Chapter
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1 | 4318-4321 | 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 Solution From the Fig 8 |
1 | 4319-4322 | 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 Solution From the Fig 8 22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD |
1 | 4320-4323 | 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 Solution From the Fig 8 22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD Thus, we have the required area
=
3π
π
2π
2
2
3π
π
0
2
2
cos
cos
cos
xdx
xdx
x dx
+
+
∫
∫
∫
= [
]
[
]
[
]
3
2
2
2
3
0
2
2
sin
sin
sin
x
x
x
π
π
π
π
π
+
+
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts |
1 | 4321-4324 | Solution From the Fig 8 22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD Thus, we have the required area
=
3π
π
2π
2
2
3π
π
0
2
2
cos
cos
cos
xdx
xdx
x dx
+
+
∫
∫
∫
= [
]
[
]
[
]
3
2
2
2
3
0
2
2
sin
sin
sin
x
x
x
π
π
π
π
π
+
+
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as
Fig 8 |
1 | 4322-4325 | 22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD Thus, we have the required area
=
3π
π
2π
2
2
3π
π
0
2
2
cos
cos
cos
xdx
xdx
x dx
+
+
∫
∫
∫
= [
]
[
]
[
]
3
2
2
2
3
0
2
2
sin
sin
sin
x
x
x
π
π
π
π
π
+
+
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as
Fig 8 22
374
MATHEMATICS
Y'
R
O
X
X'
x =�2
T
S
P�(0,1) x =�1
Y
Q(1,2)
shown in the Fig 8 |
1 | 4323-4326 | Thus, we have the required area
=
3π
π
2π
2
2
3π
π
0
2
2
cos
cos
cos
xdx
xdx
x dx
+
+
∫
∫
∫
= [
]
[
]
[
]
3
2
2
2
3
0
2
2
sin
sin
sin
x
x
x
π
π
π
π
π
+
+
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as
Fig 8 22
374
MATHEMATICS
Y'
R
O
X
X'
x =�2
T
S
P�(0,1) x =�1
Y
Q(1,2)
shown in the Fig 8 23 |
1 | 4324-4327 | Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as
Fig 8 22
374
MATHEMATICS
Y'
R
O
X
X'
x =�2
T
S
P�(0,1) x =�1
Y
Q(1,2)
shown in the Fig 8 23 Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y |
1 | 4325-4328 | 22
374
MATHEMATICS
Y'
R
O
X
X'
x =�2
T
S
P�(0,1) x =�1
Y
Q(1,2)
shown in the Fig 8 23 Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y =
2
4
0
2
x4
x
dx
−
∫
=
4
3
3
2
0
2
2
3
x12
x
×
−
= 32
16
16
3
3
3
−
= |
1 | 4326-4329 | 23 Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y =
2
4
0
2
x4
x
dx
−
∫
=
4
3
3
2
0
2
2
3
x12
x
×
−
= 32
16
16
3
3
3
−
= (1)
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
=
2
4
4
3
0
0
1
16
4
12
3
x
dx
x
=
=
∫ |
1 | 4327-4330 | Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y =
2
4
0
2
x4
x
dx
−
∫
=
4
3
3
2
0
2
2
3
x12
x
×
−
= 32
16
16
3
3
3
−
= (1)
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
=
2
4
4
3
0
0
1
16
4
12
3
x
dx
x
=
=
∫ (2)
Similarly, the area of the region OBQRO bounded by the curve y2 = 4x, y-axis,
y = 0 and y = 4
=
2
4
4
4
3
0
0
0
1
16
4
12
3
y
xdy
dy
y
=
=
=
∫
∫ |
1 | 4328-4331 | =
2
4
0
2
x4
x
dx
−
∫
=
4
3
3
2
0
2
2
3
x12
x
×
−
= 32
16
16
3
3
3
−
= (1)
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
=
2
4
4
3
0
0
1
16
4
12
3
x
dx
x
=
=
∫ (2)
Similarly, the area of the region OBQRO bounded by the curve y2 = 4x, y-axis,
y = 0 and y = 4
=
2
4
4
4
3
0
0
0
1
16
4
12
3
y
xdy
dy
y
=
=
=
∫
∫ (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i |
1 | 4329-4332 | (1)
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
=
2
4
4
3
0
0
1
16
4
12
3
x
dx
x
=
=
∫ (2)
Similarly, the area of the region OBQRO bounded by the curve y2 = 4x, y-axis,
y = 0 and y = 4
=
2
4
4
4
3
0
0
0
1
16
4
12
3
y
xdy
dy
y
=
=
=
∫
∫ (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i e |
1 | 4330-4333 | (2)
Similarly, the area of the region OBQRO bounded by the curve y2 = 4x, y-axis,
y = 0 and y = 4
=
2
4
4
4
3
0
0
0
1
16
4
12
3
y
xdy
dy
y
=
=
=
∫
∫ (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i e , area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts |
1 | 4331-4334 | (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i e , area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts Example 14 Find the area of the region
{(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Solution Let us first sketch the region whose area is to
be found out |
1 | 4332-4335 | e , area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts Example 14 Find the area of the region
{(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Solution Let us first sketch the region whose area is to
