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1 | 3318-3321 | But these integrals are very similar geometrically Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals By
assigning different values to C, we get different members of the family These together
constitute the indefinite integral |
1 | 3319-3322 | Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals By
assigning different values to C, we get different members of the family These together
constitute the indefinite integral In this case, each integral represents a parabola with
its axis along y-axis |
1 | 3320-3323 | By
assigning different values to C, we get different members of the family These together
constitute the indefinite integral In this case, each integral represents a parabola with
its axis along y-axis Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin |
1 | 3321-3324 | These together
constitute the indefinite integral In this case, each integral represents a parabola with
its axis along y-axis Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin The
curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along
y-axis in positive direction |
1 | 3322-3325 | In this case, each integral represents a parabola with
its axis along y-axis Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin The
curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along
y-axis in positive direction For C = – 1, y = x2 – 1 is obtained by shifting the parabola
y = x2 one unit along y-axis in the negative direction |
1 | 3323-3326 | Clearly, for C = 0, we obtain y = x2, a parabola with its vertex on the origin The
curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along
y-axis in positive direction For C = – 1, y = x2 – 1 is obtained by shifting the parabola
y = x2 one unit along y-axis in the negative direction Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis |
1 | 3324-3327 | The
curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along
y-axis in positive direction For C = – 1, y = x2 – 1 is obtained by shifting the parabola
y = x2 one unit along y-axis in the negative direction Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis Some of
these have been shown in the Fig 7 |
1 | 3325-3328 | For C = – 1, y = x2 – 1 is obtained by shifting the parabola
y = x2 one unit along y-axis in the negative direction Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis Some of
these have been shown in the Fig 7 1 |
1 | 3326-3329 | Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis Some of
these have been shown in the Fig 7 1 Let us consider the intersection of all these parabolas by a line x = a |
1 | 3327-3330 | Some of
these have been shown in the Fig 7 1 Let us consider the intersection of all these parabolas by a line x = a In the Fig 7 |
1 | 3328-3331 | 1 Let us consider the intersection of all these parabolas by a line x = a In the Fig 7 1,
we have taken a > 0 |
1 | 3329-3332 | Let us consider the intersection of all these parabolas by a line x = a In the Fig 7 1,
we have taken a > 0 The same is true when a < 0 |
1 | 3330-3333 | In the Fig 7 1,
we have taken a > 0 The same is true when a < 0 If the line x = a intersects the
parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0, P1, P2, P–1, P–2 etc |
1 | 3331-3334 | 1,
we have taken a > 0 The same is true when a < 0 If the line x = a intersects the
parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0, P1, P2, P–1, P–2 etc ,
respectively, then dy
dx at these points equals 2a |
1 | 3332-3335 | The same is true when a < 0 If the line x = a intersects the
parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0, P1, P2, P–1, P–2 etc ,
respectively, then dy
dx at these points equals 2a This indicates that the tangents to the
curves at these points are parallel |
1 | 3333-3336 | If the line x = a intersects the
parabolas y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1, y = x2 – 2 at P0, P1, P2, P–1, P–2 etc ,
respectively, then dy
dx at these points equals 2a This indicates that the tangents to the
curves at these points are parallel Thus,
2
C
2
C
F ( )
x dx
x
x
=
+
=
∫
(say), implies that
292
MATHEMATICS
the tangents to all the curves y = F C (x), C ∈ R, at the points of intersection of the
