Numerical Integration
AP Calculus
Numerical Integration
* Used when normal definite integration is not possible.
a). When there is no elementary function for the anti-derivative;
i.e.:

1  x dx
3
or

x cos xdx
b). Data is given in tabular or graphical form and it is too much
effort to find the representative function.
I. RIEMANN’S SUMS
REM: Riemann’s Sum uses Rectangles to approximate the
accumulation.

b
a
n
f ( x)dx  Lim f (ci ) xi
n
i 1
A = bh
=> h - Left Endpoints
- Right Endpoints
- Midpoint
The more accurate is the Midpoint Sum
(must remember how to use all three – Left, Right, and Midpoint)
..................................................................
n
b
Midpoint Rule:  f ( x)dx   f  xi  xi1  x
with x   b  a 
a
2
 n 


i 1
 xn 1  xn  
 x1  x2 
 b  a    x0  x1 
A
 f 
  ...  f 
 f 

n
2
2
2

 





In Words: The width of the subinterval times the sum of the heights
AT THE MIDPOINT of each subinterval.
Illustration:
Function:
f ( x)  4 x  x 2

4
0
(4 x  x 2 )dx 
on
[0,4]
with
n=4
Example: Graphical
Find the Average Revenue for the 5 years.
II. TRAPEZOID METHOD:
Uses Trapezoids to fill the regions
rather than rectangles:
REM:
A
1
h(b1  b2 )
2
1 ba
A (
)( f ( x0 )  f ( x1 ))
2 n
1ba
A1  
 ( f ( x0 )  f ( x1 ))
2 n 
1ba
A2  

2 n 
1ba
A3  

2 n 
( f ( x1)  f ( x2 ))
( f ( x2 )  f ( x3 ))
---------------------------------------------------------------
1ba
A 
 ( f ( x0 )  2 f ( x1 )  2 f ( x2 )  f ( x3 ))
2 n 
(Notice this is the average of the Left and Right Riemann's Sums)
Trapezoid Rule:
........................................................................

b
a
 1  b  a 
f ( x ) dx     
  f ( xi )  f ( xi 1 ) 
n 
i 1  2  
n
 1  b  a 
  
 ( f ( x0 )  2 f ( x1 )  2 f ( x2 )  ...  2 f ( xn1 )  f ( xn ))
 2  n 
In Words: One half * width of subinterval * the ( 1 , 2 , 2 , … , 2 , 1 ) pattern
of the heights found at the points of the subinterval.
........................................................................
Illustration: (Trapezoid)
Function:
f ( x)  4 x  x 2

4
0
(4 x  x 2 )dx 
on
[0,4]
with
n=4
Example: Data
The data for the acceleration a(t) of a car from 1 to 15
seconds are given in the table below. If the velocity at t = 0 is
5 ft/sec, which of the following gives the approximate
velocity at t = 15 using the Trapezoidal Rule?
t (sec)
0
3
6
9
12
15
a(t)
(ft/sec2)
4
8
6
9
10
10
A lot is bounded by a stream. and two straight roads that meet at right angles. Use
the Trapezoid Rule to approximate the area of the lot (x and y are measured in
square meters)
III. SIMPSON’S METHOD
Built on:

b
a
( Ax2  Bx  C )dx
The area of the region below a
quadratic function.
REM: Three points are required to write a quadratic equation
since the equation has 3 variables; A,B,C in
y  Ax 2  Bx  C
Therefore, to get the 3 points needed, Simpson’s uses double subintervals to
approximate the accumulation.
............................
THEOREM:
1 ba 
a ( Ax  Bx  C )dx  3  n  ( f ( x1 )  4 f ( x2 )  f ( x3 ))
with n even
b
2
Simpson’s: (cont)
EXTENDED:
1 ba 

 ( f ( x1 )  4 f ( x2 )  f ( x3 ))
3 n 
1ba 
( f ( x3 )  4 f ( x4 )  f ( x5 ))


3  __________________________________________________
n 
1 ba 

 [ f ( x1 )  4 f ( x2 )  2 f ( x3 )  4 f ( x4 )  f ( x5 )]
3
n 
.......................................................
Simpson’s Formula:

b
a
 xi1

2
f ( x)dx     ( Ax  Bx  C )dx 


i 1  xi

1ba 
 
 [ f ( x1 )  4 f ( x2 )  2 f ( x3 )  4 f ( x4 )  f ( x5 )]
3 n 
n
Note the pattern:
1,4,2,4,1
1,4,2,4,2,4,1 etc
Illustration: (Simpson’s)
Function:
f ( x)  4 x  x 2

4
0
(4 x  x 2 )dx 
on
[0,4]
with
n=4
Although the economy is continuously changing, we analyze it with discrete
measurements. The following table records the annual inflation rate as
measured each month for 13 consecutive months. Use Simpson’s Rule with
n = 12 to find the overall inflation rate for the year.
Month
Annual Rate
January
0.04
February
0.04
March
0.05
April
0.06
May
0.05
June
0.04
July
0.04
August
0.05
September
0.04
October
0.06
November
0.06
December
0.05
January
0.05
Example: Graphical - all three
Error:
Approximation gives rise to two questions>>>>
1) How close are we to the actual answer? and
2) How do we get close enough?
Error: MIDPOINT
E
M
n
Error using Midpoint with n partitions
(b  a)
E 
2
24n
M
n
3
 f (c) 
C is an un-findable number in [a,b] but
whose existence is guaranteed; therefore
do “ERROR BOUND”
(b  a)
E 
Mi 
2 
24n
3
M
n
Where Mi  the MAX of
f ( x) on i[a,b]
Error: TRAPEZIOD
E
T
n
Error using Trapezoid with n partitions
(b  a)
E 
2
12n
T
n
3
 f (c) 
C is an un-findable number in [a,b] but
whose existence is guaranteed; therefore
do “ERROR BOUND”
(b  a)
E 
Mi 
2 
12n
3
T
n
Where Mi  the MAX of
f ( x) on i[a,b]
Example: How close are we?
1
Approximate

(1  x 2 ) dx
using Trapezoid Method
0
with 4 intervals and find the Error bound.
Example: How many intervals are required?
1
Approximate
4
5x
 dx
0
to within .
1
1000
using Midpoint Rule
Error: SIMPSONS
E
S
n
Error using Simpson’s with n partitions
(b  a)
E 
4
180n
S
n
5
f
IV
(c ) 
C is an un-findable number in [a,b] but
whose existence is guaranteed; therefore
do “ERROR BOUND”
(b  a)
E 
4
180n
S
n
5
 Mi 
Where Mi  the MAX of
f IV ( x) on i[a,b]
Last Update:
• 02/05/10