be found out This region is the intersection of the
following regions |
1 | 4333-4336 | , area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts Example 14 Find the area of the region
{(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Solution Let us first sketch the region whose area is to
be found out This region is the intersection of the
following regions A1 = {(x, y) : 0 ≤ y ≤ x2 + 1},
A2 = {(x, y) : 0 ≤ y ≤ x + 1}
and
A3 = {(x, y) : 0 ≤ x ≤ 2}
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2) |
1 | 4334-4337 | Example 14 Find the area of the region
{(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Solution Let us first sketch the region whose area is to
be found out This region is the intersection of the
following regions A1 = {(x, y) : 0 ≤ y ≤ x2 + 1},
A2 = {(x, y) : 0 ≤ y ≤ x + 1}
and
A3 = {(x, y) : 0 ≤ x ≤ 2}
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2) From the Fig 8 |
1 | 4335-4338 | This region is the intersection of the
following regions A1 = {(x, y) : 0 ≤ y ≤ x2 + 1},
A2 = {(x, y) : 0 ≤ y ≤ x + 1}
and
A3 = {(x, y) : 0 ≤ x ≤ 2}
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2) From the Fig 8 24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
=
1
2
2
0
1
(
1)
(
1)
x
dx
x
dx
+
+
+
∫
∫
(Why |
1 | 4336-4339 | A1 = {(x, y) : 0 ≤ y ≤ x2 + 1},
A2 = {(x, y) : 0 ≤ y ≤ x + 1}
and
A3 = {(x, y) : 0 ≤ x ≤ 2}
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2) From the Fig 8 24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
=
1
2
2
0
1
(
1)
(
1)
x
dx
x
dx
+
+
+
∫
∫
(Why )
Fig 8 |
1 | 4337-4340 | From the Fig 8 24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
=
1
2
2
0
1
(
1)
(
1)
x
dx
x
dx
+
+
+
∫
∫
(Why )
Fig 8 24
APPLICATION OF INTEGRALS 375
=
1
2
3
2
1
0
3
2
x
x
x
x
+
+
+
=
(
)
1
1
1
0
2
2
1
3
2
+
−
+
+
−
+
= 23
6
Miscellaneous Exercise on Chapter 8
1 |
1 | 4338-4341 | 24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
=
1
2
2
0
1
(
1)
(
1)
x
dx
x
dx
+
+
+
∫
∫
(Why )
Fig 8 24
APPLICATION OF INTEGRALS 375
=
1
2
3
2
1
0
3
2
x
x
x
x
+
+
+
=
(
)
1
1
1
0
2
2
1
3
2
+
−
+
+
−
+
= 23
6
Miscellaneous Exercise on Chapter 8
1 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
2 |
1 | 4339-4342 | )
Fig 8 24
APPLICATION OF INTEGRALS 375
=
1
2
3
2
1
0
3
2
x
x
x
x
+
+
+
=
(
)
1
1
1
0
2
2
1
3
2
+
−
+
+
−
+
= 23
6
Miscellaneous Exercise on Chapter 8
1 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
2 Find the area between the curves y = x and y = x2 |
1 | 4340-4343 | 24
APPLICATION OF INTEGRALS 375
=
1
2
3
2
1
0
3
2
x
x
x
x
+
+
+
=
(
)
1
1
1
0
2
2
1
3
2
+
−
+
+
−
+
= 23
6
Miscellaneous Exercise on Chapter 8
1 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
2 Find the area between the curves y = x and y = x2 3 |
1 | 4341-4344 | 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
2 Find the area between the curves y = x and y = x2 3 Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4 |
1 | 4342-4345 | Find the area between the curves y = x and y = x2 3 Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4 4 |
1 | 4343-4346 | 3 Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4 4 Sketch the graph of y =
x +3
and evaluate
0
6
3
−
+
∫
x
dx |
1 | 4344-4347 | Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4 4 Sketch the graph of y =
x +3
and evaluate
0
6
3
−
+
∫
x
dx 5 |
1 | 4345-4348 | 4 Sketch the graph of y =
x +3
and evaluate
0
6
3
−
+
∫
x
dx 5 Find the area bounded by the curve y = sin x between x = 0 and x = 2π |
1 | 4346-4349 | Sketch the graph of y =
x +3
and evaluate
0
6
3
−
+
∫
x
dx 5 Find the area bounded by the curve y = sin x between x = 0 and x = 2π 6 |
1 | 4347-4350 | 5 Find the area bounded by the curve y = sin x between x = 0 and x = 2π 6 Find the area enclosed between the parabola y2 = 4ax and the line y = mx |
1 | 4348-4351 | Find the area bounded by the curve y = sin x between x = 0 and x = 2π 6 Find the area enclosed between the parabola y2 = 4ax and the line y = mx 7 |
1 | 4349-4352 | 6 Find the area enclosed between the parabola y2 = 4ax and the line y = mx 7 Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12 |
1 | 4350-4353 | Find the area enclosed between the parabola y2 = 4ax and the line y = mx 7 Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12 8 |