curves by the line x = a, (a ∈ R), are parallel |
1 | 3334-3337 | ,
respectively, then dy
dx at these points equals 2a This indicates that the tangents to the
curves at these points are parallel Thus,
2
C
2
C
F ( )
x dx
x
x
=
+
=
∫
(say), implies that
292
MATHEMATICS
the tangents to all the curves y = F C (x), C ∈ R, at the points of intersection of the
curves by the line x = a, (a ∈ R), are parallel Further, the following equation (statement)
( )
F ( )
C
(say)
f x dx
x
y
=
+
=
∫
,
represents a family of curves |
1 | 3335-3338 | This indicates that the tangents to the
curves at these points are parallel Thus,
2
C
2
C
F ( )
x dx
x
x
=
+
=
∫
(say), implies that
292
MATHEMATICS
the tangents to all the curves y = F C (x), C ∈ R, at the points of intersection of the
curves by the line x = a, (a ∈ R), are parallel Further, the following equation (statement)
( )
F ( )
C
(say)
f x dx
x
y
=
+
=
∫
,
represents a family of curves The different values of C will correspond to different
members of this family and these members can be obtained by shifting any one of the
curves parallel to itself |
1 | 3336-3339 | Thus,
2
C
2
C
F ( )
x dx
x
x
=
+
=
∫
(say), implies that
292
MATHEMATICS
the tangents to all the curves y = F C (x), C ∈ R, at the points of intersection of the
curves by the line x = a, (a ∈ R), are parallel Further, the following equation (statement)
( )
F ( )
C
(say)
f x dx
x
y
=
+
=
∫
,
represents a family of curves The different values of C will correspond to different
members of this family and these members can be obtained by shifting any one of the
curves parallel to itself This is the geometrical interpretation of indefinite integral |
1 | 3337-3340 | Further, the following equation (statement)
( )
F ( )
C
(say)
f x dx
x
y
=
+
=
∫
,
represents a family of curves The different values of C will correspond to different
members of this family and these members can be obtained by shifting any one of the
curves parallel to itself This is the geometrical interpretation of indefinite integral 7 |
1 | 3338-3341 | The different values of C will correspond to different
members of this family and these members can be obtained by shifting any one of the
curves parallel to itself This is the geometrical interpretation of indefinite integral 7 2 |
1 | 3339-3342 | This is the geometrical interpretation of indefinite integral 7 2 2 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals |
1 | 3340-3343 | 7 2 2 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals (I)
The process of differentiation and integration are inverses of each other in the
sense of the following results :
( )
d
f x dx
dx ∫
= f (x)
and
f( )
∫′x dx
= f (x) + C, where C is any arbitrary constant |
1 | 3341-3344 | 2 2 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals (I)
The process of differentiation and integration are inverses of each other in the
sense of the following results :
( )
d
f x dx
dx ∫
= f (x)
and
f( )
∫′x dx
= f (x) + C, where C is any arbitrary constant Fig 7 |
1 | 3342-3345 | 2 Some properties of indefinite integral
In this sub section, we shall derive some properties of indefinite integrals (I)
The process of differentiation and integration are inverses of each other in the
sense of the following results :
( )
d
f x dx
dx ∫
= f (x)
and
f( )
∫′x dx
= f (x) + C, where C is any arbitrary constant Fig 7 1
INTEGRALS 293
Proof Let F be any anti derivative of f, i |
1 | 3343-3346 | (I)
The process of differentiation and integration are inverses of each other in the
sense of the following results :
( )
d
f x dx
dx ∫
= f (x)
and
f( )
∫′x dx
= f (x) + C, where C is any arbitrary constant Fig 7 1
INTEGRALS 293
Proof Let F be any anti derivative of f, i e |
1 | 3344-3347 | Fig 7 1
INTEGRALS 293
Proof Let F be any anti derivative of f, i e ,
dF( )
x
dx
= f (x)
Then
( )
∫f x dx
= F(x) + C
Therefore
( )
d
dx ∫f x dx
=
(
F ( ) + C)
d
x
dx
=
F ( ) =
( )
d
x
f x
dx
Similarly, we note that
f ′(x) =
( )
d f x
dx
and hence
f( )
∫′x dx
= f (x) + C
where C is arbitrary constant called constant of integration |
1 | 3345-3348 | 1
INTEGRALS 293
Proof Let F be any anti derivative of f, i e ,
dF( )
x
dx
= f (x)
Then
( )
∫f x dx
= F(x) + C
Therefore
( )
d
dx ∫f x dx
=
(
F ( ) + C)
d
x
dx
=
F ( ) =
( )
d
x
f x
dx
