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1 | 3718-3721 | Therefore, the proper choice of the first
function and the second function is significant Remarks
(i)
It is worth mentioning that integration by parts is not applicable to product of
functions in all cases For instance, the method does not work for
xsin
x dx
∫ The reason is that there does not exist any function whose derivative is
x sin x |
1 | 3719-3722 | Remarks
(i)
It is worth mentioning that integration by parts is not applicable to product of
functions in all cases For instance, the method does not work for
xsin
x dx
∫ The reason is that there does not exist any function whose derivative is
x sin x (ii)
Observe that while finding the integral of the second function, we did not add
any constant of integration |
1 | 3720-3723 | For instance, the method does not work for
xsin
x dx
∫ The reason is that there does not exist any function whose derivative is
x sin x (ii)
Observe that while finding the integral of the second function, we did not add
any constant of integration If we write the integral of the second function cos x
INTEGRALS 325
as sin x + k, where k is any constant, then
xcos
x dx
∫
=
(sin
)
(sin
)
x
x
k
x
k dx
+
−
+
∫
=
(sin
)
(sin
x
x
k
x dx
k dx
+
−
−
∫
∫
=
(sin
)
cos
C
x
x
k
x – kx
+
−
+
=
sin
cos
C
x
x
x
+
+
This shows that adding a constant to the integral of the second function is
superfluous so far as the final result is concerned while applying the method of
integration by parts |
1 | 3721-3724 | The reason is that there does not exist any function whose derivative is
x sin x (ii)
Observe that while finding the integral of the second function, we did not add
any constant of integration If we write the integral of the second function cos x
INTEGRALS 325
as sin x + k, where k is any constant, then
xcos
x dx
∫
=
(sin
)
(sin
)
x
x
k
x
k dx
+
−
+
∫
=
(sin
)
(sin
x
x
k
x dx
k dx
+
−
−
∫
∫
=
(sin
)
cos
C
x
x
k
x – kx
+
−
+
=
sin
cos
C
x
x
x
+
+
This shows that adding a constant to the integral of the second function is
superfluous so far as the final result is concerned while applying the method of
integration by parts (iii)
Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function |
1 | 3722-3725 | (ii)
Observe that while finding the integral of the second function, we did not add
any constant of integration If we write the integral of the second function cos x
INTEGRALS 325
as sin x + k, where k is any constant, then
xcos
x dx
∫
=
(sin
)
(sin
)
x
x
k
x
k dx
+
−
+
∫
=
(sin
)
(sin
x
x
k
x dx
k dx
+
−
−
∫
∫
=
(sin
)
cos
C
x
x
k
x – kx
+
−
+
=
sin
cos
C
x
x
x
+
+
This shows that adding a constant to the integral of the second function is
superfluous so far as the final result is concerned while applying the method of
integration by parts (iii)
Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function |
1 | 3723-3726 | If we write the integral of the second function cos x
INTEGRALS 325
as sin x + k, where k is any constant, then
xcos
x dx
∫
=
(sin
)
(sin
)
x
x
k
x
k dx
+
−
+
∫
=
(sin
)
(sin
x
x
k
x dx
k dx
+
−
−
∫
∫
=
(sin
)
cos
C
x
x
k
x – kx
+
−
+
=
sin
cos
C
x
x
x
+
+
This shows that adding a constant to the integral of the second function is
superfluous so far as the final result is concerned while applying the method of
integration by parts (iii)
Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function Example 18 Find log x dx
∫
Solution To start with, we are unable to guess a function whose derivative is log x |
1 | 3724-3727 | (iii)
Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function Example 18 Find log x dx
∫
Solution To start with, we are unable to guess a function whose derivative is log x We
take log x as the first function and the constant function 1 as the second function |
1 | 3725-3728 | However, in cases where other function is inverse trigonometric
function or logarithmic function, then we take them as first function Example 18 Find log x dx
∫
Solution To start with, we are unable to guess a function whose derivative is log x We
take log x as the first function and the constant function 1 as the second function Then,
the integral of the second function is x |
