Block 5 Stochastic &
Dynamic Systems
Lesson 16 – Methods of Integral
Calculus
The World is now a
nonlinear, dynamic, and
uncertain place.
1
Block 5 - The End of the
Road



Lesson 16 - Integral Calculus
Lesson 17 – Stochastic (Probability) Models
Lesson 18 – Differential Equations & Dynamic
Models
 f ( x)dx
Integral Calculus
on a silver platter
2
Integration
If F(x) is a function whose derivative F’(x) = f(x), then
F(x) is called the integral of f(x)
For example, F(x) = x3 is an integral of f(x) = 3x2
Note also that G(x) = x3 + 5 and H(x) = x3 – 6 are also
integrals of f(x)
I like to call F(x) the
antiderivative.
3
Indefinite Integral
The indefinite integral of f(x), denoted by
 f ( x)dx  F ( x)  C
where C is an arbitrary constant is the most general
integral of f(x)
The indefinite integral of f(x) = 3x2 is
 3x dx  x
2
3
C
4
A Strategy
Gosh, it seems
so simple.
 f ( x)dx  F ( x)  C
the integrand
First I guess at a
function whose
derivative is f(x) and
then I add a constant
of integration.
…or use a table of integrals
5
The top five
n 1
x
1.  x n dx 
 C ; n  1
n 1
1
2.  dx  ln x  C
x
3.  a dx  ax  C
4.  e ax
e ax
dx 
C
a
a  bx 

5.   a  bx  dx 
 n  1 b
n 1
n
C
6
Basic Rules of Integration
1.  c f ( x)dx  c  f ( x)dx
2.   f ( x)  g ( x)  dx   f ( x) dx   g ( x) dx
3.   af ( x)  bg ( x)  dx  a  f ( x) dx  b  g ( x) dx
7
The top four and the basic rules
in action…
10
 2

.2 x
1.2
  7 x  5e  x  2 x  25  dx
1
2
.2 x
 7  x dx  5 e dx  10 dx  2 x 1.2 dx  25 dx
x
7 x3 5e.2 x
2 x .2


 10 ln x 
 25 x  C
3
.2
.2
8
Initial Conditions
The rate at which annual income (y) changes with
respect to years of education (x) is given by
dy
 100 x3/ 2 ; 4  x  16
dx
where y = 28,720 when x = 9. Find y.
y   100 x 3/ 2 dx  100
28, 720  40  9 
5/ 2
x
3/ 2  1
 3/ 2  1
 C  40 x 5 / 2  C
 C  9720  C
C  19, 000
y  40 x 5 / 2  19, 000
9
Integrating au , a > 0
u
a
 du  ?
A trick
let a  e
ln a
 a du    e 
ln a u
u

du   e
 ln a u

 ln a u
e
du 
C
ln a
au
e
using  e du 
C
a
au
10
More Tricks of the Trade
It’s Magic
11
Use some algebra
2
4
3


2
2
y
y
2
y

2
3
y
y

dy

y
  3     3  dy  4  9  C

 2 x  1 x  3 dx  1
6

x
3
 1
x2
3
2
x
5
x
x
2
2 x  5 x  3 dx  
 C


6
9 12 2
2
 x3 1 
x
1
2
dx    2  2  dx    x  x  dx    C
x 
2 x
x
12
Adjusting for “du” – method of
substitution
1/ 2
1
2
 x x  5 dx  2  2 x  x  5 dx 
2
1/ 2
1
1  x  5
2
   x  5   2 xdx  
2
2
3/ 2
2
3/ 2
3/ 2
1 2
  x  5  C
3
du
u   x  5  and
 2x
dx
2
n 1
u
 u du  n  1
n
13
More du’s
3 x2
 3xe dx  2  e  2 xdx
2 du
where u  x ;
 2x
dx
3 u
3 u
e du  e