Similarly, we note that
f ′(x) =
( )
d f x
dx
and hence
f( )
∫′x dx
= f (x) + C
where C is arbitrary constant called constant of integration (II)
Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent |
1 | 3346-3349 | e ,
dF( )
x
dx
= f (x)
Then
( )
∫f x dx
= F(x) + C
Therefore
( )
d
dx ∫f x dx
=
(
F ( ) + C)
d
x
dx
=
F ( ) =
( )
d
x
f x
dx
Similarly, we note that
f ′(x) =
( )
d f x
dx
and hence
f( )
∫′x dx
= f (x) + C
where C is arbitrary constant called constant of integration (II)
Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent Proof Let f and g be two functions such that
( )
d
f x dx
dx ∫
=
( )
d
g x dx
dx ∫
or
( )
( )
d
f x dx –
g x dx
dx
∫
∫
= 0
Hence
( )
( )
f x dx –
g x dx
∫
∫
= C, where C is any real number
(Why |
1 | 3347-3350 | ,
dF( )
x
dx
= f (x)
Then
( )
∫f x dx
= F(x) + C
Therefore
( )
d
dx ∫f x dx
=
(
F ( ) + C)
d
x
dx
=
F ( ) =
( )
d
x
f x
dx
Similarly, we note that
f ′(x) =
( )
d f x
dx
and hence
f( )
∫′x dx
= f (x) + C
where C is arbitrary constant called constant of integration (II)
Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent Proof Let f and g be two functions such that
( )
d
f x dx
dx ∫
=
( )
d
g x dx
dx ∫
or
( )
( )
d
f x dx –
g x dx
dx
∫
∫
= 0
Hence
( )
( )
f x dx –
g x dx
∫
∫
= C, where C is any real number
(Why )
or
( )
∫f x dx
=
( )
g x dx +C
∫
So the families of curves {
}
1
1
( )
C ,C
R
f x dx +
∈
∫
and
{
}
2
2
( )
C , C
R
g x dx +
∈
∫
are identical |
1 | 3348-3351 | (II)
Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent Proof Let f and g be two functions such that
( )
d
f x dx
dx ∫
=
( )
d
g x dx
dx ∫
or
( )
( )
d
f x dx –
g x dx
dx
∫
∫
= 0
Hence
( )
( )
f x dx –
g x dx
∫
∫
= C, where C is any real number
(Why )
or
( )
∫f x dx
=
( )
g x dx +C
∫
So the families of curves {
}
1
1
( )
C ,C
R
f x dx +
∈
∫
and
{
}
2
2
( )
C , C
R
g x dx +
∈
∫
are identical Hence, in this sense,
( )
and
( )
f x dx
g x dx
∫
∫
are equivalent |
1 | 3349-3352 | Proof Let f and g be two functions such that
( )
d
f x dx
dx ∫
=
( )
d
g x dx
dx ∫
or
( )
( )
d
f x dx –
g x dx
dx
∫
∫
= 0
Hence
( )
( )
f x dx –
g x dx
∫
∫
= C, where C is any real number
(Why )
or
( )
∫f x dx
=
( )
g x dx +C
∫
So the families of curves {
}
1
1
( )
C ,C
R
f x dx +
∈
∫
and
{
}
2
2
( )
C , C
R
g x dx +
∈
∫
are identical Hence, in this sense,
( )
and
( )
f x dx
g x dx
∫
∫
are equivalent 294
MATHEMATICS
� Note The equivalence of the families {
}
1
1
( )
f x dx+ C ,C
∈
∫
R and
{
}
2
2
( )
g x dx+ C ,C
∈
∫
R is customarily expressed by writing
( )
=
( )
f x dx
g x dx
∫
∫
,
without mentioning the parameter |
1 | 3350-3353 | )
or
( )
∫f x dx
=
( )
g x dx +C
∫
So the families of curves {
}
1
1
( )
C ,C
R
f x dx +
∈
∫
and
{
}
2
2
( )
C , C
R
g x dx +
∈
∫
are identical Hence, in this sense,
( )
and
( )
f x dx
g x dx
∫
∫
are equivalent 294
MATHEMATICS
� Note The equivalence of the families {
}
1
1
( )
f x dx+ C ,C
∈
∫
R and
{
}
2
2
( )
g x dx+ C ,C
∈
∫
R is customarily expressed by writing
( )
=
( )
f x dx
g x dx
∫
∫
,
without mentioning the parameter (III)
[
]
( ) +
( )
( )
+
( )
f x
g x
dx
f x dx
g x dx
=
∫
∫
∫
Proof
By Property (I), we have
[ ( ) +
( )]
d
f x
g x
dx
dx
∫
= f (x) + g(x) |
1 | 3351-3354 | Hence, in this sense,
( )
and
( )
f x dx
g x dx
∫
∫
are equivalent 294
MATHEMATICS
� Note The equivalence of the families {
}
1
1
( )
f x dx+ C ,C
∈
∫
R and
{
}
2
2
( )
g x dx+ C ,C
∈
∫
R is customarily expressed by writing
( )
=
( )
f x dx
g x dx
∫
∫
,
without mentioning the parameter (III)
[
]
( ) +
( )
( )
+
( )
f x
g x
dx
f x dx
g x dx
=
∫
∫
∫
Proof
By Property (I), we have
[ ( ) +
( )]
d
f x
g x
dx
dx
∫
= f (x) + g(x) (1)
On the otherhand, we find that
( )
+
( )
d
f x dx
g x dx
dx
∫
∫
=
( )
+
( )
d
d
f x dx
g x dx
dx
dx
∫
∫
= f (x) + g(x) |
1 | 3352-3355 | 294
MATHEMATICS
� Note The equivalence of the families {
}
1
1
( )
f x dx+ C ,C
∈
∫
R and
{
}
2
2
( )
g x dx+ C ,C
∈
∫
R is customarily expressed by writing
( )
=
( )
f x dx
g x dx
∫
∫
,
without mentioning the parameter (III)
[
]
( ) +
( )
( )
+
( )
f x
g x
dx
f x dx
g x dx
=
∫
∫
∫
Proof
By Property (I), we have
[ ( ) +
( )]
d
f x
g x
dx
dx
∫
= f (x) + g(x) (1)
On the otherhand, we find that
( )
+
( )
d
f x dx
g x dx
dx
∫
∫
=
( )
+
( )
d
d
f x dx
g x dx
dx
dx
∫
∫
= f (x) + g(x) (2)
Thus, in view of Property (II), it follows by (1) and (2) that
