Civil Engineering
Computation
Numerical Integration
I always wanted to be somebody, but I
should have been more specific.
Jane Wagner
Contents
  Basic
approach
  Trapezoid rule
  Truncation errors
  Simpson s 1/3 rule
  Simpson s 3/8 rule
  Richardson extrapolation
  Romberg integration
  Integration of data
2
Numerical Methods of Integration
1
The Problem

Consider definite integration
b
∫ f ( x)dx
a

Basic idea: integral represents area under the
curve

Approximate integral by approximating area
Numerical Methods of Integration
3
Geometry of Integration Area
4
Numerical Methods of Integration
2
Rectangular Approximation

If n rectangular areas are placed under
the curve:
Δx = (b-a) / n

5
Individual area defined by Ai = f(xi)* Δx
Numerical Methods of Integration
Trapezoid Rule
6

But rectangles miss curved area tops

So add triangle to top of each rectangle
Numerical Methods of Integration
3

Then Ai = 0.5*(f(xi+1) + f(xi))* Δx

Fill area under curve with these areas and add

For example if we wanted to find the area under
the curve shown below bounded by the limits
x=3 and x=6
y = 3.2x 2 + 2.5x + 3
7
Numerical Methods of Integration
a b n Δx 3 6 4 0.75 x f(x) 3 39.3 3.75 57.375 4.5 79.05 5.25 104.325 6 133.2 y = 3.2x 2 + 2.5x + 3
8
Numerical Methods of Integration
4
Ai = 0.5*(f(xi+1) + f(xi))* Δx
a b n Δx 3 6 4 0.75 x 3 3.75 4.5 5.25 6 f(x) 39.3 57.375 79.05 104.325 133.2 Ai 36.253125 51.159375 68.765625 89.071875 y = 3.2x 2 + 2.5x + 3
9
Numerical Methods of Integration
Ai = 0.5*(f(xi+1) + f(xi))* Δx
a b n Δx 3 6 4 0.75 x 3 3.75 f(x) 39.3 57.375 Ai 36.253125 51.159375 4.5 5.25 6 79.05 104.325 133.2 68.765625 89.071875 245.25 The area under the curve is the
sum of the areas under all the
segments (delta x).
y = 3.2x 2 + 2.5x + 3
10
Numerical Methods of Integration
5
  If
we consider n rectangles, we can
develop a general formula for using
the trapezoid rule.
f ( a ) + f ( a + !x )
!x
2
f ( a + !x ) + f ( a + 2!x )
A2 =
!x
2
f ( a + 2!x ) + f ( a + 3!x )
A3 =
!x
2
!
A1 =
An =
11
f ( a + ( n "1) !x ) + f ( a + n!x )
2
!x
Numerical Methods of Integration
  And
with some manipulation
f ( a ) + f ( a + !x )
!x
2
f ( a + !x ) + f ( a + 2!x )
A2 =
!x
2
f ( a + 2!x ) + f ( a + 3!x )
A3 =
!x
2
!
A1 =
An =
12
f ( a + ( n "1) !x ) + f ( b)
2
!x
Numerical Methods of Integration
6
  Summing
up the incremental areas
f ( a ) + f ( a + !x )
f ( a + !x ) + f ( a + 2!x )
!x +
!x
2
2
f ( a + ( n "1) !x ) + f ( b)
f ( a + 2!x ) + f ( a + 3!x )
+
!x +! +
!x
2
2
Area =
13
Numerical Methods of Integration
  Some
of the terms appear twice so we
can rewrite the expression
Area =
f ( a ) + 2 f ( a + !x ) + 2 f ( a + 2!x ) +!2 f ( a + ( n "1) !x ) + f ( b)
2
f ( a ) + f ( b) $ n"1
'
Area =
!x & # f ( a + i!x ))
2
% i=1
(
14
!x
Numerical Methods of Integration
7
  The
summation expression is very
easy to program using a loop
f (a ) + 2 f (a + Δx ) + 2 f (a + 2Δx ) +  2 f (a + (n − 1)Δx ) + f (b )
Δx
2
a 3 f (a ) + f (b )
⎡ n −1
⎤
b 6 Area =
Δx + ⎢∑ f (a + iΔx )⎥ Δx
n 4 2
⎣ i =1
⎦
Δx 0.75 Area =
x 3 3.75 4.5 5.25 6 f(x) 39.3 57.375 79.05 104.325 133.2 f(a)+f(b) /2 Sum 172.5 86.25 240.75 327 245.25 Ai 36.253125 51.159375 68.765625 89.071875 245.25 15
Numerical Methods of Integration
<mes delta x Two forms of 2nd Derivative

Estimate the 2nd derivative in 2 ways:
1. maximum value, in magnitude
2. average value:
f"avg = [f'(b) - f'(a)] / (b-a)
(This can be justified by applying the Mean
Value Theorem to the second derivative
or by integrating it over the interval)
16
Numerical Methods of Integration
8
Maximum Error
f ( x ) = 3.2x 2 + 2.5x + 3
f ! ( x ) = 6.4x + 2.5
f !! ( x ) = 6.4
For 3 <= x <= 6 max f"(x) is the same
across the interval
f"(3) =6.4
If the second derivative was a function
rather than a number, you would need to
find the x value where the second
derivative was a maximum.
17
Numerical Methods of Integration
Maximum Error
( "x )2 * ( b ! a) * f ## ( x )
=!
ETrun
12
where x is a value between a and b such that f ##(x) is a maximum
ETrun
( 0.75)2 * ( 6 ! 3) * 6.4 = !0.9
=!
f ( x ) = 3.2x 2 + 2.5x + 3
f ! ( x ) = 6.4x + 2.5
f !! ( x ) = 6.4
12
Minus sign means estimate greater than
correct value
  Correct value is 244.35 and our estimate
is 245.25