1 | 3726-3729 | Example 18 Find log x dx
∫
Solution To start with, we are unable to guess a function whose derivative is log x We
take log x as the first function and the constant function 1 as the second function Then,
the integral of the second function is x Hence,
(log |
1 | 3727-3730 | We
take log x as the first function and the constant function 1 as the second function Then,
the integral of the second function is x Hence,
(log 1)
x
dx
∫
= log
1
[
(log ) 1
]
d
x
dx
x
dx dx
dx
−
∫
∫
∫
=
1
(log )
–
log
C
x
x
x dx
x
x – x
x
⋅
=
+
∫ |
1 | 3728-3731 | Then,
the integral of the second function is x Hence,
(log 1)
x
dx
∫
= log
1
[
(log ) 1
]
d
x
dx
x
dx dx
dx
−
∫
∫
∫
=
1
(log )
–
log
C
x
x
x dx
x
x – x
x
⋅
=
+
∫ Example 19 Find
x
x e dx
∫
Solution Take first function as x and second function as ex |
1 | 3729-3732 | Hence,
(log 1)
x
dx
∫
= log
1
[
(log ) 1
]
d
x
dx
x
dx dx
dx
−
∫
∫
∫
=
1
(log )
–
log
C
x
x
x dx
x
x – x
x
⋅
=
+
∫ Example 19 Find
x
x e dx
∫
Solution Take first function as x and second function as ex The integral of the second
function is ex |
1 | 3730-3733 | 1)
x
dx
∫
= log
1
[
(log ) 1
]
d
x
dx
x
dx dx
dx
−
∫
∫
∫
=
1
(log )
–
log
C
x
x
x dx
x
x – x
x
⋅
=
+
∫ Example 19 Find
x
x e dx
∫
Solution Take first function as x and second function as ex The integral of the second
function is ex Therefore,
x
∫x e dx
=
1
x
x
x e
e dx
−
∫⋅
= xex – ex + C |
1 | 3731-3734 | Example 19 Find
x
x e dx
∫
Solution Take first function as x and second function as ex The integral of the second
function is ex Therefore,
x
∫x e dx
=
1
x
x
x e
e dx
−
∫⋅
= xex – ex + C Example 20 Find
1
2
sin
1
–
x
x dx
−x
∫
Solution Let first function be sin – 1x and second function be
2
1
x
x
− |
1 | 3732-3735 | The integral of the second
function is ex Therefore,
x
∫x e dx
=
1
x
x
x e
e dx
−
∫⋅
= xex – ex + C Example 20 Find
1
2
sin
1
–
x
x dx
−x
∫
Solution Let first function be sin – 1x and second function be
2
1
x
x
− First we find the integral of the second function, i |
1 | 3733-3736 | Therefore,
x
∫x e dx
=
1
x
x
x e
e dx
−
∫⋅
= xex – ex + C Example 20 Find
1
2
sin
1
–
x
x dx
−x
∫
Solution Let first function be sin – 1x and second function be
2
1
x
x
− First we find the integral of the second function, i e |
1 | 3734-3737 | Example 20 Find
1
2
sin
1
–
x
x dx
−x
∫
Solution Let first function be sin – 1x and second function be
2
1
x
x
− First we find the integral of the second function, i e ,
2
1
x dx
−x
∫ |
1 | 3735-3738 | First we find the integral of the second function, i e ,
2
1
x dx
−x
∫ Put t =1 – x2 |
1 | 3736-3739 | e ,
2
1
x dx
−x
∫ Put t =1 – x2 Then dt = – 2x dx
326
MATHEMATICS
Therefore,
2
1
x dx
−x
∫
=
1
2
dt
–
∫t
=
2
–
1
t
x
= −
−
Hence,
1
2
sin
1
–
x
x dx
−x
∫
=
(
)
1
2
2
12
(sin
)
1
(
1
)
1
– x
–
x
–
x
dx
x
−
−
−
−
∫
=
2
1
1
sin
C
–
x
x
x
−
−
+
+
=
2
1
1
sin
C
x –
x
x
−
−
+
Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts |
1 | 3737-3740 | ,
2
1
x dx
−x
∫ Put t =1 – x2 Then dt = – 2x dx
326
MATHEMATICS
Therefore,
2
1
x dx
−x
∫
=
1
2
dt
–
∫t
=
2
–
1
t
x
= −
−
Hence,
1
2
sin
1
–
x
x dx
−x
∫
=
(
)
1
2
2
12
(sin
)
1
(
1
)
1
– x
–
x
–
x
dx
x
−
−
−
−
∫
=
2
1
1
sin
C
–
x
x
x
−
−
+
+
=
2
1
1
sin
C
x –
x
x
−
−
+
Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts Example 21 Find
exsin
x dx
∫
Solution Take ex as the first function and sin x as second function |
1 | 3738-3741 | Put t =1 – x2 Then dt = – 2x dx
326
MATHEMATICS
Therefore,
2
1
x dx
−x
∫
=
1
2
dt
–
∫t
=
2
–
1
t
x
= −
−
Hence,
1
2
sin
1
–
x
x dx
−x
∫
=
(
)
1
2
2
12
(sin
)
1
(
1
)
1
– x
–
x
–
x
dx
x
−
−
−
−
∫
=
2
1
1
sin
C
–
x
x
x
−
−
+
+
=
2
1
1
sin
C
x –
x
x
−
−
+
Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts Example 21 Find
exsin
x dx
∫
Solution Take ex as the first function and sin x as second function Then, integrating
by parts, we have
I
sin
(
cos )
cos
x
x
x
e
x dx
e
–
x
e
x dx
=
=
+
∫
∫
= – e x cos x + I1 (say) |
1 | 3739-3742 | Then dt = – 2x dx
326
MATHEMATICS
Therefore,
2
1
x dx
−x
∫
=
1
2
dt
–
∫t
=
2
–
1
t
x
= −
−
Hence,
1
2
sin
1
–
x
x dx
−x
∫
=
(
)
1
2
2
12
(sin
)
1
(
1
)
1
– x
–
x
–
x
dx
x
−
−
−
−
∫
=
2
1
1
sin
C
–
x
x
x
−
−
+
+
=
2
1
1
sin
C
x –
x
x
−
−
+
Alternatively, this integral can also be worked out by making substitution sin–1 x = θ and