(
)
( )
( )
f x
g x
dx
+
∫
=
( )
( )
f x dx
g x dx
+
∫
∫ |
1 | 3353-3356 | (III)
[
]
( ) +
( )
( )
+
( )
f x
g x
dx
f x dx
g x dx
=
∫
∫
∫
Proof
By Property (I), we have
[ ( ) +
( )]
d
f x
g x
dx
dx
∫
= f (x) + g(x) (1)
On the otherhand, we find that
( )
+
( )
d
f x dx
g x dx
dx
∫
∫
=
( )
+
( )
d
d
f x dx
g x dx
dx
dx
∫
∫
= f (x) + g(x) (2)
Thus, in view of Property (II), it follows by (1) and (2) that
(
)
( )
( )
f x
g x
dx
+
∫
=
( )
( )
f x dx
g x dx
+
∫
∫ (IV) For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
Proof By the Property (I),
( )
( )
d
k f x dx
k f x
dx
=
∫ |
1 | 3354-3357 | (1)
On the otherhand, we find that
( )
+
( )
d
f x dx
g x dx
dx
∫
∫
=
( )
+
( )
d
d
f x dx
g x dx
dx
dx
∫
∫
= f (x) + g(x) (2)
Thus, in view of Property (II), it follows by (1) and (2) that
(
)
( )
( )
f x
g x
dx
+
∫
=
( )
( )
f x dx
g x dx
+
∫
∫ (IV) For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
Proof By the Property (I),
( )
( )
d
k f x dx
k f x
dx
=
∫ Also
( )
d
k
f x dx
dx
∫
=
( )
=
( )
kd
f x dx
k f x
dx ∫
Therefore, using the Property (II), we have
( )
( )
k f x dx
k
f x dx
=
∫
∫ |
1 | 3355-3358 | (2)
Thus, in view of Property (II), it follows by (1) and (2) that
(
)
( )
( )
f x
g x
dx
+
∫
=
( )
( )
f x dx
g x dx
+
∫
∫ (IV) For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
Proof By the Property (I),
( )
( )
d
k f x dx
k f x
dx
=
∫ Also
( )
d
k
f x dx
dx
∫
=
( )
=
( )
kd
f x dx
k f x
dx ∫
Therefore, using the Property (II), we have
( )
( )
k f x dx
k
f x dx
=
∫
∫ (V)
Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2, |
1 | 3356-3359 | (IV) For any real number k,
( )
( )
k f x dx
k
f x dx
=
∫
∫
Proof By the Property (I),
( )
( )
d
k f x dx
k f x
dx
=
∫ Also
( )
d
k
f x dx
dx
∫
=
( )
=
( )
kd
f x dx
k f x
dx ∫
Therefore, using the Property (II), we have
( )
( )
k f x dx
k
f x dx
=
∫
∫ (V)
Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2, , fn and the real numbers, k1, k2, |
1 | 3357-3360 | Also
( )
d
k
f x dx
dx
∫
=
( )
=
( )
kd
f x dx
k f x
dx ∫
Therefore, using the Property (II), we have
( )
( )
k f x dx
k
f x dx
=
∫
∫ (V)
Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2, , fn and the real numbers, k1, k2, , kn giving
[
]
1 1
2 2
( )
( )
( )
n
n
k f x
k f
x |
1 | 3358-3361 | (V)
Properties (III) and (IV) can be generalised to a finite number of functions
f1, f2, , fn and the real numbers, k1, k2, , kn giving
[
]
1 1
2 2
( )
( )
( )
n
n
k f x
k f
x k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( )
( )
n
n
k
f x dx
k
f
x dx |
1 | 3359-3362 | , fn and the real numbers, k1, k2, , kn giving
[
]
1 1
2 2
( )
( )
( )
n
n
k f x
k f
x k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( )
( )
n
n
k
f x dx
k
f
x dx k
f
x dx
+
+
+
∫
∫
∫ |
1 | 3360-3363 | , kn giving
[
]
1 1
2 2
( )
( )
( )
n
n
k f x
k f
x k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( )
( )
n
n
k
f x dx
k
f
x dx k
f
x dx
+
+
+
∫
∫
∫ To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function |
1 | 3361-3364 | k f
x
dx
+
+
+
∫
=
1
1
2
2
( )
( )
( )
n
n
k
f x dx
k
f
x dx k
f
x dx
+
+
+
∫
∫
∫ To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection |
1 | 3362-3365 | k
f
x dx
+
+
+
∫
∫
∫ To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection We illustrate it
through some examples |
1 | 3363-3366 | To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection We illustrate it
through some examples INTEGRALS 295
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i)
cos 2x
(ii)
3x2 + 4x3
(iii)
1
x , x ≠ 0
Solution
(i)
We look for a function whose derivative is cos 2x |
1 | 3364-3367 | The search for the requisite function for finding
an anti derivative is known as integration by the method of inspection We illustrate it
through some examples INTEGRALS 295
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i)
cos 2x
(ii)
3x2 + 4x3
(iii)
1
x , x ≠ 0
Solution
(i)
We look for a function whose derivative is cos 2x Recall that
d
dx sin 2x = 2 cos 2x
or
cos 2x = 1
2
d
dx (sin 2x) =
21 sin 2
d
x
dx
Therefore, an anti derivative of cos 2x is
1 sin 2
2
x |
1 | 3365-3368 | We illustrate it
through some examples INTEGRALS 295
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i)
cos 2x
(ii)
3x2 + 4x3
(iii)
1
x , x ≠ 0
Solution
