
t
t
+
−
+
×
×
=
2
–1
32
4
1
0
4
sin 1
3
2
2
+
×
+
×
×
32
32
0
8
4
3
2
3
π
=
+
+
×
=
+ π
= 4(8 3 )
3
+ π
Example 8 In Fig 8 17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6 Find the area between the arc AB and the
chord AB Solution Given equation of the ellipse 9x2 + y2 = 36 can be expressed as
2
2
1
4
36
x
+y
= or
2
2
2
2
1
2
6
x
+y
= and hence, its shape is as given in Fig 8 |
1 | 4272-4275 | 17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6 Find the area between the arc AB and the
chord AB Solution Given equation of the ellipse 9x2 + y2 = 36 can be expressed as
2
2
1
4
36
x
+y
= or
2
2
2
2
1
2
6
x
+y
= and hence, its shape is as given in Fig 8 17 |
1 | 4273-4276 | Find the area between the arc AB and the
chord AB Solution Given equation of the ellipse 9x2 + y2 = 36 can be expressed as
2
2
1
4
36
x
+y
= or
2
2
2
2
1
2
6
x
+y
= and hence, its shape is as given in Fig 8 17 Accordingly, the equation of the chord AB is
y – 0 = 6
0 (
2)
0
−2 x
−
−
or
y = – 3(x – 2)
or
y = – 3x + 6
Area of the shaded region as shown in the Fig 8 |
1 | 4274-4277 | Solution Given equation of the ellipse 9x2 + y2 = 36 can be expressed as
2
2
1
4
36
x
+y
= or
2
2
2
2
1
2
6
x
+y
= and hence, its shape is as given in Fig 8 17 Accordingly, the equation of the chord AB is
y – 0 = 6
0 (
2)
0
−2 x
−
−
or
y = – 3(x – 2)
or
y = – 3x + 6
Area of the shaded region as shown in the Fig 8 17 |
1 | 4275-4278 | 17 Accordingly, the equation of the chord AB is
y – 0 = 6
0 (
2)
0
−2 x
−
−
or
y = – 3(x – 2)
or
y = – 3x + 6
Area of the shaded region as shown in the Fig 8 17 =
2
2
2
0
0
3
4
(6
3 )
x dx
x dx
−
−
−
∫
∫
(Why |
1 | 4276-4279 | Accordingly, the equation of the chord AB is
y – 0 = 6
0 (
2)
0
−2 x
−
−
or
y = – 3(x – 2)
or
y = – 3x + 6
Area of the shaded region as shown in the Fig 8 17 =
2
2
2
0
0
3
4
(6
3 )
x dx
x dx
−
−
−
∫
∫
(Why )
=
2
2
2
2
–1
0
0
4
3
3
4
sin
6
2
2
2
2
x
x
x
x
x
−
+
−
−
=
–1
2
12
3
0
2sin
(1)
12
2
2
×
+
−
−
π
3
2
6
2
=
×
×
−
= 3π – 6
Fig 8 |
1 | 4277-4280 | 17 =
2
2
2
0
0
3
4
(6
3 )
x dx
x dx
−
−
−
∫
∫
(Why )
=
2
2
2
2
–1
0
0
4
3
3
4
sin
6
2
2
2
2
x
x
x
x
x
−
+
−
−
=
–1
2
12
3
0
2sin
(1)
12
2
2
×
+
−
−
π
3
2
6
2
=
×
×
−
= 3π – 6
Fig 8 17
370
MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1) |
1 | 4278-4281 | =
2
2
2
0
0
3
4
(6
3 )
x dx
x dx
−
−
−
∫
∫
(Why )
=
2
2
2
2
–1
0
0
4
3
3
4
sin
6
2
2
2
2
x
x
x
x
x
−
+
−
−
=
–1
2
12
3
0
2sin
(1)
12
2
2
×
+
−
−
π
3
2
6
2
=
×
×
−
= 3π – 6
Fig 8 17
370
MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1) Solution Let A(1, 0), B(2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8 |
1 | 4279-4282 | )
=
2
2
2
2
–1
0
0
4
3
3
4
sin
6
2
2
2
2
x
x
x
x
x
−
+
−
−
=
–1
2
12
3
0
2sin
(1)
12
2
2
×
+
−
−
π
3
2
6
2
=
×
×
−
= 3π – 6
Fig 8 17
370
MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1) Solution Let A(1, 0), B(2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8 18) |
1 | 4280-4283 | 17
370
MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1) Solution Let A(1, 0), B(2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8 18) Area of ∆ABC
= Area of ∆ABD + Area of trapezium
BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2(x – 1), y = 4 – x, y = 1
2 (x – 1), respectively |
1 | 4281-4284 | Solution Let A(1, 0), B(2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8 18) Area of ∆ABC
= Area of ∆ABD + Area of trapezium
BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2(x – 1), y = 4 – x, y = 1
2 (x – 1), respectively Hence,
area of ∆ ABC =
2
3
3
1
2
1
1
2 (
1)
(4
)
2
x
x
dx
x dx
−dx
−
+
−
−
∫
∫
∫
=
2
3
3
2
2
2
1
2
1
1
2
4
2
2
2
2
x
x
x
x
x
x
−
+
−
−
−
=
2
2
2
2
1
3
2
2
2
1
4 3
4 2
2
2
2
2
−
−
−
+
× −
−
×
−
–
2
1
3
1
3
1
2
2
2
−
−
−
= 3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4 |
1 | 4282-4285 | 18) Area of ∆ABC
= Area of ∆ABD + Area of trapezium
BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2(x – 1), y = 4 – x, y = 1
2 (x – 1), respectively Hence,
area of ∆ ABC =
2
3
3
1
2
1
1
2 (
1)
(4
)
2
x
x
dx
x dx
−dx
−
+
−
−
∫
∫
∫
=
2
3
3
2
2
2
1
2
1
1
2
4
2
2
2
2
x
x
x
x
x
x
−
+
−
−
−
=
2
2
2
2
1
3
2
2
2
1
4 3
4 2
2
2
2
2
−
−
−
+
× −
−
×
−
–
2
1
3
1
3
1
2
2
2
−
−
−
= 3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4 Solution Equations of the given circles are
x2 + y2 = 4 |
1 | 4283-4286 | Area of ∆ABC
= Area of ∆ABD + Area of trapezium
BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2(x – 1), y = 4 – x, y = 1
2 (x – 1), respectively Hence,
area of ∆ ABC =
2
3
3
1
2
1
1
2 (
1)
(4
)
2
x
x
dx
x dx
−dx
−
+
−
−
∫
∫
∫
=
2
3
3
2
2
2
1
2
1
1
2
4
2
2
2
2
x
x
x
x
x
x
−
+
−
−
−
=
2
2
2
2
1
3
2
2
2
1
4 3
4 2
2
2
2
2
−
−
−
+
× −
−
×
−
–
2
1
3
1
3
1
2
2
2
−
−
−
= 3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4 Solution Equations of the given circles are
x2 + y2 = 4 (1)
and
(x – 2)2 + y2 = 4 |
1 | 4284-4287 | Hence,
area of ∆ ABC =
2
3
3
1
2
1
1
2 (
1)
(4
)
2
x
x
dx
x dx
−dx
−
+
−
−
∫
∫
∫
=
2
3
3
2
2
2
1
2
1
1
2
4
2
2
2
2
x
x
x
x
x
x
−
+
−
−
−
=
2
2
2
2
1
3
2
2
2
1
4 3
4 2
2
2
2
2
−
−
−
+
× −
−
×
−
–
2
1
3
1
3
1
2
2
2
−
−
−
= 3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4 Solution Equations of the given circles are
x2 + y2 = 4 (1)
and
(x – 2)2 + y2 = 4 (2)
Equation (1) is a circle with centre O at the
origin and radius 2 |
1 | 4285-4288 | Solution Equations of the given circles are
x2 + y2 = 4 (1)
and
(x – 2)2 + y2 = 4 (2)
Equation (1) is a circle with centre O at the
origin and radius 2 Equation (2) is a circle with
centre C (2, 0) and radius 2 |
1 | 4286-4289 | (1)
and
(x – 2)2 + y2 = 4 (2)
Equation (1) is a circle with centre O at the
origin and radius 2 Equation (2) is a circle with
centre C (2, 0) and radius 2 Solving equations
(1) and (2), we have
(x –2)2 + y2 = x2 + y2
or
x2 – 4x + 4 + y2 = x2 + y2
or
x = 1 which gives y =
3
±
Thus, the points of intersection of the given
circles are A(1,
3 ) and A′(1, –
3 ) as shown in
the Fig 8 |
1 | 4287-4290 | (2)