then integrating by parts Example 21 Find
exsin
x dx
∫
Solution Take ex as the first function and sin x as second function Then, integrating
by parts, we have
I
sin
(
cos )
cos
x
x
x
e
x dx
e
–
x
e
x dx
=
=
+
∫
∫
= – e x cos x + I1 (say) (1)
Taking ex
and cos x as the first and second functions, respectively, in I1, we get
I1 =
sin
sin
x
x
e
x – e
x dx
∫
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
Hence,
I =
sin
(sin
cos ) + C
2
x
x
e
e
x dx
x –
x
=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function |
1 | 3740-3743 | Example 21 Find
exsin
x dx
∫
Solution Take ex as the first function and sin x as second function Then, integrating
by parts, we have
I
sin
(
cos )
cos
x
x
x
e
x dx
e
–
x
e
x dx
=
=
+
∫
∫
= – e x cos x + I1 (say) (1)
Taking ex
and cos x as the first and second functions, respectively, in I1, we get
I1 =
sin
sin
x
x
e
x – e
x dx
∫
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
Hence,
I =
sin
(sin
cos ) + C
2
x
x
e
e
x dx
x –
x
=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function 7 |
1 | 3741-3744 | Then, integrating
by parts, we have
I
sin
(
cos )
cos
x
x
x
e
x dx
e
–
x
e
x dx
=
=
+
∫
∫
= – e x cos x + I1 (say) (1)
Taking ex
and cos x as the first and second functions, respectively, in I1, we get
I1 =
sin
sin
x
x
e
x – e
x dx
∫
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
Hence,
I =
sin
(sin
cos ) + C
2
x
x
e
e
x dx
x –
x
=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function 7 6 |
1 | 3742-3745 | (1)
Taking ex
and cos x as the first and second functions, respectively, in I1, we get
I1 =
sin
sin
x
x
e
x – e
x dx
∫
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
Hence,
I =
sin
(sin
cos ) + C
2
x
x
e
e
x dx
x –
x
=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function 7 6 1 Integral of the type
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
=
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫ |
1 | 3743-3746 | 7 6 1 Integral of the type
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
=
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫ (1)
Taking f(x) and ex as the first function and second function, respectively, in I 1 and
integrating it by parts, we have I1 = f (x) ex –
( )
C
x
f
′x e dx
+
∫
Substituting I1 in (1), we get
I =
( )
( )
( )
C
x
x
x
e f x
f
x e dx
e f
x dx
′
′
−
+
+
∫
∫
= ex f (x) + C
INTEGRALS 327
Thus,
′
∫
[
( )
( )]
ex
f x + f
x
dx =
( )
C
x
e f x +
Example 22 Find (i)
1
2
1
(tan
)
1
x
–
e
x
x
+
+
∫
dx (ii)
2
2
(
+ 1)
( + 1)
x
x
e
∫x
dx
Solution
(i) We have I =
1
12
(tan
)
1
x
–
e
x
dx
x
+
+
∫
Consider f (x) = tan– 1x, then f ′(x) =
2
1
1
x
+
Thus, the given integrand is of the form ex [ f (x) + f ′(x)] |
1 | 3744-3747 | 6 1 Integral of the type
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
=
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫ (1)
Taking f(x) and ex as the first function and second function, respectively, in I 1 and
integrating it by parts, we have I1 = f (x) ex –
( )
C
x
f
′x e dx
+
∫
Substituting I1 in (1), we get
I =
( )
( )
( )
C
x
x
x
e f x
f
x e dx
e f
x dx
′
′
−
+
+
∫
∫
= ex f (x) + C
INTEGRALS 327
Thus,
′
∫
[
( )
( )]
ex
f x + f
x
dx =
( )
C
x
e f x +
Example 22 Find (i)
1
2
1
(tan
)
1
x
–
e
x
x
+
+
∫
dx (ii)
2
2
(
+ 1)
( + 1)
x
x
e
∫x
dx
Solution
(i) We have I =
1
12
(tan
)
1
x
–
e
x
dx
x
+
+
∫
Consider f (x) = tan– 1x, then f ′(x) =
2
1
1
x
+
Thus, the given integrand is of the form ex [ f (x) + f ′(x)] Therefore,
1
12
I
(tan
)
1
x
–
e
x
dx
x
=
+
+
∫
= ex tan– 1x + C
(ii) We have
2
2
(
+1)
I
( +1)
x
x
e
= ∫x
dx
2
2
1 +1+1)
[
]
( +1)
x
x –
e
dx
x
= ∫
2
2
2
1
2
[
]
( +1)
( +1)
x
x –
e
dx
x
x
=
+
∫
2
1
2
[
+
]
+1
( +1)
x
x –
e
dx
x
x
= ∫
Consider
1
( )
1
x
f x
x
−
=
+
, then
2
2
( )
(
1)
f
x
x
′
=
+
Thus, the given integrand is of the form ex [f (x) + f ′(x)] |
1 | 3745-3748 | 1 Integral of the type
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
=
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫ (1)
Taking f(x) and ex as the first function and second function, respectively, in I 1 and
integrating it by parts, we have I1 = f (x) ex –
( )
C
x
f
′x e dx
+
∫
Substituting I1 in (1), we get