(i)
We look for a function whose derivative is cos 2x Recall that
d
dx sin 2x = 2 cos 2x
or
cos 2x = 1
2
d
dx (sin 2x) =
21 sin 2
d
x
dx
Therefore, an anti derivative of cos 2x is
1 sin 2
2
x (ii)
We look for a function whose derivative is 3x2 + 4x3 |
1 | 3366-3369 | INTEGRALS 295
Example 1 Write an anti derivative for each of the following functions using the
method of inspection:
(i)
cos 2x
(ii)
3x2 + 4x3
(iii)
1
x , x ≠ 0
Solution
(i)
We look for a function whose derivative is cos 2x Recall that
d
dx sin 2x = 2 cos 2x
or
cos 2x = 1
2
d
dx (sin 2x) =
21 sin 2
d
x
dx
Therefore, an anti derivative of cos 2x is
1 sin 2
2
x (ii)
We look for a function whose derivative is 3x2 + 4x3 Note that
(
)
3
4
d
x
x
dx
+
= 3x2 + 4x3 |
1 | 3367-3370 | Recall that
d
dx sin 2x = 2 cos 2x
or
cos 2x = 1
2
d
dx (sin 2x) =
21 sin 2
d
x
dx
Therefore, an anti derivative of cos 2x is
1 sin 2
2
x (ii)
We look for a function whose derivative is 3x2 + 4x3 Note that
(
)
3
4
d
x
x
dx
+
= 3x2 + 4x3 Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4 |
1 | 3368-3371 | (ii)
We look for a function whose derivative is 3x2 + 4x3 Note that
(
)
3
4
d
x
x
dx
+
= 3x2 + 4x3 Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4 (iii)
We know that
1
1
1
(log )
0 and
[log (
)]
(
1)
0
d
d
x
,x
– x
–
,x
dx
x
dx
– x
x
=
>
=
=
<
Combining above, we get
(
)
1
log
0
d
x
, x
dx
=x
≠
Therefore,
1
log
dx
x
x
=
∫
is one of the anti derivatives of 1
x |
1 | 3369-3372 | Note that
(
)
3
4
d
x
x
dx
+
= 3x2 + 4x3 Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4 (iii)
We know that
1
1
1
(log )
0 and
[log (
)]
(
1)
0
d
d
x
,x
– x
–
,x
dx
x
dx
– x
x
=
>
=
=
<
Combining above, we get
(
)
1
log
0
d
x
, x
dx
=x
≠
Therefore,
1
log
dx
x
x
=
∫
is one of the anti derivatives of 1
x Example 2 Find the following integrals:
(i)
3
2
1
x – dx
∫x
(ii)
(32
1)
x
dx
+
∫
(iii) ∫
23
1
(
2
–
)
+
∫
x
x
e
dx
x
Solution
(i)
We have
3
2
2
1
–
x –
dx
x dx –
x
dx
x
=
∫
∫
∫
(by Property V)
296
MATHEMATICS
=
1
1
2
1
1
2
C
C
1 1
2
1
–
x
x
–
–
+
+
+
+
+
+
; C1, C2 are constants of integration
=
2
1
1
2
C
C
2
1
–
x
–x
–
–
+
=
2
1
2
1 + C
C
x2
–
x
+
=
2
1 + C
2
x
x
+
, where C = C 1 – C2 is another constant of integration |
1 | 3370-3373 | Therefore, an anti derivative of 3x2 + 4x3 is x3 + x4 (iii)
We know that
1
1
1
(log )
0 and
[log (
)]
(
1)
0
d
d
x
,x
– x
–
,x
dx
x
dx
– x
x
=
>
=
=
<
Combining above, we get
(
)
1
log
0
d
x
, x
dx
=x
≠
Therefore,
1
log
dx
x
x
=
∫
is one of the anti derivatives of 1
x Example 2 Find the following integrals:
(i)
3
2
1
x – dx
∫x
(ii)
(32
1)
x
dx
+
∫
(iii) ∫
23
1
(
2
–
)
+
∫
x
x
e
dx
x
Solution
(i)
We have
3
2
2
1
–
x –
dx
x dx –
x
dx
x
=
∫
∫
∫
(by Property V)
296
MATHEMATICS
=
1
1
2
1
1
2
C
C
1 1
2
1
–
x
x
–
–
+
+
+
+
+
+
; C1, C2 are constants of integration
=
2
1
1
2
C
C
2
1
–
x
–x
–
–
+
=
2
1
2
1 + C
C
x2
–
x
+
=
2
1 + C
2
x
x
+
, where C = C 1 – C2 is another constant of integration �Note From now onwards, we shall write only one constant of integration in the
final answer |
1 | 3371-3374 | (iii)
We know that
1
1
1
(log )
0 and
[log (
)]
(
1)
0
d
d
x
,x
– x
–
,x
dx
x
dx
– x
x
=
>
=
=
<
Combining above, we get
(
)
1
log
0
d
x
, x
dx
=x
≠
Therefore,
1
log
dx
x
x
=
∫
is one of the anti derivatives of 1
x Example 2 Find the following integrals:
(i)
3
2
1
x – dx
∫x
(ii)
(32
1)
x
dx
+
∫
(iii) ∫
23
1
(
2
–
)
+
∫
x
x
e
dx
x
Solution
(i)
We have
3
2
2
1
–
x –
dx
x dx –
x
dx
x
=
∫
∫
∫
(by Property V)
296
MATHEMATICS
=
1
1
2
1
1
2
C
C
1 1
2
1
–
x
x
–
–
+
+
+
+
+
+
; C1, C2 are constants of integration
=
2
1
1
2
C
C
2
1
–
x
–x
–
–
+
=
2
1
2
1 + C
C
x2
–
x
+
=
2
1 + C
2
x
x
+
, where C = C 1 – C2 is another constant of integration �Note From now onwards, we shall write only one constant of integration in the
final answer (ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
=
335
C
5
x
+x
+
(iii) We have
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS 297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
=
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
= 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant |
1 | 3372-3375 | Example 2 Find the following integrals:
(i)
3
2
1
x – dx
∫x
(ii)
(32
1)
x
dx
+
∫
(iii) ∫
23
1
(
2
–
)
+
∫
x
x
e
dx
x
Solution
(i)
We have
3
2
2
1
–
x –
dx
x dx –
x
dx
x
=
