Mathematics for Engineering: Calculus
Chapter 3: TECHNIQUES OF INTEGRATION
Department of Mathematics, FPT University
Content
3.1 Integration by Parts
3.6 Numerical Intrgration
3.7 Improper Integrals
TECHNIQUES OF INTEGRATION
3.1
Integration by Parts
In this section, we will learn:
How to integrate complex functions by parts.
INTEGRATION BY PARTS
ò f ( x) g '( x) dx = f ( x) g( x) - ò g( x) f '( x) dx
 Let u = f(x) and v = g(x).
 Then, the differentials are:
du = f’(x) dx and dv = g’(x) dx
Thus, by the Substitution Rule, the formula for
integration by parts becomes:
udv
=
uv
vdu
ò
ò
INTEGRATION BY PARTS
Evaluating both sides of Formula 1 between a and b,
assuming f’ and g’ are continuous, and using the FTC,
we obtain:
ò
b
a
f ( x) g '( x) dx = f ( x) g( x)]a
b
- ò g( x) f '( x) dx
b
a
INTEGRATION BY PARTS
Example 1
Find ∫ x sin x dx
INTEGRATION BY PARTS
Let
u= x
Then, du = dx
Example 1
dv = sin xdx
v = - cos x
Using Formula 2, we have:
INTEGRATION BY PARTS
Example 2
Evaluate ∫ ex sinx dx
 ex does not become simpler when differentiated.
 Neither does sin x become simpler.
INTEGRATION BY PARTS
Example 2
Nevertheless, we try choosing
u = ex and dv = sin x
 Then, du = ex dx and v = – cos x.
INTEGRATION BY PARTS
Example 2
So, integration by parts gives:
e
sin
xdx
=
e
cos
x
+
e
cos
xdx
ò
ò
x
x
x
INTEGRATION BY PARTS
Example 2
The integral we have obtained, ∫ excos x dx,
is no simpler than the original one.
 At least, it’s no more difficult.
 Having had success in the preceding example
integrating by parts twice, we do it again.
INTEGRATION BY PARTS
Example 2
This time, we use
u = ex and dv = cos x dx
Then, du = ex dx, v = sin x, and
e
cos
xdx
=
e
sin
x
e
sin
xdx
ò
ò
x
x
x
INTEGRATION BY PARTS
Example 2
we get:
ò e sin xdx = -e cos x + e sin x
- ò e sin xdx
x
x
x
x
 This can be regarded as an equation to be
solved for the unknown integral.
INTEGRATION BY PARTS
Example 2
Adding to both sides ∫ ex sin x dx,
we obtain:
2ò e sin xdx = -e cos x + e sin x
x
x
x
INTEGRATION BY PARTS
Example 2
Dividing by 2 and adding the constant
of integration, we get:
e
sin
xdx
=
e
(sin
x
cos
x
)
+
C
ò
x
1
2
x
INTEGRATION BY PARTS
Example 3
Suppose f(x) is continuous and
differentiable, f(1)=4 and
1
Find
 xf '( x)dx
0
a
4/5
b
5/4
c
1
d
None of the others
e
-1
1
 f ( x)dx  5
0
INTEGRATION BY PARTS
Example 4
Suppose f(x) is continuous and differentiable,
f(1)=3, f(3)=1 and 3
 xf '( x)dx  13
1
What is the average value of f on the interval
[1,3]?
TECHNIQUES OF INTEGRATION
3.6
Numerical Integration
Numerical Integration
Left endpoint Method
b
 f ( x)dx  x[ f ( x )  f ( x )  ...  f ( x
0
a
1
n 1
)]
Numerical Integration
Right endpoint Method
b
 f ( x)dx  x[ f ( x )  f ( x )  ...  f ( x )]
1
a
2
n
Numerical Integration
Midpoint Method
f(x)
∆x
x1
x2
x3 x n
b
 f ( x)dx  x[ f ( x )  f ( x
1
a
2
)  ...  f ( x n )]
Numerical Integration
Trapezoidal Method
b