I =
( )
( )
( )
C
x
x
x
e f x
f
x e dx
e f
x dx
′
′
−
+
+
∫
∫
= ex f (x) + C
INTEGRALS 327
Thus,
′
∫
[
( )
( )]
ex
f x + f
x
dx =
( )
C
x
e f x +
Example 22 Find (i)
1
2
1
(tan
)
1
x
–
e
x
x
+
+
∫
dx (ii)
2
2
(
+ 1)
( + 1)
x
x
e
∫x
dx
Solution
(i) We have I =
1
12
(tan
)
1
x
–
e
x
dx
x
+
+
∫
Consider f (x) = tan– 1x, then f ′(x) =
2
1
1
x
+
Thus, the given integrand is of the form ex [ f (x) + f ′(x)] Therefore,
1
12
I
(tan
)
1
x
–
e
x
dx
x
=
+
+
∫
= ex tan– 1x + C
(ii) We have
2
2
(
+1)
I
( +1)
x
x
e
= ∫x
dx
2
2
1 +1+1)
[
]
( +1)
x
x –
e
dx
x
= ∫
2
2
2
1
2
[
]
( +1)
( +1)
x
x –
e
dx
x
x
=
+
∫
2
1
2
[
+
]
+1
( +1)
x
x –
e
dx
x
x
= ∫
Consider
1
( )
1
x
f x
x
−
=
+
, then
2
2
( )
(
1)
f
x
x
′
=
+
Thus, the given integrand is of the form ex [f (x) + f ′(x)] Therefore,
2
12
1
C
1
(
1)
x
x
x
x
e dx
e
x
x
+
−
=
+
+
+
∫
EXERCISE 7 |
1 | 3746-3749 | (1)
Taking f(x) and ex as the first function and second function, respectively, in I 1 and
integrating it by parts, we have I1 = f (x) ex –
( )
C
x
f
′x e dx
+
∫
Substituting I1 in (1), we get
I =
( )
( )
( )
C
x
x
x
e f x
f
x e dx
e f
x dx
′
′
−
+
+
∫
∫
= ex f (x) + C
INTEGRALS 327
Thus,
′
∫
[
( )
( )]
ex
f x + f
x
dx =
( )
C
x
e f x +
Example 22 Find (i)
1
2
1
(tan
)
1
x
–
e
x
x
+
+
∫
dx (ii)
2
2
(
+ 1)
( + 1)
x
x
e
∫x
dx
Solution
(i) We have I =
1
12
(tan
)
1
x
–
e
x
dx
x
+
+
∫
Consider f (x) = tan– 1x, then f ′(x) =
2
1
1
x
+
Thus, the given integrand is of the form ex [ f (x) + f ′(x)] Therefore,
1
12
I
(tan
)
1
x
–
e
x
dx
x
=
+
+
∫
= ex tan– 1x + C
(ii) We have
2
2
(
+1)
I
( +1)
x
x
e
= ∫x
dx
2
2
1 +1+1)
[
]
( +1)
x
x –
e
dx
x
= ∫
2
2
2
1
2
[
]
( +1)
( +1)
x
x –
e
dx
x
x
=
+
∫
2
1
2
[
+
]
+1
( +1)
x
x –
e
dx
x
x
= ∫
Consider
1
( )
1
x
f x
x
−
=
+
, then
2
2
( )
(
1)
f
x
x
′
=
+
Thus, the given integrand is of the form ex [f (x) + f ′(x)] Therefore,
2
12
1
C
1
(
1)
x
x
x
x
e dx
e
x
x
+
−
=
+
+
+
∫
EXERCISE 7 6
Integrate the functions in Exercises 1 to 22 |
1 | 3747-3750 | Therefore,
1
12
I
(tan
)
1
x
–
e
x
dx
x
=
+
+
∫
= ex tan– 1x + C
(ii) We have
2
2
(
+1)
I
( +1)
x
x
e
= ∫x
dx
2
2
1 +1+1)
[
]
( +1)
x
x –
e
dx
x
= ∫
2
2
2
1
2
[
]
( +1)
( +1)
x
x –
e
dx
x
x
=
+
∫
2
1
2
[
+
]
+1
( +1)
x
x –
e
dx
x
x
= ∫
Consider
1
( )
1
x
f x
x
−
=
+
, then
2
2
( )
(
1)
f
x
x
′
=
+
Thus, the given integrand is of the form ex [f (x) + f ′(x)] Therefore,
2
12
1
C
1
(
1)
x
x
x
x
e dx
e
x
x
+
−
=
+
+
+
∫
EXERCISE 7 6
Integrate the functions in Exercises 1 to 22 1 |
1 | 3748-3751 | Therefore,
2
12
1
C
1
(
1)
x
x
x
x
e dx
e
x
x
+
−
=
+
+
+
∫
EXERCISE 7 6
Integrate the functions in Exercises 1 to 22 1 x sin x
2 |
1 | 3749-3752 | 6
Integrate the functions in Exercises 1 to 22 1 x sin x
2 x sin 3x
3 |
1 | 3750-3753 | 1 x sin x
2 x sin 3x
3 x2 ex
4 |
1 | 3751-3754 | x sin x
2 x sin 3x
3 x2 ex
4 x log x
5 |
1 | 3752-3755 | x sin 3x
3 x2 ex
4 x log x
5 x log 2x
6 |
1 | 3753-3756 | x2 ex
4 x log x
5 x log 2x
6 x2 log x
7 |
1 | 3754-3757 | x log x
5 x log 2x
6 x2 log x
7 x sin– 1x
8 |
1 | 3755-3758 | x log 2x
6 x2 log x
7 x sin– 1x
8 x tan–1 x
9 |
1 | 3756-3759 | x2 log x
7 x sin– 1x
8 x tan–1 x
9 x cos–1 x
10 |
1 | 3757-3760 | x sin– 1x
8 x tan–1 x
9 x cos–1 x
10 (sin–1x)2
11 |
1 | 3758-3761 | x tan–1 x
9 x cos–1 x
10 (sin–1x)2
11 1
2
cos
1
x
x
x
−
−
12 |
1 | 3759-3762 | x cos–1 x
10 (sin–1x)2
11 1
2
cos
1
x
x
x
−
−
12 x sec2 x
13 |
1 | 3760-3763 | (sin–1x)2
11 1
2
cos
1
x
x
x
−
−
12 x sec2 x
13 tan –1x
14 |
1 | 3761-3764 | 1
2
cos
1
x
x
x
−
−
12 x sec2 x
13 tan –1x
14 x (log x)2
15 |
1 | 3762-3765 | x sec2 x
13 tan –1x
14 x (log x)2
15 (x2 + 1) log x
328
MATHEMATICS
16 |
1 | 3763-3766 | tan –1x
14 x (log x)2
15 (x2 + 1) log x
328
MATHEMATICS
16 ex (sinx + cosx) 17 |
1 | 3764-3767 | x (log x)2
15 (x2 + 1) log x
328
MATHEMATICS
16 ex (sinx + cosx) 17 2
(1
)
x
x e
+x
18 |
1 | 3765-3768 | (x2 + 1) log x
328
MATHEMATICS
16 ex (sinx + cosx) 17 2
(1
)
x
x e
+x
18 1
sin
1
cos
x
x
e
x
+
+
19 |
1 | 3766-3769 | ex (sinx + cosx) 17 2
(1
)
x
x e
+x
18 1
sin
1
cos
x
x
e
x
+
+
19 2
1
–1
xe
x
x
20 |
1 | 3767-3770 | 2
(1
)
x
x e
+x
18 1
sin
1
cos
x
x
e
x
+
+
19 2
1
–1
xe
x
x
20 3
(
3)
(
1)
x
x
e
−x
−
21 |
1 | 3768-3771 | 1
sin
1
cos
x
x
e
x
+
+
19 2
1
–1
xe
x
x
20 3
(
3)
(
1)
x
x
e
−x