18
Numerical Methods of Integration
9
Average Error
If we were not so lucky with the second
derivative, we could use the average of
the second derivative across the interval
This would be the change in the first
derivative across the interval divided by
the length of the interval
19
Numerical Methods of Integration
Average Error
So the calculation for the average 2nd
derivative would be
f !! ( x )avg =
20
f ! ( b) " f ! ( a)
b"a
Numerical Methods of Integration
10
Average Error
And then you can substitute this
expression for the second derivative into
the expression for the truncation error
ETrun = !
f ## ( x )avg
ETrun = !
ETrun = !
ETrun = !
21
( "x )2 * ( b ! a) * f ## ( x )
12
f # ( b) ! f # ( a)
=
b!a
f# b ! f# a
( "x )2 * ( b ! a) * ( ) ( )
b!a
( "x )2
12
12
( f # ( b ) ! f # ( a ))
( 0.75)2 * ( 40.9 ! 21.7) = !2.4
12
Numerical Methods of Integration
Average Error
The value found for the this typically is less
conservative than that found using the
maximum second derivative over the
interval
ETrun = !
f ## ( x )avg
ETrun = !
ETrun = !
ETrun = !
22
( "x )2 * ( b ! a) * f ## ( x )
12
f # ( b) ! f # ( a)
=
b!a
f# b ! f# a
( "x )2 * ( b ! a) * ( ) ( )
b!a
( "x )2
12
12
( f # ( b ) ! f # ( a ))
( 0.75)2 * ( 40.9 ! 21.7) = !2.4
12
Numerical Methods of Integration
11
Trapezoid Rule Error Form
Reasonable
  Trapezoid
rule based on connecting
adjacent points with straight line
  Resulting
error formula supports this
—exact for linear f(x) since f"(x) zero
for linear f(x)
  Better
result expected if a quadratic fit
used for adjacent points
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Numerical Methods of Integration
Simpson s 1/3 Rule

I = Δx*[f(a) + 4*f(a+Δx) +2*f(a+2*Δx) +
4*f(a+3*Δx) …+2*f(b-2*Δx) + 4*f(bΔx)+f(b)]/3
First and last function values multiplied
by 1, interior function values multiplied
by 4 and 2, alternating
n must be even for Simpson s 1/3 rule
24
Numerical Methods of Integration
12
Simpson s 1/3 Rule
Truncation Error

ET = 53.28214 – 53.27472 = 0.00742

Estimate Simpson s 1/3 rule truncation error


ETRUN =
-
Δx 4 * (b - a) * f IV (x)
180
Simpson s 1/3 rule was derived with quadratic fit
between adjacent points on the curve
But error formula is exact for a cubic integrand
since fIV(x) would be 0
  Thus have better result than expected

25
Numerical Methods of Integration
Simpson s 3/8 Rule

Use Simpson s 3/8 rule (and 1/3 rule) for
an odd number of intervals
Simpson s 3/8 rule is used when n = 3,
Not any odd number, only 3
  But any odd number 3 or greater can be
broken into 3 + an even number

(b − a) * [ f (a) + 3 * f (a + Δx) + 3 * f (a + 2Δx) + f (b)]
8
26
Numerical Methods of Integration
13
Richardson Extrapolation,
a Clever Idea
  Truncation
error can be reduced by
increasing n
  But increasing n increases number of
calculations and hence round-off error
  Eventually round-off error increase
can overwhelm truncation error
decrease
  Want method to improve accuracy
without increasing n too much
27
Numerical Methods of Integration

Let Δx1 = 2* Δx2 i.e. interval is halved

Solving for C and substituting obtain:
I = [4*I2-I1] / 3
a weighted average of two
evaluations of integral using
trapezoid rule
  I2 is more accurate since uses smaller
Δx
28
Numerical Methods of Integration
14
Richardson Example

With n = 4, Δx = 1 I = 54.95879

With n = 8, Δx= 0.5 I = 53.70091

Then I = [4*53.70091 – 54.95879]/3 =
53.28162

ET = 53.28214 – 53.28162 = 0.00052
a very good result!
29
Numerical Methods of Integration
Romberg Integration

30
Can combine multiple Richardson
evaluations
k=1
k=2
n Δx trapezoid
Richardson
1 4
80.47190
53.19066
2 2
60.01097
53.27473
4 1
54.95879
53.28160
8 0.5 53.70091
Numerical Methods of Integration
15

Richardson results combined further
using:
Ir,k = [16*Ir+1,k-1 – Ir,k-1] / 15
I1,3 = [16*53.27473 – 53.19066]/15 =
53.28034

Generalize the formula as
Ir,k = [4k-1*Ir+1,k-1 – Ir,k-1]/(4k-1 -1)

31
Truncation error changes as Δ2k
Numerical Methods of Integration
16