The integration corresponds to the inverse operation of the differentiation:
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 differentiation, 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 definition of the integral
3.1 Interpretation of the integral
As explained on fig.(6), the Riemann definition 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 infinitesimal 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 definition based on the derivative
. We show here that theRiemann definition of the integral, as a surface area, is equivalent to the definition 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
φ
,for 0
≤
φ
≤
π/
2:
10
du
√
1
−
u
2
=
π/
20
cos
φ dφ
1
−
sin
2
φ
=
π/
20
dφ
=
π
2
.
Note that, in the interval [0
,π/
2], we have
cos
2
φ
=
|
cos
φ
|
= cos
φ,
since cos
φ >
0.
3.4 Improper integrals
The domain of integration of an integral might either contain a singular point, wherethe function to integrate is not defined, or might not be bounded. In both cases, thecorresponding integral is said to be convergent if the result of the integration is finite, anddivergent if the result of the integration is infinite. We describe here this situation for theintegration of power law functions.
Case of a non-compact domain of integration
We first 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 satisfies
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 infinity when
N
goes to infinity.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 finite;
•
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 finite;
•
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
No comments:
Post a Comment