−
21 e2x sin x
22 |
1 | 3769-3772 | 2
1
–1
xe
x
x
20 3
(
3)
(
1)
x
x
e
−x
−
21 e2x sin x
22 1
2
2
sin
1
–
x
x
+
Choose the correct answer in Exercises 23 and 24 |
1 | 3770-3773 | 3
(
3)
(
1)
x
x
e
−x
−
21 e2x sin x
22 1
2
2
sin
1
–
x
x
+
Choose the correct answer in Exercises 23 and 24 23 |
1 | 3771-3774 | e2x sin x
22 1
2
2
sin
1
–
x
x
+
Choose the correct answer in Exercises 23 and 24 23 3
2
x
∫x e dx
equals
(A)
3
1
C
3
ex
+
(B)
2
1
C
3
ex
+
(C)
3
1
C
2
ex
+
(D)
2
1
C
2
ex
+
24 |
1 | 3772-3775 | 1
2
2
sin
1
–
x
x
+
Choose the correct answer in Exercises 23 and 24 23 3
2
x
∫x e dx
equals
(A)
3
1
C
3
ex
+
(B)
2
1
C
3
ex
+
(C)
3
1
C
2
ex
+
(D)
2
1
C
2
ex
+
24 sec
(1
tan )
ex
x
x dx
+
∫
equals
(A) ex cos x + C
(B) ex sec x + C
(C) ex sin x + C
(D) ex tan x + C
7 |
1 | 3773-3776 | 23 3
2
x
∫x e dx
equals
(A)
3
1
C
3
ex
+
(B)
2
1
C
3
ex
+
(C)
3
1
C
2
ex
+
(D)
2
1
C
2
ex
+
24 sec
(1
tan )
ex
x
x dx
+
∫
equals
(A) ex cos x + C
(B) ex sec x + C
(C) ex sin x + C
(D) ex tan x + C
7 6 |
1 | 3774-3777 | 3
2
x
∫x e dx
equals
(A)
3
1
C
3
ex
+
(B)
2
1
C
3
ex
+
(C)
3
1
C
2
ex
+
(D)
2
1
C
2
ex
+
24 sec
(1
tan )
ex
x
x dx
+
∫
equals
(A) ex cos x + C
(B) ex sec x + C
(C) ex sin x + C
(D) ex tan x + C
7 6 2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :
(i)
2
2
x
a
dx
−
∫
(ii)
2
2
x
+a dx
∫
(iii)
2
2
a
−x dx
∫
(i)
Let
2
2
I
x
a dx
=
−
∫
Taking constant function 1 as the second function and integrating by parts, we
have
I =
2
2
2
2
1
2
2
x
x
x
a
x dx
x
a
−
−
−
∫
=
2
2
2
2
2
x
x
x
a
dx
x
a
−
−
−
∫
=
2
2
2
2
2
2
2
x
a
a
x
x
a
dx
x
a
−
+
−
−
−
∫
INTEGRALS 329
=
2
2
2
2
2
2
2
dx
x
x
a
x
a dx
a
x
a
−
−
−
−
−
∫
∫
=
2
2
2
2
2
I
dx
x
x
a
a
x
a
−
− −
−
∫
or
2I =
2
2
2
2
2
dx
x
x
a
a
x
a
−
−
−
∫
or
I = ∫
2
2
x – a dx =
2
2
2
2
2
–
–
log
+
–
+ C
2
2
x
a
x
a
x
x
a
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
(ii) ∫
2
2
2
2
2
2
2
1
+
=
+
+
log
+
+
+ C
2
a2
x
a dx
x
x
a
x
x
a
(iii) ∫
2
2
2
2
2
–1
1
–
=
–
+
sin
+ C
2
a2
x
a
x dx
x a
x
a
Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively |
1 | 3775-3778 | sec
(1
tan )
ex
x
x dx
+
∫
equals
(A) ex cos x + C
(B) ex sec x + C
(C) ex sin x + C
(D) ex tan x + C
7 6 2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :
(i)
2
2
x
a
dx
−
∫
(ii)
2
2
x
+a dx
∫
(iii)
2
2
a
−x dx
∫
(i)
Let
2
2
I
x
a dx
=
−
∫
Taking constant function 1 as the second function and integrating by parts, we
have
I =
2
2
2
2
1
2
2
x
x
x
a
x dx
x
a
−
−
−
∫
=
2
2
2
2
2
x
x
x
a
dx
x
a
−
−
−
∫
=
2
2
2
2
2
2
2
x
a
a
x
x
a
dx
x
a
−
+
−
−
−
∫
INTEGRALS 329
=
2
2
2
2
2
2
2
dx
x
x
a
x
a dx
a
x
a
−
−
−
−
−
∫
∫
=
2
2
2
2
2
I
dx
x
x
a
a
x
a
−
− −
−
∫
or
2I =
2
2
2
2
2
dx
x
x
a
a
x
a
−
−
−
∫
or
I = ∫
2
2
x – a dx =
2
2
2
2
2
–
–
log
+
–
+ C
2
2
x
a
x
a
x
x
a
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
(ii) ∫
2
2
2
2
2
2
2
1
+
=
+
+
log
+
+
+ C
2
a2
x
a dx
x
x
a
x
x
a
(iii) ∫
2
2
2
2
2
–1
1
–
=
–
+
sin
+ C
2
a2
x
a
x dx
x a
x
a
Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively Example 23 Find
2
2
5
x
x
dx
+
+
∫
Solution Note that
2
2
5
x
x
dx
+
+
∫
=
2
(
1)
4
x
dx
+
+
∫
Put x + 1 = y, so that dx = dy |
1 | 3776-3779 | 6 2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :
(i)
2
2
x
a
dx
−
∫
(ii)
2
2
x
+a dx
∫
(iii)
2
2
a
−x dx
∫
(i)
Let
2
2
I
x
a dx
=
−
∫
Taking constant function 1 as the second function and integrating by parts, we
have
I =
2
2
2
2
1
2
2
x
x
x
a
x dx
x
a
−
−
−
∫
=
2
2
2
2
2
x
x
x
a
dx
x
a
−
−
−
∫
=
2
2
2
2
2
2
2
x
a
a
x
x
a
dx
x
a
−
+
−
−
−
∫
INTEGRALS 329
=
2
2
2
2
2
2
2
dx
x
x
a
x
a dx
a
x
a
−
−
−
−
−
∫
∫
=
2
2
2
2
2
I
dx
x
x
a
a
x
a
−
− −
−
∫
or
2I =
2
2
2
2
2
dx
x
x
a
a
x
a
−
−
−
∫
or
I = ∫
2
2
x – a dx =
2
2
2
2
2
–
–
log
+
–
+ C
2
2
x
a
x
a
x
x
a
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
(ii) ∫
2
2
2
2
2
2
2
1
+
=
+
+
log
+
+
+ C
2
a2
x
a dx
x
x
a
x
x
a
(iii) ∫
2
2
2
2
2
–1
1
–
=
–
+
sin
+ C
2
a2
x
a
x dx
x a
x
a
Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively Example 23 Find
2
2
5
x
x
dx
+
+
∫
Solution Note that
2
2
5
x
x
dx
+
+
∫
=
2
(
1)
4
x
dx
+
+
∫
Put x + 1 = y, so that dx = dy Then
2
2
5
x
x
dx
+
+
∫
=
2
22
y
dy
+
∫
=
2
2
1
4
4
log
4
C
2
2
y
y
y
y
+
+
+
+
+
[using 7 |
1 | 3777-3780 | 2 Integrals of some more types