a
x
x
f ( x)dx 
[ f ( x0 )  f ( x1 )]  ...  [ f ( xn 1 )  f ( xn )]
2
2
x

[ f ( x0 )  2 f ( x1 )  ...  2 f ( xn 1 )  f ( xn )]
2
Numerical Integration
Simpson Method
x
x
a f ( x)dx  3 [ f ( x0 )  4 f ( x1 )  f ( x2 )]  ...  3 [ f ( xn2 )  4 f ( xn1 )  f ( xn )]
b
x
 [ f ( x0 )  4 f ( x1 )  2 f ( x2 )  ...  4 f ( xn 1 )  f ( xn )]
3
Numerical Integration
Example
Approximate the integral
with n = 8, using:
a. Left/Right endpoints
b. Midpoints
c. Trapezoidal method
d. Simpson method
ò
2
1
(1/ x) dx
Numerical Integration
Estimate error for Midpoint and Trapezoidal method
• Suppose | f’’(x) | ≤ K for a ≤ x ≤ b.
• If ET and EM are the errors in the Trapezoidal and
Midpoint Rules, then
K (b - a)
ET £
12n2
3
and
K (b - a)
EM £
24n2
3
Numerical Integration
Estimate error for Simpson method
• Suppose | f(4)(x) | ≤ K for a ≤ x ≤ b.
• If ES is the error in the Simpson method, then
K (b - a)5
Es £
4
180n
Numerical Integration
Example
How large should we take n in order to guarantee
that the Trapezoidal, Midpoint Rule, Simpson rule
approximations for
ò
2
1
(1/ x) dx
are accurate to within 0.0001?
Numerical Integration
| f’’(x) | ≤ 2 for 1 ≤ x ≤ 2
Accuracy to within 0.0001 means that error < 0.0001
3
2(1)
Trapezoidal: Choose smallest n so that:
< 0.0001
2
12n
 n = 41
3
2(1)
Midpoint:
< 0.0001  n = 30
24n2
f
Simpson:

(4)
24
( x) = 5 £ 24
x
24(1)5
< 0.0001
4
180n

n=8
TECHNIQUES OF INTEGRATION
3.7
Improper Integrals
In this section, we will learn:
How to solve definite integrals
where the interval is infinite and
where the function has an infinite discontinuity.
IMPROPER INTEGRAL OF TYPE 1
Definition
If
ò
t
a
f ( x) dx exists for every number t ≥ a, then
¥
t
¥ f ( x) dx = lim ¥ f ( x) dx
a
t ®¥
a
provided this limit exists (as a finite number).
IMPROPER INTEGRAL OF TYPE 1
Definition
If
ò
b
t
f ( x) dx exists for every number t ≤ a, then

b

b
f ( x) dx  lim  f ( x) dx
t  t
provided this limit exists (as a finite number).
IMPROPER INTEGRAL OF TYPE 1
CONVERGENT AND DIVERGENT
The improper integrals
are called:
ò
¥
a
Definition
f ( x) dx and
ò
b
-¥
f ( x) dx
 Convergent if the corresponding limit exists.
 Divergent if the limit does not exist.
IMPROPER INTEGRAL OF TYPE 1
Definition
If both
ò
¥
a
f ( x) dx and ò
a
-¥
f ( x) dx are convergent,
then we define:
ò
¥
-¥
a
¥
-¥
a
f ( x) dx = ò f ( x) dx + ò f ( x) dx
 Here, any real number a can be used.
IMPROPER INTEGRAL OF TYPE 1
Example 1
For what values of p is the integral
ò
¥
1
1
convergent?
dx
p
x
 Convergent if p > 1
 Divergent if p ≤ 1
IMPROPER INTEGRAL OF TYPE 1
Example 2
• Investigate the convergence of the improper integrals:
• (a)
0
x e
2
x3
dx

• (b)

x e
2

x3
dx
IMPROPER INTEGRAL OF TYPE 2
Definition
If f is continuous on [a, b) and is discontinuous at b, then
b
t
a
t ®b- a
¥ f ( x) dx = lim ¥ f ( x) dx
if this limit exists (as a finite number).
IMPROPER INTEGRAL OF TYPE 2
Definition
If f is continuous on (a, b] and is discontinuous at a, then
b
b
¥ f ( x) dx = lim ¥ f ( x) dx
a
t ® a+ t
if this limit exists (as a finite number).
IMPROPER INTEGRAL OF TYPE 2
Definition
The improper integral
ò
b
a
f ( x) dx is called:
 Convergent if the corresponding limit exists.
 Divergent if the limit does not exist.
IMPROPER INTEGRAL OF TYPE 2
Definition
If f has a discontinuity at c, where a < c < b, and
both
ò
c
a
f ( x) dx and
ò
b
c
f ( x) dx are convergent, then
we define:
ò
b
a
f ( x) dx = ò f ( x) dx + ò f ( x) dx
c
b
a
c
IMPROPER INTEGRAL OF TYPE 2
Example 1
Let b> 2. Investigate the convergence of the improper integrals:
b
dx
a ( x  a) p
Answer: diverges if p
 1 and converges if p<1.
IMPROPER INTEGRAL OF TYPE 2
Example 2
Investigate the convergence of the improper integral:
3

0
dx
x 1
COMPARISON THEOREM
Suppose f and g are continuous functions with f(x) ≥ g(x) ≥ 0
for x ≥ a.
a. If
ò
¥
a
f ( x) dxis convergent, then
ò
¥
a
g( x) dx
is convergent.
b. If
ò
¥
a
¥
g( x) dx is divergent, then ò f ( x) dx
is divergent.
Note: A similar theorem is true for Type II integrals
a
COMPARISON THEOREM
Example

Does I 
| cos( x ) | dx
converge?
2
1
x
We have,
| cos( x ) |
1
0
 2
2
x
x

dx
and  2 converges
x
1
→ I converges.
COMPARISON THEOREM
Example
Investigate the convergence of the improper integrals

(a)


1

(b)
1 
(c)
x2
dx
x
x2
dx
3
x
e
0
 x2
dx
Thanks