1 | 4351-4354 | 7 Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12 8 Find the area of the smaller region bounded by the ellipse
2
2
1
9
4
x
+y
= and the
line
1
3
2
x
+y
= |
1 | 4352-4355 | Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12 8 Find the area of the smaller region bounded by the ellipse
2
2
1
9
4
x
+y
= and the
line
1
3
2
x
+y
= 9 |
1 | 4353-4356 | 8 Find the area of the smaller region bounded by the ellipse
2
2
1
9
4
x
+y
= and the
line
1
3
2
x
+y
= 9 Find the area of the smaller region bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
=
and the
line
1
x
y
a
b
+
= |
1 | 4354-4357 | Find the area of the smaller region bounded by the ellipse
2
2
1
9
4
x
+y
= and the
line
1
3
2
x
+y
= 9 Find the area of the smaller region bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
=
and the
line
1
x
y
a
b
+
= 10 |
1 | 4355-4358 | 9 Find the area of the smaller region bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
=
and the
line
1
x
y
a
b
+
= 10 Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis |
1 | 4356-4359 | Find the area of the smaller region bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
=
and the
line
1
x
y
a
b
+
= 10 Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis 11 |
1 | 4357-4360 | 10 Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis 11 Using the method of integration find the area bounded by the curve
1
x
+y
= |
1 | 4358-4361 | Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis 11 Using the method of integration find the area bounded by the curve
1
x
+y
= [Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1] |
1 | 4359-4362 | 11 Using the method of integration find the area bounded by the curve
1
x
+y
= [Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1] 376
MATHEMATICS
12 |
1 | 4360-4363 | Using the method of integration find the area bounded by the curve
1
x
+y
= [Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1] 376
MATHEMATICS
12 Find the area bounded by curves {(x, y) : y ≥ x2 and y = | x|} |
1 | 4361-4364 | [Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1] 376
MATHEMATICS
12 Find the area bounded by curves {(x, y) : y ≥ x2 and y = | x|} 13 |
1 | 4362-4365 | 376
MATHEMATICS
12 Find the area bounded by curves {(x, y) : y ≥ x2 and y = | x|} 13 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) |
1 | 4363-4366 | Find the area bounded by curves {(x, y) : y ≥ x2 and y = | x|} 13 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) 14 |
1 | 4364-4367 | 13 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) 14 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
15 |
1 | 4365-4368 | 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) 14 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
15 Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer in the following Exercises from 16 to 20 |
1 | 4366-4369 | 14 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
15 Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer in the following Exercises from 16 to 20 16 |
1 | 4367-4370 | 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
15 Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer in the following Exercises from 16 to 20 16 Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9
(B)
415
−
(C)
415
(D)
17
4
17 |
1 | 4368-4371 | Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}
Choose the correct answer in the following Exercises from 16 to 20 16 Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9
(B)
415
−
(C)
415
(D)
17
4
17 The area bounded by the curve y = x | x |, x-axis and the ordinates x = – 1 and
x = 1 is given by
(A) 0
(B) 1
3
(C)
32
(D)
4
3
[Hint : y = x2 if x > 0 and y = – x2 if x < 0] |
1 | 4369-4372 | 16 Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9
(B)
415
−
(C)
415
(D)
17
4
17 The area bounded by the curve y = x | x |, x-axis and the ordinates x = – 1 and
x = 1 is given by
(A) 0
(B) 1
3
(C)
32
(D)
4
3
[Hint : y = x2 if x > 0 and y = – x2 if x < 0] 18 |
1 | 4370-4373 | Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9
(B)
415
−
(C)
415
(D)
17
4
17 The area bounded by the curve y = x | x |, x-axis and the ordinates x = – 1 and
x = 1 is given by
(A) 0
(B) 1
3
(C)
32
(D)
4
3
[Hint : y = x2 if x > 0 and y = – x2 if x < 0] 18 The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A)