Here, we discuss some special types of standard integrals based on the technique of
integration by parts :
(i)
2
2
x
a
dx
−
∫
(ii)
2
2
x
+a dx
∫
(iii)
2
2
a
−x dx
∫
(i)
Let
2
2
I
x
a dx
=
−
∫
Taking constant function 1 as the second function and integrating by parts, we
have
I =
2
2
2
2
1
2
2
x
x
x
a
x dx
x
a
−
−
−
∫
=
2
2
2
2
2
x
x
x
a
dx
x
a
−
−
−
∫
=
2
2
2
2
2
2
2
x
a
a
x
x
a
dx
x
a
−
+
−
−
−
∫
INTEGRALS 329
=
2
2
2
2
2
2
2
dx
x
x
a
x
a dx
a
x
a
−
−
−
−
−
∫
∫
=
2
2
2
2
2
I
dx
x
x
a
a
x
a
−
− −
−
∫
or
2I =
2
2
2
2
2
dx
x
x
a
a
x
a
−
−
−
∫
or
I = ∫
2
2
x – a dx =
2
2
2
2
2
–
–
log
+
–
+ C
2
2
x
a
x
a
x
x
a
Similarly, integrating other two integrals by parts, taking constant function 1 as the
second function, we get
(ii) ∫
2
2
2
2
2
2
2
1
+
=
+
+
log
+
+
+ C
2
a2
x
a dx
x
x
a
x
x
a
(iii) ∫
2
2
2
2
2
–1
1
–
=
–
+
sin
+ C
2
a2
x
a
x dx
x a
x
a
Alternatively, integrals (i), (ii) and (iii) can also be found by making trigonometric
substitution x = a secθ in (i), x = a tanθ in (ii) and x = a sinθ in (iii) respectively Example 23 Find
2
2
5
x
x
dx
+
+
∫
Solution Note that
2
2
5
x
x
dx
+
+
∫
=
2
(
1)
4
x
dx
+
+
∫
Put x + 1 = y, so that dx = dy Then
2
2
5
x
x
dx
+
+
∫
=
2
22
y
dy
+
∫
=
2
2
1
4
4
log
4
C
2
2
y
y
y
y
+
+
+
+
+
[using 7 6 |
1 | 3778-3781 | Example 23 Find
2
2
5
x
x
dx
+
+
∫
Solution Note that
2
2
5
x
x
dx
+
+
∫
=
2
(
1)
4
x
dx
+
+
∫
Put x + 1 = y, so that dx = dy Then
2
2
5
x
x
dx
+
+
∫
=
2
22
y
dy
+
∫
=
2
2
1
4
4
log
4
C
2
2
y
y
y
y
+
+
+
+
+
[using 7 6 2 (ii)]
=
2
2
1 (
1)
2
5
2 log
1
2
5
C
2 x
x
x
x
x
x
+
+
+
+
+ +
+
+
+
Example 24 Find
2
3
2x
x dx
−
−
∫
Solution Note that
2
2
3
2
4
(
1)
x
x
dx
x
dx
−
−
=
−
+
∫
∫
330
MATHEMATICS
Put x + 1 = y so that dx = dy |
1 | 3779-3782 | Then
2
2
5
x
x
dx
+
+
∫
=
2
22
y
dy
+
∫
=
2
2
1
4
4
log
4
C
2
2
y
y
y
y
+
+
+
+
+
[using 7 6 2 (ii)]
=
2
2
1 (
1)
2
5
2 log
1
2
5
C
2 x
x
x
x
x
x
+
+
+
+
+ +
+
+
+
Example 24 Find
2
3
2x
x dx
−
−
∫
Solution Note that
2
2
3
2
4
(
1)
x
x
dx
x
dx
−
−
=
−
+
∫
∫
330
MATHEMATICS
Put x + 1 = y so that dx = dy Thus
2
3
2x
x dx
−
−
∫
=
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 |
1 | 3780-3783 | 6 2 (ii)]
=
2
2
1 (
1)
2
5
2 log
1
2
5
C
2 x
x
x
x
x
x
+
+
+
+
+ +
+
+
+
Example 24 Find
2
3
2x
x dx
−
−
∫
Solution Note that
2
2
3
2
4
(
1)
x
x
dx
x
dx
−
−
=
−
+
∫
∫
330
MATHEMATICS
Put x + 1 = y so that dx = dy Thus
2
3
2x
x dx
−
−
∫
=
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 6 |
1 | 3781-3784 | 2 (ii)]
=
2
2
1 (
1)
2
5
2 log
1
2
5
C
2 x
x
x
x
x
x
+
+
+
+
+ +
+
+
+
Example 24 Find
2
3
2x
x dx
−
−
∫
Solution Note that
2
2
3
2
4
(
1)
x
x
dx
x
dx
−
−
=
−
+
∫
∫
330
MATHEMATICS
Put x + 1 = y so that dx = dy Thus
2
3
2x
x dx
−
−
∫
=
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 6 2 (iii)]
=
2
1
1
1
(
1)
3
2
2 sin
C
2
2
–
x
x
x
x
+
+
−
−
+
+
EXERCISE 7 |
1 | 3782-3785 | Thus
2
3
2x
x dx
−
−
∫
=
2
4
−y dy
∫
=
2
1
1
4
4
sin
C
2
2
2
– y
y
−y
+
+
[using 7 6 2 (iii)]
=
2
1
1
1
(
1)
3
2
2 sin
C
2
2
–
x
x
x
x
+
+
−
−
+
+
EXERCISE 7 7
Integrate the functions in Exercises 1 to 9 |
1 | 3783-3786 | 6 2 (iii)]
=
2
1
1
1
(
1)
3
2
2 sin
C
2
2
–
x
x
x
x
+
+
−
−
+
+
EXERCISE 7 7
Integrate the functions in Exercises 1 to 9 1 |
1 | 3784-3787 | 2 (iii)]
=
2
1
1
1
(
1)
3
2
2 sin
C
2
2
–
x
x
x
x
+
+
−
−
+
+
EXERCISE 7 7
Integrate the functions in Exercises 1 to 9 1 2
4
−x
2 |
1 | 3785-3788 | 7
Integrate the functions in Exercises 1 to 9 1 2
4
−x
2 2
1
−4x
3 |
1 | 3786-3789 | 1 2
4
−x
2 2
1
−4x
3 2
4
6
x
x
+
+
4 |
1 | 3787-3790 | 2
4
−x
2 2
1
−4x
3 2
4
6
x
x
+
+
4 2
4
1
x
x
+
+
5 |
1 | 3788-3791 | 2
1
−4x
3 2
4
6
x
x
+
+
4 2
4
1
x
x
+
+
5 2
1
4x
x
−
−
6 |
1 | 3789-3792 | 2
4
6
x
x
+
+
4 2
4
1
x
x
+
+
5 2
1
4x
x
−
−
6 2
4
5
x
x
+
−
7 |
1 | 3790-3793 | 2
4
1
x
x
+
+
5 2
1
4x
x
−
−
6 2
4
5
x
x
+
−
7 2
1
3x
x
+
−
8 |
1 | 3791-3794 | 2
1
4x
x
−
−
6 2
4
5
x
x
+
−
7 2
1
3x
x
+
−
8 2
3
x
+x
9 |
1 | 3792-3795 | 2
4
5
x
x
+
−
7 2
1
3x
x
+
−
8 2
3
x
+x
9 2
1
x9
+
Choose the correct answer in Exercises 10 to 11 |
1 | 3793-3796 | 2
1
3x
x
+
−
8 2
3
x
+x
9 2
1
x9
+
Choose the correct answer in Exercises 10 to 11 10 |
1 | 3794-3797 | 2
3
x
+x
9 2
1
x9
+
Choose the correct answer in Exercises 10 to 11 10 2
1
+x dx
∫
is equal to
(A)