2
2
3 x2
 e C
2
x2
2x
1
 x 2  5 dx  x 2  5  2 xdx 
du
2
where u  x  5;
 2x
dx
1
2
du

ln
u

ln
x
  5  C
u
14
Integration by Parts
derived from the
product rule for derivatives
 u dv  uv   v du
Slick Harry
is trying to sell
an ENM student
a simple
integration formula.
 xe dx  ?
x
let u  x; du  dx and dv  e x dx; v   e x dx  e x
 xe dx  xe   e dx  xe
x
x
x
x
 e  e  x  1  C
x
x
15
Another one?
 u dv  uv   v du
 ln y dy  ?
dy
let u  ln y ; du  and v  y; dv  dy
y
 dy 
 ln y dy   ln y  y   y  y   y ln y  y  C
 y  ln y  1  C
16
Integration by Tables
A favorite integration formula of engineering students is:
dx
x 1
px


ln
a

be


 a  be px a pa
dx
x
1
.1x
find : 


ln
5

2
e
C


.1x
5  2e
5 .1 5
There is no shame in
using a Table of
Integrals.
17
Check it out!
.1 x



2
e
.1

d x
1
1
.1 x
ln  5  2e   C   
 
.1 x
dx  5 .1 5 
5
.5
5

2
e




.5  5  2e .1x    5  .2e .1x 
.5  5   5  2e .1x 
2.5
1


.1 x
.1 x
5

2
e
2.5
5

2
e
 

dx
x
1
.1x
find : 


ln
5

2
e
C


.1x
5  2e
5 .1 5
18
Another Table Problem
find  7 x 2 ln  4 x  dx
n 1
n 1
u
ln
u
u
Table :  u n ln u du 

C
2
n 1
 n  1
set n  2, u  4 x, du  4dx
7
2
 7 x ln  4 x  dx  43   4 x  ln  4 x   4dx 
3
3

7  4 x  ln  4 x   4 x  
1
3  ln  4 x 


 C

  C  7x 
64 
3
9 
9
 3
2
19
An Engineer’s Favorite Table
ax
e
ax
1.  xe dx  2  ax  1
a
dx
ex
2. 
 ln
x
1 e
1  ex
3.  ln x dx  x ln x  x
2
2
x
x
4.  x ln x dx  ln x 
2
4
no, this one
The motivated student may wish to verify the above by showing that
the derivatives of the right hand functions results in the corresponding integrands.
20
b
The Definite Integral
 f ( x)dx  F (b)  F (a) where F '( x)  f ( x)
a
Areas under the curve
21
Definite Integral
Given a function f(x) that is continuous on the interval
[a,b] we divide the interval into n subintervals of equal
width, x, and from each interval choose a point,
xi*. Then the definite integral of f(x) from a to b is
22
Area under the curve
f(x)
x
x
23
The Fundamental Theorem of
Calculus
Let f be a continuous real-valued function defined
on a closed interval [a, b]. Let F be a function such that
for all x in [a, b]
then
Several engineering management
students informally meeting after class
to discuss the implications of the
fundamental theorem
24
Fundamental Theorem
25
Evaluating a definite integral
2
 4 x dx  F (2)  F (0) 
2
0
4  23 
3
32
0 
3
3
4
x
where F ( x)   4 x 2 dx 
3
2
  x  1
2
x  1

dx 
0
33 1 26
  
3 3 3
3
3 2
0
2  1


3
3
0  1


3
3
26
The Area under a curve
The area under the curve of a probability density
function over its entire domain is always equal to
one. Verify that the following function is a probability
density function:
2
3t
f (t )  9 , 0  t  1000
10
1000

