The integration corresponds to the inverse operation of the diļ¬erentiation:
F
is a primitiveof
f
if
F
(
x
) =
f
(
x
)
dx
⇒
F
′
(
x
) =
f
(
x
)
.
We have
ba
f
(
x
)
dx
=
F
(
b
)
−
F
(
a
)
,
and therefore
x
0
f
(
u
)
du
=
F
(
x
)
−
F
(0)
.
Make sure never to use the same name for the variable of integration and the limit of theintegral.
From the linearity of the diļ¬erentiation, integrals have the following properties:
•
ab
f
(
x
)
dx
=
−
ba
f
(
x
)
dx
•
ca
f
(
x
)
dx
=
ba
f
(
x
)
dx
+
cb
f
(
x
)
dx
•
ba
[
c
1
f
1
(
x
) +
c
2
f
2
(
x
)]
dx
=
c
1
ba
f
1
(
x
)
dx
+
c
2
ba
f
2
(
x
)
dx
, where
c
1
,c
2
are constants.12
abdxxf(x )
k k
Figure 6: Riemann deļ¬nition of the integral
3.1 Interpretation of the integral
As explained on ļ¬g.(6), the Riemann deļ¬nition of
ba
f
(
x
)
dx
,
b > a
, corresponds to thesurface area between the line
y
=
f
(
x
) and the straight line
y
= 0, from
x
=
a
to
x
=
b
.Indeed, this area can be seen as the sum of the inļ¬nitesimal areas
dx
×
f
(
x
), and we have
ba
f
(
x
)
dx
= lim
n
→∞
b
−
an
n
−
1
k
=0
f
(
x
k
)
where
x
k
=
a
+
kb
−
an, k
= 0
,...,n
−
1
,
Equivalence with the deļ¬nition based on the derivative
. We show here that theRiemann deļ¬nition of the integral, as a surface area, is equivalent to the deļ¬nition givenpreviously. From the Riemann interpretation, the quantity
F
(
x
) =
xa
du f
(
u
)corresponds to the surface area of between the lines
y
=
f
(
x
) and
y
= 0, from
a
to
x
. Theintegral from
a
to
x
+ ∆
x
is then
F
(
x
+ ∆
x
) =
x
+∆
xa
du f
(
u
)
,
13
and the derivative of the function
F
is
F
′
(
x
) = lim
∆
x
→
0
1∆
x
x
+∆
xa
du f
(
u
)
−
xa
du f
(
u
)
= lim
∆
x
→
0
1∆
x
x
+∆
xx
du f
(
u
)
.
The latter expression corresponds to the surface area between the lines
y
=
f
(
x
) and
y
= 0from
x
to
x
+ ∆
x
, which is equal to ∆
x
×
f
(
x
)+ higher powers in ∆
x
. As a consequence,we obtain the expected result:
F
′
(
x
) = lim
∆
x
→
0
1∆
x
∆
xf
(
x
) + (∆
x
)
2
···
=
f
(
x
)
.
As a consequence of this interpretation of the integral, if two functions
f,g
satisfy
f
(
x
)
≤
g
(
x
) for
a
≤
x
≤
b,
then
ba
f
(
x
)
dx
≤
ba
g
(
x
)
dx.
3.2 Integration by part
The derivative of the product of two functions
f,g
is (
fg
)
′
=
f
′
g
+
fg
′
, such that we obtain,after integration
f
(
x
)
g
(
x
) =
f
′
(
x
)
g
(
x
)
dx
+
f
(
x
)
g
′
(
x
)
dx,
which can be helpful to calculate one of the integrals on the right hand side, if we knowthe other:
ba
f
′
(
x
)
g
(
x
)
dx
= [
f
(
x
)
g
(
x
)]
ba
−
ba
f
(
x
)
g
′
(
x
)
dx
Example
Integration by part is very useful for the integration of trigonometric functionsmultiplied by power law functions, as
dx x
cos
x
=
dx x
(sin
x
)
′
=
x
sin
x
−
dx
sin
x
=
x
sin
x
+ cos
x
+
c,
where
c
is a constant.