(
)
2
2
1
1
log
1
C
2
2
x
x
x
x
+
+
+
+
+
(B)
2 23
2(1
)
C
3
+x
+
(C)
2 23
2
(1
)
C
3 x
+x
+
(D)
2
2
2
2
1
1
log
1
C
2
2
x
x
x
x
x
+
+
+
+
+
11 |
1 | 3795-3798 | 2
1
x9
+
Choose the correct answer in Exercises 10 to 11 10 2
1
+x dx
∫
is equal to
(A)
(
)
2
2
1
1
log
1
C
2
2
x
x
x
x
+
+
+
+
+
(B)
2 23
2(1
)
C
3
+x
+
(C)
2 23
2
(1
)
C
3 x
+x
+
(D)
2
2
2
2
1
1
log
1
C
2
2
x
x
x
x
x
+
+
+
+
+
11 2
8
7
x
x
dx
−
+
∫
is equal to
(A)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
+
−
+
−
+
+
(B)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
+
−
+
+
+ +
−
+
+
(C)
2
2
1 (
4)
8
7
3
2log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
(D)
2
2
1
9
(
4)
8
7
log
4
8
7
C
2
2
x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
INTEGRALS 331
7 |
1 | 3796-3799 | 10 2
1
+x dx
∫
is equal to
(A)
(
)
2
2
1
1
log
1
C
2
2
x
x
x
x
+
+
+
+
+
(B)
2 23
2(1
)
C
3
+x
+
(C)
2 23
2
(1
)
C
3 x
+x
+
(D)
2
2
2
2
1
1
log
1
C
2
2
x
x
x
x
x
+
+
+
+
+
11 2
8
7
x
x
dx
−
+
∫
is equal to
(A)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
+
−
+
−
+
+
(B)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
+
−
+
+
+ +
−
+
+
(C)
2
2
1 (
4)
8
7
3
2log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
(D)
2
2
1
9
(
4)
8
7
log
4
8
7
C
2
2
x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
INTEGRALS 331
7 7 Definite Integral
In the previous sections, we have studied about the indefinite integrals and discussed
few methods of finding them including integrals of some special functions |
1 | 3797-3800 | 2
1
+x dx
∫
is equal to
(A)
(
)
2
2
1
1
log
1
C
2
2
x
x
x
x
+
+
+
+
+
(B)
2 23
2(1
)
C
3
+x
+
(C)
2 23
2
(1
)
C
3 x
+x
+
(D)
2
2
2
2
1
1
log
1
C
2
2
x
x
x
x
x
+
+
+
+
+
11 2
8
7
x
x
dx
−
+
∫
is equal to
(A)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
+
−
+
−
+
+
(B)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
+
−
+
+
+ +
−
+
+
(C)
2
2
1 (
4)
8
7
3
2log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
(D)
2
2
1
9
(
4)
8
7
log
4
8
7
C
2
2
x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
INTEGRALS 331
7 7 Definite Integral
In the previous sections, we have studied about the indefinite integrals and discussed
few methods of finding them including integrals of some special functions In this
section, we shall study what is called definite integral of a function |
1 | 3798-3801 | 2
8
7
x
x
dx
−
+
∫
is equal to
(A)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
+
−
+
−
+
+
(B)
2
2
1 (
4)
8
7
9log
4
8
7
C
2 x
x
x
x
x
x
+
−
+
+
+ +
−
+
+
(C)
2
2
1 (
4)
8
7
3
2log
4
8
7
C
2 x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
(D)
2
2
1
9
(
4)
8
7
log
4
8
7
C
2
2
x
x
x
x
x
x
−
−
+
−
−
+
−
+
+
INTEGRALS 331
7 7 Definite Integral
In the previous sections, we have studied about the indefinite integrals and discussed
few methods of finding them including integrals of some special functions In this
section, we shall study what is called definite integral of a function The definite integral
has a unique value |
1 | 3799-3802 | 7 Definite Integral
In the previous sections, we have studied about the indefinite integrals and discussed
few methods of finding them including integrals of some special functions In this
section, we shall study what is called definite integral of a function The definite integral
has a unique value A definite integral is denoted by
( )
b
∫a f x dx
, where a is called the
lower limit of the integral and b is called the upper limit of the integral |
1 | 3800-3803 | In this
section, we shall study what is called definite integral of a function The definite integral
has a unique value A definite integral is denoted by
( )
b
∫a f x dx
, where a is called the
lower limit of the integral and b is called the upper limit of the integral The definite
integral is introduced either as the limit of a sum or if it has an anti derivative F in the
interval [a, b], then its value is the difference between the values of F at the end
points, i |
1 | 3801-3804 | The definite integral
has a unique value A definite integral is denoted by
( )
b
∫a f x dx
, where a is called the
lower limit of the integral and b is called the upper limit of the integral The definite