0
2
3
3t
t
dt  9
9
10
10
1000
0
10 


3 3
10
9
1
27
Improper Integrals

r
 f ( x) dx  lim  f ( x) dx
a
r 
a
b

b
f ( x) dx  lim
r 




Otherwise it is
divergent, right?
0
f ( x) dx 


 f ( x) dx
r

f ( x) dx   f ( x) dx
0
and if the limit exists,
the integral is said to be
convergent.
28
Example – an Improper
Integral

r
2 r
1
1
x
dx  lim 
3
1 x3 dx  lim

r 
r 
x
2
1
1
1 1
 1 1 
 lim  2    0  
r  2r
2
2 2

29
Let’s do another one…

 ke
0
x
r
dx  lim  ke dx  lim ke
r 
x
0
r 
x r
0
 lim  ke  k   k
r 
r
An engineering professor
caught doing some improper
integration.
30
The Engineers Little Table of
Improper Definite Integrals

1.  e
 ax
0

2.  xe
 ax
0

1
dx 
a
3.  x e
n  ax
0

4.  x e
0
A little table for engineers
1
dx  2
a
2  x2
n!
dx  n 1
a
dx 

4
31
Some Modeling with Integrals
Taking it to the limit…
32
The Crime Rate
The total number of crimes is increasing at the rate
of 8t + 10 where t = months from the start of the
year. How many crimes will be committed during the
last 6 months of the year?
given :
d  crime
dt
 8t  10
12
 8t  10  dt  4t
2
 10t
12
6
  4 144   120   4  36   60   492
6
33
Learning Curves
Cumulative Cost
Y (i )  ai
b
x
hours to produce ith unit
x
T ( x)   Y (i )   ai
i 1
T ( x)
V ( x) 
x
i 1
b
cumulative direct labor
hrs to produce x units
average unit hours to
produce x units
34
Learning Curves
Approximate Cumulative Cost
x
b 1
x
ax
T ( x)   Y (i ) di   a i di 
b 1
0
0
b
b 1
b
ax
ax
V ( x) 

(b  1) x b  1
35
Learning Curves - example
Production of the first 10 F-222’s, the Air Force’s new
steam driven fighter, resulted in a 71 percent learning
curve in dollar cost where the first aircraft cost $18
million. What will be cost of the second lot of 10 aircraft?
20
T ( x)   18 i
10
.4941
.5059 20
18 x
di 
.5059
10
18
 20.5059  10.5059   $47.903 million

.5059
ln(71/100)
b
47.903
V ( x) 
 $4.790 million
ln 2
10
36
 .4941
The Average of a Function
The average or mean value of a function y = f(x)
over the interval [a,b] is given by:
b
1
y
f ( x)dx

ba a
Find the average of the function y = x2 over the interval
[1,3]:
3
3
1
x
27  1 26
1
2
y
x dx 


4

3 1 1
6
6
3
 3 2  1
3
37
Average profit
An oil company’s profit in dollars for the qth million gallons
sold is given by
P = P(q) = 369q – 2.1q2 – 400
If the company sells 100 million gallons this year, what is
the average profit per gallon sold?
100
  369q  2.1q
0
100
2
 400  dq

100

1  369q 2.1q


400
q

100  2
3
0
2
3
 10 4  369 210 
1  369(104 ) 2.1(106 )
2


 400(10 )  0  

 4


100 
2
3
3

 100  2
 100 184.5  74  11, 050 or
$11,050
 $.01105
6
10
38
An Inventory Problem
Demand for an item is constant over time at the rate of 720 per
year. Whenever the on-hand inventory reaches zero, a shipment of
60 units is received. The inventory holding cost is based upon the
average on-hand inventory.
Let y = 60 – 720t be the on-hand inventory as a function of time
where t is in years. It takes 60/720 = 1/12 yr to go from an
inventory of 60 to 0.
1
y
1/12  0
1/12

0
1/12

720t 
 60  720t  dt  12  60t 

2

0
2
2
60
 60 
1
 12    6  720     60 
 30
2
 12 
 12 
60
t
39

V
f ( x, y, z) x y z
Bring on the Integrals!!!!
An engineering student
searching for answers…
40