∫
∫
∫
(by Property V)
296
MATHEMATICS
=
1
1
2
1
1
2
C
C
1 1
2
1
–
x
x
–
–
+
+
+
+
+
+
; C1, C2 are constants of integration
=
2
1
1
2
C
C
2
1
–
x
–x
–
–
+
=
2
1
2
1 + C
C
x2
–
x
+
=
2
1 + C
2
x
x
+
, where C = C 1 – C2 is another constant of integration �Note From now onwards, we shall write only one constant of integration in the
final answer (ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
=
335
C
5
x
+x
+
(iii) We have
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS 297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
=
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
= 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3 |
1 | 3373-3376 | �Note From now onwards, we shall write only one constant of integration in the
final answer (ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
=
335
C
5
x
+x
+
(iii) We have
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS 297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
=
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
= 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3 Remarks
(i)
We see that if F is an anti derivative of f, then so is F + C, where C is any
constant |
1 | 3374-3377 | (ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
=
335
C
5
x
+x
+
(iii) We have
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS 297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
=
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
= 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3 Remarks
(i)
We see that if F is an anti derivative of f, then so is F + C, where C is any
constant Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ∈ R |
1 | 3375-3378 | Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3 Remarks
(i)
We see that if F is an anti derivative of f, then so is F + C, where C is any
constant Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ∈ R In applications, it is often necessary to satisfy an
additional condition which then determines a specific value of C giving unique
anti derivative of the given function |
1 | 3376-3379 | Remarks
(i)
We see that if F is an anti derivative of f, then so is F + C, where C is any
constant Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ∈ R In applications, it is often necessary to satisfy an
additional condition which then determines a specific value of C giving unique
anti derivative of the given function (ii)
Sometimes, F is not expressible in terms of elementary functions viz |
1 | 3377-3380 | Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ∈ R In applications, it is often necessary to satisfy an
additional condition which then determines a specific value of C giving unique
anti derivative of the given function (ii)
Sometimes, F is not expressible in terms of elementary functions viz , polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc |
1 | 3378-3381 | In applications, it is often necessary to satisfy an
additional condition which then determines a specific value of C giving unique
anti derivative of the given function (ii)
Sometimes, F is not expressible in terms of elementary functions viz , polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc We are
therefore blocked for finding
( )
∫f x dx |
1 | 3379-3382 | (ii)
Sometimes, F is not expressible in terms of elementary functions viz , polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc We are
therefore blocked for finding
( )
∫f x dx For example, it is not possible to find
e– x2
dx
∫
by inspection since we can not find a function whose derivative is
e– x2
298
MATHEMATICS
(iii)
When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly |
1 | 3380-3383 | , polynomial,
logarithmic, exponential, trigonometric functions and their inverses etc We are
therefore blocked for finding
( )
∫f x dx For example, it is not possible to find
e– x2
dx
∫
by inspection since we can not find a function whose derivative is
e– x2
298
MATHEMATICS
(iii)
When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly For instance
4
1
4
5
1
C
C
4
1
5
y
y dy
y
+
=
+
=
+
+
∫
7 |
1 | 3381-3384 | We are
therefore blocked for finding
( )
∫f x dx For example, it is not possible to find
e– x2
dx
∫
by inspection since we can not find a function whose derivative is
e– x2