integral is introduced either as the limit of a sum or if it has an anti derivative F in the
interval [a, b], then its value is the difference between the values of F at the end
points, i e |
1 | 3802-3805 | A definite integral is denoted by
( )
b
∫a f x dx
, where a is called the
lower limit of the integral and b is called the upper limit of the integral The definite
integral is introduced either as the limit of a sum or if it has an anti derivative F in the
interval [a, b], then its value is the difference between the values of F at the end
points, i e , F(b) – F(a) |
1 | 3803-3806 | The definite
integral is introduced either as the limit of a sum or if it has an anti derivative F in the
interval [a, b], then its value is the difference between the values of F at the end
points, i e , F(b) – F(a) Here, we shall consider these two cases separately as discussed
below:
7 |
1 | 3804-3807 | e , F(b) – F(a) Here, we shall consider these two cases separately as discussed
below:
7 7 |
1 | 3805-3808 | , F(b) – F(a) Here, we shall consider these two cases separately as discussed
below:
7 7 1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b] |
1 | 3806-3809 | Here, we shall consider these two cases separately as discussed
below:
7 7 1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b] Assume that all the
values taken by the function are non negative, so the graph of the function is a curve
above the x-axis |
1 | 3807-3810 | 7 1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b] Assume that all the
values taken by the function are non negative, so the graph of the function is a curve
above the x-axis The definite integral
( )
b
∫a f x dx
is the area bounded by the curve y = f (x), the
ordinates x = a, x = b and the x-axis |
1 | 3808-3811 | 1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b] Assume that all the
values taken by the function are non negative, so the graph of the function is a curve
above the x-axis The definite integral
( )
b
∫a f x dx
is the area bounded by the curve y = f (x), the
ordinates x = a, x = b and the x-axis To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7 |
1 | 3809-3812 | Assume that all the
values taken by the function are non negative, so the graph of the function is a curve
above the x-axis The definite integral
( )
b
∫a f x dx
is the area bounded by the curve y = f (x), the
ordinates x = a, x = b and the x-axis To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7 2) |
1 | 3810-3813 | The definite integral
( )
b
∫a f x dx
is the area bounded by the curve y = f (x), the
ordinates x = a, x = b and the x-axis To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7 2) Divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2] , |
1 | 3811-3814 | To evaluate this area, consider the region PRSQP
between this curve, x-axis and the ordinates x = a and x = b (Fig 7 2) Divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2] , ,
[xr – 1, xr], |
1 | 3812-3815 | 2) Divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2] , ,
[xr – 1, xr], , [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h, |
1 | 3813-3816 | Divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2] , ,
[xr – 1, xr], , [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h, , xr = a + rh and
xn = b = a + nh or |
1 | 3814-3817 | ,
[xr – 1, xr], , [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h, , xr = a + rh and
xn = b = a + nh or b
a
n
−h
=
We note that as n → ∞, h → 0 |
1 | 3815-3818 | , [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h, , xr = a + rh and
xn = b = a + nh or b
a
n
−h
=
We note that as n → ∞, h → 0 Fig 7 |
1 | 3816-3819 | , xr = a + rh and
xn = b = a + nh or b
a
n
−h
=
We note that as n → ∞, h → 0 Fig 7 2
O
Y
X
X'
Y'
Q
P
C
M
D
L
S
A
B
R
a = x0 x1 x2
xr-1 xr
x =b
n
y
f x
= ( )
332
MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [xr – 1, xr], r = 1, 2, 3, …, n |
1 | 3817-3820 | b
a
n
−h
=
We note that as n → ∞, h → 0 Fig 7 2
O
Y
X
X'
Y'
Q
P
C
M
D
L
S
A
B
R
a = x0 x1 x2
xr-1 xr
x =b
n
y
f x
= ( )
332
MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [xr – 1, xr], r = 1, 2, 3, …, n From Fig 7 |
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