Download Block 5 Stochastic & Dynamic Systems Lesson 14 – Integral Calculus

Block 5 Stochastic &
Dynamic Systems
Lesson 14 – Integral Calculus
The World is now a nonlinear,
dynamic, and uncertain place.
Block 5 - The End of the
Road
 Lesson 14 - Integral Calculus
 Lesson 15 – Stochastic (Probability) Models
 Lesson 16 – Differential Equations
 Lesson 17 – Dynamic Models

f ( x)dx
Integral Calculus
on a silver platter
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.
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
2
3
3x
dx

x
C

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
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
n
C
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
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
3
.2 x
.2
7 x 5e
2x


 10 ln x 
 25 x  C
3
.2
.2
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 x3/ 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 x5/ 2  19, 000
Integrating au , a > 0
 a du  ?
u
let a  eln a
 a du    e 
ln a u
u

du   e
au
e
using  e du 
C
a
au
 ln a u

 ln a u
e
du 
C
ln a
Tricks of the Trade
It’s Magic
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
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
u
 u du  n  1
n
3/ 2
1 2
  x  5  C
3
du
u   x  5  and
 2x
dx
2
n 1
3/ 2
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
 x2  5 dx  x2  5  2 xdx
du
2
where u  x  5;
 2x
dx
1
2
du

ln
u

ln
x
  5  C
u
Integration by Parts
derived from the
product rule for derivatives
Slick Harry
is trying to sell
an ENM student
a simple
integration formula.
 u dv  uv   v du
x
xe
 dx  ?
let u  x; du  dx and dv  e dx; v   e dx  e
x
x
x
x
x
x
x
x
xe
dx

xe

e
dx

xe

e

e
 x  1  C


x
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
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.
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
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
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
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
x2
x2
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.
The Definite Integral
b
 f ( x)dx  F (b)  F (a) where F '( x)  f ( x)
a
Areas under the curve
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
Area under the curve
f(x)
x
x
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
Fundamental Theorem
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
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
Area between Curves
Find the area bounded by y = 4 – 4x2 and
y = x2 - 1
y1 = 4 – 4x2
y1 - y2 = 5 – 5x2

1
1
(5  5 x ) dx 
(-1, 0)
20
3
(1, 0)
y2 = x 2 - 1
Improper Integrals

r
 f ( x) dx  lim  f ( x) dx
a
r 
a
b

b
f ( x) dx  lim
r 




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.
Otherwise it is
divergent, right?
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 
 lim  2    0  
r  2r
2
2 2

1
Let’s do another one…

 ke
0
x
r
dx  lim  ke dx  lim ke
r 
x
0
 lim  ke  k   k
r 
r
An engineering professor
caught doing some improper
integration.
r 
x r
0
The Engineers Little Table of
Improper Definite Integrals

1.  e
 ax
0

2.  xe
 ax
0

1
dx 
a
1
dx  2
a
3.  x e
n  ax
0

4.  x e
0
2  x2
A little table for engineers
n!
dx  n 1
a
dx 

4
Some Applications
Taking it to the limit…
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
6
2
 10t
12
6
  4 144   120   4  36   60   492
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
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
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? (sim-lc 18e6
20 71)
20
.5059 20
T ( x)   18 i
10
.4941
18 x
di 
.5059
10
18
.5059
.5059


20
 10   $47.903 million

.5059
ln(71/100)
47.903
b
V ( x) 
 $4.790 million
ln 2
10
 .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
1
x
27  1 26
1
2
y
x dx 


4

3 1 1
6
6
3
 3 2  1
3
3
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


 400q 

100  2
3
0
2
3
 104  369 210 
1  369(104 ) 2.1(106 )
2


 400(10 )  0  

 4


100 
2
3
3

 100  2
$11,050
 100 184.5  74  11, 050 or
 $.01105
6
10
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 2 
 60  720t  dt  12  60t 

2

0
2
60
 60 
1
 12    6  720     60 
 30
2
 12 
 12 
60
t
Annuities
Let A = the present value of a continuous
annuity at an annual rate r (compounded
continuously) for T years if a payment at time t
is at the rate of f(t) per year. Then
T
A   f (t )e dt
 rt
0
A is the present value of a continuous income stream
Annuity Example
Determine the present value of a continuous annuity
at an annual rate of 8% for 10 years if the payment at
time t is at the rate of 1000t dollars per year.
10
A   (1000t )e
.08t
0
10
dt  1000  te
.08t
dt  1000
0
 e.08(10)
1 
 1000 
.08(10)  1 
2 
2
 .08 
  .08 
 1000  126.37377  156.25  $29,876
ax
e
ax
xe
 dx  a 2  ax  1
e
10
.08t
 .08
2
 .08t  1
0
Annuities
Let S = the accumulated amount of a continuous
annuity at an annual rate r (compounded continuously)
for T years if a payment at time t is at the rate of f(t) per
year. Then
T
S   f (t )e
0
r (T  t )
dt
Back to the example
10
S   1000t  e
0
.08(10 t )
10
dt  1000e
.8
 te
.08t
dt
0
10
 e .08t

 2225.541 
.08t  1 
2 
  .08 
 0
 e .8
1 
 2225.541 
.8  1 
2 
2
 .08 
  .08 
 2225.541 126.3737  156.25  $66, 491
More of that darn example
Recall continuous compounding
S  Pe
rt
.0810 
S  $29,876e
 $66, 490
Iterated Integrals
Evaluate
1
1
0
0

1

0
3 y 2 x3dxdy
x2
0
( x 2  xy  y 2 )dydx
Double Integral
Evaluate the integral over R where R is the
triangle formed by y = x, y = 0, x = 1.
 ( x  y )dA  
2
2
1 x

0 0
R
( x  y )dydx  
2
2
1 1

0 y
( x 2  y 2 )dxdy