298
MATHEMATICS
(iii)
When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly For instance
4
1
4
5
1
C
C
4
1
5
y
y dy
y
+
=
+
=
+
+
∫
7 2 |
1 | 3382-3385 | For example, it is not possible to find
e– x2
dx
∫
by inspection since we can not find a function whose derivative is
e– x2
298
MATHEMATICS
(iii)
When the variable of integration is denoted by a variable other than x, the integral
formulae are modified accordingly For instance
4
1
4
5
1
C
C
4
1
5
y
y dy
y
+
=
+
=
+
+
∫
7 2 3 Comparison between differentiation and integration
1 |
1 | 3383-3386 | For instance
4
1
4
5
1
C
C
4
1
5
y
y dy
y
+
=
+
=
+
+
∫
7 2 3 Comparison between differentiation and integration
1 Both are operations on functions |
1 | 3384-3387 | 2 3 Comparison between differentiation and integration
1 Both are operations on functions 2 |
1 | 3385-3388 | 3 Comparison between differentiation and integration
1 Both are operations on functions 2 Both satisfy the property of linearity, i |
1 | 3386-3389 | Both are operations on functions 2 Both satisfy the property of linearity, i e |
1 | 3387-3390 | 2 Both satisfy the property of linearity, i e ,
(i)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
d
d
d
k f
x
k
f
x
k
f
x
k
f
x
dx
dx
dx
+
=
+
(ii)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
k f
x
k
f
x
dx
k
f
x dx
k
f
x dx
+
=
+
∫
∫
∫
Here k1 and k2 are constants |
1 | 3388-3391 | Both satisfy the property of linearity, i e ,
(i)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
d
d
d
k f
x
k
f
x
k
f
x
k
f
x
dx
dx
dx
+
=
+
(ii)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
k f
x
k
f
x
dx
k
f
x dx
k
f
x dx
+
=
+
∫
∫
∫
Here k1 and k2 are constants 3 |
1 | 3389-3392 | e ,
(i)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
d
d
d
k f
x
k
f
x
k
f
x
k
f
x
dx
dx
dx
+
=
+
(ii)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
k f
x
k
f
x
dx
k
f
x dx
k
f
x dx
+
=
+
∫
∫
∫
Here k1 and k2 are constants 3 We have already seen that all functions are not differentiable |
1 | 3390-3393 | ,
(i)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
d
d
d
k f
x
k
f
x
k
f
x
k
f
x
dx
dx
dx
+
=
+
(ii)
[
]
1
1
2
2
1
1
2
2
( )
( )
( )
( )
k f
x
k
f
x
dx
k
f
x dx
k
f
x dx
+
=
+
∫
∫
∫
Here k1 and k2 are constants 3 We have already seen that all functions are not differentiable Similarly, all functions
are not integrable |
1 | 3391-3394 | 3 We have already seen that all functions are not differentiable Similarly, all functions
are not integrable We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes |
1 | 3392-3395 | We have already seen that all functions are not differentiable Similarly, all functions
are not integrable We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes 4 |
1 | 3393-3396 | Similarly, all functions
are not integrable We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes 4 The derivative of a function, when it exists, is a unique function |
1 | 3394-3397 | We will learn more about nondifferentiable functions and
nonintegrable functions in higher classes 4 The derivative of a function, when it exists, is a unique function The integral of
a function is not so |
1 | 3395-3398 | 4 The derivative of a function, when it exists, is a unique function The integral of
a function is not so However, they are unique upto an additive constant, i |
1 | 3396-3399 | The derivative of a function, when it exists, is a unique function The integral of
a function is not so However, they are unique upto an additive constant, i e |
1 | 3397-3400 | The integral of
a function is not so However, they are unique upto an additive constant, i e , any
two integrals of a function differ by a constant |
1 | 3398-3401 | However, they are unique upto an additive constant, i e , any
two integrals of a function differ by a constant 5 |
1 | 3399-3402 | e , any
two integrals of a function differ by a constant 5 When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P |
1 | 3400-3403 | , any
two integrals of a function differ by a constant 5 When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P |
1 | 3401-3404 | 5 When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P 6 |
1 | 3402-3405 | When a polynomial function P is differentiated, the result is a polynomial whose