Equation (1) is a circle with centre O at the
origin and radius 2 Equation (2) is a circle with
centre C (2, 0) and radius 2 Solving equations
(1) and (2), we have
(x –2)2 + y2 = x2 + y2
or
x2 – 4x + 4 + y2 = x2 + y2
or
x = 1 which gives y =
3
±
Thus, the points of intersection of the given
circles are A(1,
3 ) and A′(1, –
3 ) as shown in
the Fig 8 19 |
1 | 4288-4291 | Equation (2) is a circle with
centre C (2, 0) and radius 2 Solving equations
(1) and (2), we have
(x –2)2 + y2 = x2 + y2
or
x2 – 4x + 4 + y2 = x2 + y2
or
x = 1 which gives y =
3
±
Thus, the points of intersection of the given
circles are A(1,
3 ) and A′(1, –
3 ) as shown in
the Fig 8 19 Fig 8 |
1 | 4289-4292 | Solving equations
(1) and (2), we have
(x –2)2 + y2 = x2 + y2
or
x2 – 4x + 4 + y2 = x2 + y2
or
x = 1 which gives y =
3
±
Thus, the points of intersection of the given
circles are A(1,
3 ) and A′(1, –
3 ) as shown in
the Fig 8 19 Fig 8 18
Fig 8 |
1 | 4290-4293 | 19 Fig 8 18
Fig 8 19
APPLICATION OF INTEGRALS 371
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO]
(Why |
1 | 4291-4294 | Fig 8 18
Fig 8 19
APPLICATION OF INTEGRALS 371
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO]
(Why )
= 2 [area of the region ODAO + area of the region DCAD]
=
1
2
0
1
2
y dx
y dx
+
∫
∫
=
1
2
2
2
0
1
2
4
(
2)
4
x
dx
x dx
−
−
+
−
∫
∫
(Why |
1 | 4292-4295 | 18
Fig 8 19
APPLICATION OF INTEGRALS 371
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO]
(Why )
= 2 [area of the region ODAO + area of the region DCAD]
=
1
2
0
1
2
y dx
y dx
+
∫
∫
=
1
2
2
2
0
1
2
4
(
2)
4
x
dx
x dx
−
−
+
−
∫
∫
(Why )
=
1
2
–1
0
1
1
2
2
(
2) 4
(
2)
4sin
2
2
x2
x
x
−
−
−
−
+
×
+
2
2
–1
1
1
1
2
4
4sin
2
2
x2
x
x
−
+
×
=
1
2
2
–1
2
–1
1
0
2
(
2) 4
(
2)
4sin
4
4sin
2
2
x
x
x
x
x
x
−
−
−
−
+
+
−
+
=
–1
1
–1
1
1
1
3
4sin
4sin ( 1)
4sin 1
3
4sin
2
2
−
−
−
−
+
−
−
+
−
−
=
3
4
4
4
3
4
6
2
2
6
π
π
π
π
−
−
×
+
×
+
×
−
−
×
=
2
2
3
2
2
3
3
3
π
π
−
−
+ π +
π −
−
= 8
2 3
3
π −
EXERCISE 8 |
1 | 4293-4296 | 19
APPLICATION OF INTEGRALS 371
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO]
(Why )
= 2 [area of the region ODAO + area of the region DCAD]
=
1
2
0
1
2
y dx
y dx
+
∫
∫
=
1
2
2
2
0
1
2
4
(
2)
4
x
dx
x dx
−
−
+
−
∫
∫
(Why )
=
1
2
–1
0
1
1
2
2
(
2) 4
(
2)
4sin
2
2
x2
x
x
−
−
−
−
+
×
+
2
2
–1
1
1
1
2
4
4sin
2
2
x2
x
x
−
+
×
=
1
2
2
–1
2
–1
1
0
2
(
2) 4
(
2)
4sin
4
4sin
2
2
x
x
x
x
x
x
−
−
−
−
+
+
−
+
=
–1
1
–1
1
1
1
3
4sin
4sin ( 1)
4sin 1
3
4sin
2
2
−
−
−
−
+
−
−
+
−
−
=
3
4
4
4
3
4
6
2
2
6
π
π
π
π
−
−
×
+
×
+
×
−
−
×
=
2
2
3
2
2
3
3
3
π
π
−
−
+ π +
π −
−
= 8
2 3
3
π −
EXERCISE 8 2
1 |
1 | 4294-4297 | )
= 2 [area of the region ODAO + area of the region DCAD]
=
1
2
0
1
2
y dx
y dx
+
∫
∫
=
1
2
2
2
0
1
2
4
(
2)
4
x
dx
x dx
−
−
+
−
∫
∫
(Why )
=
1
2
–1
0
1
1
2
2
(
2) 4
(
2)
4sin
2
2
x2
x
x
−
−
−
−
+
×
+
2
2
–1
1
1
1
2
4
4sin
2
2
x2
x
x
−
+
×
=
1
2
2
–1
2
–1
1
0
2
(
2) 4
(
2)
4sin
4
4sin
2
2
x
x
x
x
x
x
−
−
−
−
+
+
−
+
=
–1
1
–1
1
1
1
3
4sin
4sin ( 1)
4sin 1
3
4sin