4 (4
3)
3
π −
(B)
4 (4
3)
3
π +
(C)
4 (8
3)
3
π −
(D)
4 (8
3)
3
π +
19 |
1 | 4371-4374 | The area bounded by the curve y = x | x |, x-axis and the ordinates x = – 1 and
x = 1 is given by
(A) 0
(B) 1
3
(C)
32
(D)
4
3
[Hint : y = x2 if x > 0 and y = – x2 if x < 0] 18 The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A)
4 (4
3)
3
π −
(B)
4 (4
3)
3
π +
(C)
4 (8
3)
3
π −
(D)
4 (8
3)
3
π +
19 The area bounded by the y-axis, y = cos x and y = sin x when 0
2
x
π
≤
≤
is
(A) 2 ( 2 1)
−
(B)
2
−1
(C)
2
+1
(D)
2
Summary
� The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula: Area
( )
b
b
a
a
ydx
f x dx
=
=
∫
∫ |
1 | 4372-4375 | 18 The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A)
4 (4
3)
3
π −
(B)
4 (4
3)
3
π +
(C)
4 (8
3)
3
π −
(D)
4 (8
3)
3
π +
19 The area bounded by the y-axis, y = cos x and y = sin x when 0
2
x
π
≤
≤
is
(A) 2 ( 2 1)
−
(B)
2
−1
(C)
2
+1
(D)
2
Summary
� The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula: Area
( )
b
b
a
a
ydx
f x dx
=
=
∫
∫ � The area of the region bounded by the curve x = φ (y), y-axis and the lines
y = c, y = d is given by the formula: Area
( )
d
d
c
c
xdy
y dy
=
=
φ
∫
∫ |
1 | 4373-4376 | The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
(A)
4 (4
3)
3
π −
(B)
4 (4
3)
3
π +
(C)
4 (8
3)
3
π −
(D)
4 (8
3)
3
π +
19 The area bounded by the y-axis, y = cos x and y = sin x when 0
2
x
π
≤
≤
is
(A) 2 ( 2 1)
−
(B)
2
−1
(C)
2
+1
(D)
2
Summary
� The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula: Area
( )
b
b
a
a
ydx
f x dx
=
=
∫
∫ � The area of the region bounded by the curve x = φ (y), y-axis and the lines
y = c, y = d is given by the formula: Area
( )
d
d
c
c
xdy
y dy
=
=
φ
∫
∫ APPLICATION OF INTEGRALS 377
� The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[
]
Area
( )
( )
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], a < c < b, then
[
]
[
]
Area
( )
( )
( )
( )
c
b
a
c
f x
g x dx
g x
f x dx
=
−
+
−
∫
∫ |
1 | 4374-4377 | The area bounded by the y-axis, y = cos x and y = sin x when 0
2
x
π
≤
≤
is
(A) 2 ( 2 1)
−
(B)
2
−1
(C)
2
+1
(D)
2
Summary
� The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula: Area
( )
b
b
a
a
ydx
f x dx
=
=
∫
∫ � The area of the region bounded by the curve x = φ (y), y-axis and the lines
y = c, y = d is given by the formula: Area
( )
d
d
c
c
xdy
y dy
=
=
φ
∫
∫ APPLICATION OF INTEGRALS 377
� The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[
]
Area
( )
( )
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], a < c < b, then
[
]
[
]
Area
( )
( )
( )
( )
c
b
a
c
f x
g x dx
g x
f x dx
=
−
+
−
∫
∫ Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece |
1 | 4375-4378 | � The area of the region bounded by the curve x = φ (y), y-axis and the lines
y = c, y = d is given by the formula: Area
( )
d
d
c
c
xdy
y dy
=
=
φ
∫
∫ APPLICATION OF INTEGRALS 377
� The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[
]
Area
( )
( )
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], a < c < b, then
[
]
[
]
Area
( )
( )
( )
( )
c
b
a
c
f x
g x dx
g x
f x dx
=
−
+
−
∫
∫ Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc |
1 | 4376-4379 | APPLICATION OF INTEGRALS 377
� The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[
]
Area
( )
( )
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], a < c < b, then
[
]
[
]
Area
( )
( )
( )
( )
c
b
a
c
f x
g x dx
g x
f x dx
=
−
+
−
∫
∫ Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc In this sense, the method of exhaustion can be regarded as an early method
of integration |
1 | 4377-4380 | Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc In this sense, the method of exhaustion can be regarded as an early method
of integration The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B |
1 | 4378-4381 | This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc In this sense, the method of exhaustion can be regarded as an early method
of integration The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B C |
1 | 4379-4382 | In this sense, the method of exhaustion can be regarded as an early method
of integration The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B C ) and Archimedes
(300 B |
1 | 4380-4383 | The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B C ) and Archimedes
(300 B C |