3.3 Change of variable
Suppose that one can write
x
=
g
(
u
), where
u
represents another variable with which theintegral can be calculated. We have then
dx
=
g
′
(
u
)
du
and
ba
f
(
x
)
dx
=
g
−
1
(
b
)
g
−
1
(
a
)
f
(
g
(
u
))
g
′
(
u
)
du,
14
where
g
−
1
represents the inverse function of
g
:
x
=
g
(
u
)
⇔
u
=
g
−
1
(
x
). For the change of variable to be consistent, one must make sure that there is a one-to-one relation between
x
and
u
in the interval [
a,b
].
Example
In the following integral, one makes the change of variable
u
= sin
Ļ
,for 0
≤
Ļ
≤
Ļ/
2:
10
du
√
1
−
u
2
=
Ļ/
20
cos
Ļ dĻ
1
−
sin
2
Ļ
=
Ļ/
20
dĻ
=
Ļ
2
.
Note that, in the interval [0
,Ļ/
2], we have
cos
2
Ļ
=
|
cos
Ļ
|
= cos
Ļ,
since cos
Ļ >
0.
3.4 Improper integrals
The domain of integration of an integral might either contain a singular point, wherethe function to integrate is not deļ¬ned, or might not be bounded. In both cases, thecorresponding integral is said to be convergent if the result of the integration is ļ¬nite, anddivergent if the result of the integration is inļ¬nite. We describe here this situation for theintegration of power law functions.
Case of a non-compact domain of integration
We ļ¬rst show that the integral
I
1
=
∞
1
dxx
diverges. For this, one can see on a graph that
I
1
>
∞
n
=2
1
n,
and we show, that the sum of the inverse of the integers is divergent.
Proof - from the 14th century!
The sum of the inverses of integers, up to 2
N
, can be written:
2
N
n
=1
1
n
= 1 +
12
+
13+14
+
15+16+17+18
+
19+
···
+116
+
···
and satisļ¬es
2
N
n
=1
1
n>
1 +
12
+
14+14
+
18+18+18+18
+
116+
···
+116
+
···
15
The sum in each bracket is equal to 1/2, and there are
N
bracket, such that
2
N
n
=1
1
n>
1 +
N
2
,
which shows that the sum goes to inļ¬nity when
N
goes to inļ¬nity.As a consequence,
∞
1
dxx
is
divergent
.
Consider now the integral, for
a
= 1,
I
a
=
∞
1
dxx
a
= lim
x
→∞
x
1
−
a
−
11
−
a
As can be seen, the result depends on
a
:
•
if
a >
1 then
I
a
= 1
/
(
a
−
1) is ļ¬nite;
•
if
a <
1 then
I
a
= +
∞
.Since the integral
I
a
also diverges for
a
= 1, it converges only for
a >
1.
Case of a singular point
Consider the integral, for
b
= 1,
J
b
=
10
dxx
b
= lim
x
→
0
1
−
x
1
−
b
1
−
b
As can be seen, the result depends on the power
b
:
•
if
b <
1 then
J
b
= 1
/
(1
−
b
) is ļ¬nite;
•
if
b >
1 then
J
b
= +
∞
.The integral
J
b
also diverges for
b
= 1 (the surface area is the same as the previous case,with a non-compact domain of integration), it therefore converges only for
b <
1. Ingeneral, we have:
1
z
dx
(
x
−
z
)
b
is
convergent if
b <
1divergent if
b
≥
1
Example
Consider the integral
∞
1
dx
(
x
−
1)
b
(2
x
+ 3)
a
16
•
at
x
= 1: the integrand is equivalent to 5
−
a
/
(
x
−
1)
b
, such that there is convergenceif
b <
1;
•
at
x
=
∞
: the integrand is equivalent to 2
−
a
/x
a
+
b
, such that there is convergence if
a
+
b >
1;As a consequence, the integral is convergent only if
b <
1 and
a
+
b >
1 simultaneously.17
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