Chapter
stringclasses 18
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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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