1 | 4381-4384 | C ) and Archimedes
(300 B C )
Systematic approach to the theory of Calculus began in the 17th century |
1 | 4382-4385 | ) and Archimedes
(300 B C )
Systematic approach to the theory of Calculus began in the 17th century In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve |
1 | 4383-4386 | C )
Systematic approach to the theory of Calculus began in the 17th century In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents |
1 | 4384-4387 | )
Systematic approach to the theory of Calculus began in the 17th century In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’ |
1 | 4385-4388 | In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’ In 1696, he followed a suggestion made by J |
1 | 4386-4389 | Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’ In 1696, he followed a suggestion made by J Bernoulli and changed this article to
Calculus integrali |
1 | 4387-4390 | During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’ In 1696, he followed a suggestion made by J Bernoulli and changed this article to
Calculus integrali This corresponded to Newton’s inverse method of tangents |
1 | 4388-4391 | In 1696, he followed a suggestion made by J Bernoulli and changed this article to
Calculus integrali This corresponded to Newton’s inverse method of tangents Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different |
1 | 4389-4392 | Bernoulli and changed this article to
Calculus integrali This corresponded to Newton’s inverse method of tangents Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different However, respective theories accomplished results that
were practically identical |
1 | 4390-4393 | This corresponded to Newton’s inverse method of tangents Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different However, respective theories accomplished results that
were practically identical Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral |
1 | 4391-4394 | Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different However, respective theories accomplished results that
were practically identical Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P |
1 | 4392-4395 | However, respective theories accomplished results that
were practically identical Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P de Fermat, I |
1 | 4393-4396 | Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P de Fermat, I Newton and G |
1 | 4394-4397 | Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P de Fermat, I Newton and G Leibnitz at the end of 17th century |
1 | 4395-4398 | de Fermat, I Newton and G Leibnitz at the end of 17th century 378
MATHEMATICS
—�
�
�
�
�—
However, this justification by the concept of limit was only developed in the
works of A |
1 | 4396-4399 | Newton and G Leibnitz at the end of 17th century 378
MATHEMATICS
—�
�
�
�
�—
However, this justification by the concept of limit was only developed in the
works of A L |
1 | 4397-4400 | Leibnitz at the end of 17th century 378
MATHEMATICS
—�
�
�
�
�—
However, this justification by the concept of limit was only developed in the
works of A L Cauchy in the early 19th century |
1 | 4398-4401 | 378
MATHEMATICS
—�
�
�
�
�—
However, this justification by the concept of limit was only developed in the
works of A L Cauchy in the early 19th century Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis |
1 | 4399-4402 | L Cauchy in the early 19th century Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz” |
1 | 4400-4403 | Cauchy in the early 19th century Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz” DIFFERENTIAL EQUATIONS
379
�He who seeks for methods without having a definite problem in mind
seeks for the most part in vain |
1 | 4401-4404 | Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz” DIFFERENTIAL EQUATIONS
379
�He who seeks for methods without having a definite problem in mind
seeks for the most part in vain – D |
1 | 4402-4405 | The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz” DIFFERENTIAL EQUATIONS
379
�He who seeks for methods without having a definite problem in mind
seeks for the most part in vain – D HILBERT �
9 |
1 | 4403-4406 | DIFFERENTIAL EQUATIONS
379
�He who seeks for methods without having a definite problem in mind
seeks for the most part in vain – D HILBERT �
9 1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i |
1 | 4404-4407 | – D HILBERT �
9 1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i e |
1 | 4405-4408 | HILBERT �
9 1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i e , how to find f ′(x) for a given
function f at each x in its domain of definition |
1 | 4406-4409 | 1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i e , how to find f ′(x) for a given
function f at each x in its domain of definition Further, in
the chapter on Integral Calculus, we discussed how to find
a function f whose derivative is the function g, which may
also be formulated as follows:
For a given function g, find a function f such that
dy
dx = g(x), where y = f(x) |
1 | 4407-4410 | e , how to find f ′(x) for a given
function f at each x in its domain of definition Further, in
the chapter on Integral Calculus, we discussed how to find
a function f whose derivative is the function g, which may
also be formulated as follows:
For a given function g, find a function f such that
dy
dx = g(x), where y = f(x) (1)
An equation of the form (1) is known as a differential
equation |
1 | 4408-4411 | , how to find f ′(x) for a given
function f at each x in its domain of definition Further, in
the chapter on Integral Calculus, we discussed how to find
a function f whose derivative is the function g, which may
also be formulated as follows:
For a given function g, find a function f such that
dy
dx = g(x), where y = f(x) (1)
An equation of the form (1) is known as a differential
equation A formal definition will be given later |
1 | 4409-4412 | Further, in
the chapter on Integral Calculus, we discussed how to find
a function f whose derivative is the function g, which may
also be formulated as follows:
For a given function g, find a function f such that
dy
dx = g(x), where y = f(x) (1)
An equation of the form (1) is known as a differential
equation A formal definition will be given later These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc |
1 | 4410-4413 | (1)
An equation of the form (1) is known as a differential
equation A formal definition will be given later These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc Hence, an indepth study of differential
equations has assumed prime importance in all modern scientific investigations |
1 | 4411-4414 | A formal definition will be given later These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc Hence, an indepth study of differential
equations has assumed prime importance in all modern scientific investigations In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas |
1 | 4412-4415 | These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc Hence, an indepth study of differential
equations has assumed prime importance in all modern scientific investigations In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas 9 |
1 | 4413-4416 | Hence, an indepth study of differential
equations has assumed prime importance in all modern scientific investigations In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas 9 2 Basic Concepts
We are already familiar with the equations of the type:
x2 – 3x + 3 = 0 |
1 | 4414-4417 | In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas 9 2 Basic Concepts
We are already familiar with the equations of the type:
x2 – 3x + 3 = 0 (1)
sin x + cos x = 0 |
1 | 4415-4418 | 9 2 Basic Concepts
We are already familiar with the equations of the type:
x2 – 3x + 3 = 0 (1)
sin x + cos x = 0 (2)
x + y = 7 |
1 | 4416-4419 | 2 Basic Concepts
We are already familiar with the equations of the type:
x2 – 3x + 3 = 0 (1)
sin x + cos x = 0 (2)
x + y = 7 (3)
Chapter 9
DIFFERENTIAL EQUATIONS
Henri Poincare
(1854-1912 )
© NCERT
not to be republished
MATHEMATICS
380
Let us consider the equation:
xdy
dx +y
= 0 |
1 | 4417-4420 | (1)
sin x + cos x = 0 (2)
x + y = 7 (3)
Chapter 9
DIFFERENTIAL EQUATIONS
Henri Poincare
(1854-1912 )
© NCERT
not to be republished
MATHEMATICS
380
Let us consider the equation:
xdy
dx +y
= 0 (4)
We see that equations (1), (2) and (3) involve independent and/or dependent variable
(variables) only but equation (4) involves variables as well as derivative of the dependent
variable y with respect to the independent variable x |
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