degree is 1 less than the degree of P When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P 6 We can speak of the derivative at a point |
1 | 3403-3406 | When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P 6 We can speak of the derivative at a point We never speak of the integral at a
point, we speak of the integral of a function over an interval on which the integral
is defined as will be seen in Section 7 |
1 | 3404-3407 | 6 We can speak of the derivative at a point We never speak of the integral at a
point, we speak of the integral of a function over an interval on which the integral
is defined as will be seen in Section 7 7 |
1 | 3405-3408 | We can speak of the derivative at a point We never speak of the integral at a
point, we speak of the integral of a function over an interval on which the integral
is defined as will be seen in Section 7 7 7 |
1 | 3406-3409 | We never speak of the integral at a
point, we speak of the integral of a function over an interval on which the integral
is defined as will be seen in Section 7 7 7 The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point |
1 | 3407-3410 | 7 7 The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point Similarly, the indefinite integral of
a function represents geometrically, a family of curves placed parallel to each
other having parallel tangents at the points of intersection of the curves of the
family with the lines orthogonal (perpendicular) to the axis representing the variable
of integration |
1 | 3408-3411 | 7 The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point Similarly, the indefinite integral of
a function represents geometrically, a family of curves placed parallel to each
other having parallel tangents at the points of intersection of the curves of the
family with the lines orthogonal (perpendicular) to the axis representing the variable
of integration 8 |
1 | 3409-3412 | The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point Similarly, the indefinite integral of
a function represents geometrically, a family of curves placed parallel to each
other having parallel tangents at the points of intersection of the curves of the
family with the lines orthogonal (perpendicular) to the axis representing the variable
of integration 8 The derivative is used for finding some physical quantities like the velocity of a
moving particle, when the distance traversed at any time t is known |
1 | 3410-3413 | Similarly, the indefinite integral of
a function represents geometrically, a family of curves placed parallel to each
other having parallel tangents at the points of intersection of the curves of the
family with the lines orthogonal (perpendicular) to the axis representing the variable
of integration 8 The derivative is used for finding some physical quantities like the velocity of a
moving particle, when the distance traversed at any time t is known Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known |
1 | 3411-3414 | 8 The derivative is used for finding some physical quantities like the velocity of a
moving particle, when the distance traversed at any time t is known Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known 9 |
1 | 3412-3415 | The derivative is used for finding some physical quantities like the velocity of a
moving particle, when the distance traversed at any time t is known Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known 9 Differentiation is a process involving limits |
1 | 3413-3416 | Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known 9 Differentiation is a process involving limits So is integration, as will be seen in
Section 7 |
1 | 3414-3417 | 9 Differentiation is a process involving limits So is integration, as will be seen in
Section 7 7 |
1 | 3415-3418 | Differentiation is a process involving limits So is integration, as will be seen in
Section 7 7 INTEGRALS 299
10 |
1 | 3416-3419 | So is integration, as will be seen in
Section 7 7 INTEGRALS 299
10 The process of differentiation and integration are inverses of each other as
discussed in Section 7 |
1 | 3417-3420 | 7 INTEGRALS 299
10 The process of differentiation and integration are inverses of each other as
discussed in Section 7 2 |
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