2
2
−
−
−
−
+
−
−
+
−
−
=
3
4
4
4
3
4
6
2
2
6
π
π
π
π
−
−
×
+
×
+
×
−
−
×
=
2
2
3
2
2
3
3
3
π
π
−
−
+ π +
π −
−
= 8
2 3
3
π −
EXERCISE 8 2
1 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y |
1 | 4295-4298 | )
=
1
2
–1
0
1
1
2
2
(
2) 4
(
2)
4sin
2
2
x2
x
x
−
−
−
−
+
×
+
2
2
–1
1
1
1
2
4
4sin
2
2
x2
x
x
−
+
×
=
1
2
2
–1
2
–1
1
0
2
(
2) 4
(
2)
4sin
4
4sin
2
2
x
x
x
x
x
x
−
−
−
−
+
+
−
+
=
–1
1
–1
1
1
1
3
4sin
4sin ( 1)
4sin 1
3
4sin
2
2
−
−
−
−
+
−
−
+
−
−
=
3
4
4
4
3
4
6
2
2
6
π
π
π
π
−
−
×
+
×
+
×
−
−
×
=
2
2
3
2
2
3
3
3
π
π
−
−
+ π +
π −
−
= 8
2 3
3
π −
EXERCISE 8 2
1 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y 2 |
1 | 4296-4299 | 2
1 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y 2 Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1 |
1 | 4297-4300 | Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y 2 Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1 3 |
1 | 4298-4301 | 2 Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1 3 Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3 |
1 | 4299-4302 | Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1 3 Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3 4 |
1 | 4300-4303 | 3 Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3 4 Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2) |
1 | 4301-4304 | Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3 4 Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2) 5 |
1 | 4302-4305 | 4 Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2) 5 Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4 |
1 | 4303-4306 | Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2) 5 Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4 372
MATHEMATICS
Choose the correct answer in the following exercises 6 and 7 |
1 | 4304-4307 | 5 Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4 372
MATHEMATICS
Choose the correct answer in the following exercises 6 and 7 6 |
1 | 4305-4308 | Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4 372
MATHEMATICS
Choose the correct answer in the following exercises 6 and 7 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)
7 |
1 | 4306-4309 | 372
MATHEMATICS
Choose the correct answer in the following exercises 6 and 7 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)
7 Area lying between the curves y2 = 4x and y = 2x is
(A)
32
(B) 1
3
(C)
41
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum |
1 | 4307-4310 | 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)
7 Area lying between the curves y2 = 4x and y = 2x is
(A)
32
(B) 1
3
(C)
41
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum Solution From Fig 8 |
1 | 4308-4311 | 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)
7 Area lying between the curves y2 = 4x and y = 2x is
(A)
32
(B) 1
3
(C)
41
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum Solution From Fig 8 20, the vertex of the parabola
y2 = 4ax is at origin (0, 0) |
