1
NUMERICAL INTEGRATION
• Motivation:
Most such integrals cannot be evaluated explicitly. Many others it is often faster to integrate them numerically rather than evaluating them exactly using a complicated antiderivative of f(x)
• Example:
The solution of this integral equation with Matlab is 1/2*2^(1/2)*pi^(1/2)*FresnelS(2^(1/2)/pi^(1/2)*x)
we cannot find this solution analytically by techniques in calculus.
dxx )sin( 2
2
• Methods of Numerical Integration– Trapezoidal Rule’s– 1/3 Simpson’s method– 3/8 Simpson’s method
• Applied in two dimensional domain
Course content
3
Trapezoidal Rule’s
f
fp
4
• Function f approximately by function fp. Then,
where fp is a linear polynomial interpolation, that is
By substitution u=x-x0 we have
where
dxffdx p
)()( bf
ab
axaf
ba
bxf p
)()(2
bfafh
dxffdx p
abh
5
Trapezoidal Rule’s
f
fp
6
• For two interval, we can use summation operation to derive the formula of two interval trapezoidal that is
where
)()(2
)()(2 2110 xfxf
hxfxf
hfdx
202 xx
h
)()(2)(2 210 xfxfxfh
7
Trapezoidal Rule’s
f
fp
8
• Similar to two interval trapezoidal, we can derive three interval trapezoidal formula that is
where
• Thus, for n interval we have
where and
for
)()(2)(2)(2 3210 xfxfxfxfh
fdx
303 xx
h
)()(2)(2
1
10 n
n
ii xfxfxf
hfdx
n
xxh n 0
)(12
)( ''20t
nr xfh
xxE
21 ii
t
xxx
9
1/3 Simpson’s
f
fp
10
• Function f approximately by function fp. Then,
where fp is a quadratic polynomial interpolation, that is
By substitution u=x-x0 we have
where
dxffdx p
)()()( 2
1202
101
2101
200
2010
21 xfxxxx
xxxxxf
xxxx
xxxxxf
xxxx
xxxxf p
)()(4)(3 210 xfxfxfh
dxffdx p
202 xx
h
11
f
fp
1/3 Simpson’s
12
• For 4 subinterval we have
where
• Thus, for n subinterval we have
where and
)()(4)(2)(4)(3 43210 xfxfxfxfxfh
fdx
404 xx
h
)()(2)(4)(3
2
6,4,2
1
5,3,10 n
n
jj
n
ii xfxfxfxf
hfdx
n
xxh n 0
)(180
)( )(40t
ivnr xfh
xxE
13
3/8 Simpson’s
ffp
14
• Similar to 1/3 Simpson’s method, f approximately by function fp where fp is a cubic polynomial interpolation, that is
By substitution u=x-x0 we have
where and
)()(
)()(
3231303
2102
321202
310
1312101
3200
302010
321
xfxxxxxx
xxxxxxxf
xxxxxx
xxxxxx
xfxxxxxx
xxxxxxxf
xxxxxx
xxxxxxf p
)()(3)(3)(8
33210 xfxfxfxf
hdxffdx p
303 xx
h
)(80
3 )(5t
ivr xfhE
15
Numerical Integration in a Two Dimensional Domain
c(x)
d(x)
=a b=0x 1Nx1x Nx
0,0y
2,1y
1,1y
0,1y
Ny ,0
2,0y
1,0y
Ny ,1NNy ,
2,Ny
1,Ny
0,Ny
16
• A double integration in the domain is written as
• The numerical integration of above equation is to reduce to a combination of one-dimensional problems
dxdyyxfIb
a
xd
xc
)(
)(
),(
17
Procedure: • Step 1: Define
So, the solution is
• Step 2: Divided the range of integration [a,b] into
N equispaced intervals with the interval size
So, the grid points will be denoted by
and then we have
)(
)(
),()(xd
xc
dyyxfxG
b
a
dxxGI )(
N
abhx
Nxxx ,,, 10
,),()()(
)(i
i
xd
xc
ii dyyxfxG
18
• Step 3: Divided the domain of integration
into N equispaced intervals with the interval size
So, the grid points denoted by
• Step 4: By Applying numerical integration for one-dimensional (for example the trapezoidal rule) we have
for
N
xcxdh iiy
)()(
)(),( ii xcxd
Niii yyy ,1,0, ,,,
1
1,,0, ),(),(2),(
2)(
N
jNiijiiii
yi yxfyxfyxf
hxG
Ni ,,2,1,0
19
• Step 5: By applying numerical integration (for example trapezoidal rule) in one-dimensional domain we have the solution of double integration is
)()(2)(2
1
10 N
N
ii
x xGxGxGh
I