1 | 4309-4312 | Area lying between the curves y2 = 4x and y = 2x is
(A)
32
(B) 1
3
(C)
41
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum Solution From Fig 8 20, the vertex of the parabola
y2 = 4ax is at origin (0, 0) The equation of the
latus rectum LSL′ is x = a |
1 | 4310-4313 | Solution From Fig 8 20, the vertex of the parabola
y2 = 4ax is at origin (0, 0) The equation of the
latus rectum LSL′ is x = a Also, parabola is
symmetrical about the x-axis |
1 | 4311-4314 | 20, the vertex of the parabola
y2 = 4ax is at origin (0, 0) The equation of the
latus rectum LSL′ is x = a Also, parabola is
symmetrical about the x-axis The required area of the region OLL′O
= 2(area of the region OLSO)
=
20
∫a ydx
=
20
4
a
ax dx
∫
=
0
2
2
a
a
xdx
×
∫
=
3
2
0
2
4
3
a
a
x
×
=
23
8
3
a a
=
82
3
a
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1 |
1 | 4312-4315 | The equation of the
latus rectum LSL′ is x = a Also, parabola is
symmetrical about the x-axis The required area of the region OLL′O
= 2(area of the region OLSO)
=
20
∫a ydx
=
20
4
a
ax dx
∫
=
0
2
2
a
a
xdx
×
∫
=
3
2
0
2
4
3
a
a
x
×
=
23
8
3
a a
=
82
3
a
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1 Solution As shown in the Fig 8 |
1 | 4313-4316 | Also, parabola is
symmetrical about the x-axis The required area of the region OLL′O
= 2(area of the region OLSO)
=
20
∫a ydx
=
20
4
a
ax dx
∫
=
0
2
2
a
a
xdx
×
∫
=
3
2
0
2
4
3
a
a
x
×
=
23
8
3
a a
=
82
3
a
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1 Solution As shown in the Fig 8 21, the line
y = 3x + 2 meets x-axis at x =
32
−
and its graph
lies below x-axis for
1, 32
x
−
∈ −
and above
x-axis for
32 ,1
x
−
∈
|
1 | 4314-4317 | The required area of the region OLL′O
= 2(area of the region OLSO)
=
20
∫a ydx
=
20
4
a
ax dx
∫
=
0
2
2
a
a
xdx
×
∫
=
3
2
0
2
4
3
a
a
x
×
=
23
8
3
a a
=
82
3
a
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1 Solution As shown in the Fig 8 21, the line
y = 3x + 2 meets x-axis at x =
32
−
and its graph
lies below x-axis for
1, 32
x
−
∈ −
and above
x-axis for
32 ,1
x
−
∈
Fig 8 |
1 | 4315-4318 | Solution As shown in the Fig 8 21, the line
y = 3x + 2 meets x-axis at x =
32
−
and its graph
lies below x-axis for
1, 32
x
−
∈ −
and above
x-axis for
32 ,1
x
−
∈
Fig 8 21
X′
O
Y′
X
Y
S
L
L'
(��,�0)
a
Fig 8 |
1 | 4316-4319 | 21, the line
y = 3x + 2 meets x-axis at x =
32
−
and its graph
lies below x-axis for
1, 32
x
−
∈ −
and above
x-axis for
32 ,1
x
−
∈
Fig 8 21
X′
O
Y′
X
Y
S
L
L'
(��,�0)
a
Fig 8 20
APPLICATION OF INTEGRALS 373
Fig 8 |
1 | 4317-4320 | Fig 8 21
X′
O
Y′
X
Y
S
L
L'
(��,�0)
a
Fig 8 20
APPLICATION OF INTEGRALS 373
Fig 8 23
The required area = Area of the region ACBA + Area of the region ADEA
=
2
1
3
2
1
3
(3
2)
(3
2)
x
dx
x
dx
−
−
−
+
+
+
∫
∫
=
2
1
2
2
3
2
1
3
3
3
2
2
2
2
x
x
x
x
−
−
−
+
+
+
= 1
25
13
6
6
3
+
=
Example 13 Find the area bounded by
the curve y = cos x between x = 0 and
x = 2π |
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