Numerical Methods 3

Numerical Methods - Numerical
Integration
N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science, Rajkot (Guj.)
niravbvyas@gmail.com
N. B. Vyas Numerical Methods - Numerical Integration

Numerical Integration
Let I =
b
a
y dx where y = f(x) takes the values y0, y1, . . . , yn for
x0, x1, . . . , xn

Let I =
b
a
x0, x1, . . . , xn
Let us divide the interval (a, b) into n sub-intervals of width h so
that x0 = a, x1 = a + h = x0 + h, x2 = x0 + 2h, . . .,
xn = x0 + nh = b then

Let I =
b
a
x0, x1, . . . , xn
that x0 = a, x1 = a + h = x0 + h, x2 = x0 + 2h, . . .,
I =
b
a
y dx =
x0+nh
x0
f(x) dx

Let I =
b
a
x0, x1, . . . , xn
that x0 = a, x1 = a + h = x0 + h, x2 = x0 + 2h, . . .,
I =
b
a
y dx =
x0+nh
x0
f(x) dx
Trapezoidal rule:
b=x0+nh
a=x0
f(x)dx =
h
2
[(y0 + yn) + 2 (y1 + y2 + .... + yn)]; h =
b − a
n
If the number of strips is increased; that is, h is decreased, then
the accuracy of the approximation is increased.

Simpson’s
1
3
rd rule:

Simpson’s
1
3
rd rule:
x0+nh
x0
f(x)dx = h
3 [(y0 + yn) + 4(y1 + y3 + ....)
+2(y3 + y4 + ....)]; h = b−a
n

Simpson’s
1
3
rd rule:
x0+nh
x0
f(x)dx = h
3 [(y0 + yn) + 4(y1 + y3 + ....)
+2(y3 + y4 + ....)]; h = b−a
n
while applying this rule, the given interval must be divided into
even number of equal sub-intervals. i.e. n must be even.

Simpson’s
1
3
rd rule:
x0+nh
x0
f(x)dx = h
3 [(y0 + yn) + 4(y1 + y3 + ....)
+2(y3 + y4 + ....)]; h = b−a
n
Simpson’s
3
8
th rule:

Simpson’s
1
3
rd rule:
x0+nh
x0
f(x)dx = h
3 [(y0 + yn) + 4(y1 + y3 + ....)
+2(y3 + y4 + ....)]; h = b−a
n
Simpson’s
3
8
th rule:
x0+nh
x0
f(x)dx = 3h
8 [(y0 + yn) + 3(y1 + y2 + y4 + y5 + ....)
+2(y3 + y6 + ....)]; h = b−a
n

Simpson’s
1
3
rd rule:
x0+nh
x0
f(x)dx = h
3 [(y0 + yn) + 4(y1 + y3 + ....)
+2(y3 + y4 + ....)]; h = b−a
n
Simpson’s
3
8
th rule:
x0+nh
x0
f(x)dx = 3h
8 [(y0 + yn) + 3(y1 + y2 + y4 + y5 + ....)
+2(y3 + y6 + ....)]; h = b−a
n
while applying this rule, the number of sub-intervals should be
taken as a multiple of 3 i.e. n must be multiple of 3

Gaussian Integration Formula:
1
−1
f(t)dt =
n
i=1
wif(ti)
It should be noted here that, t = ±1 is obtained by setting
x =
1
2
[(b + a) + t (b − a)]

Gaussian Integration Formula: The following table gives the
values for n = 2, 3, 4, 5

Example
Ex. Evaluate
1
0
e−x2
dx by using Gaussion integration formula for
n = 3.
Sol. Here, we have to ﬁrst convert the given integral from 0 to 1 into
an integral from −1 to 1. x = 1
2 [(b + a) + t (b − a)], a = 0 and
b = 1
∴ x =
t + 1
2
⇒ dx =
dt
2
∴
1
0
exp(−x2)dx =
1
2
1
−1
exp −
1
4
(t + 1)2 dt

Error
Error in Quadrature Formula:
If yp is a polynomial representing the function y = f(x) in the
interval [x0, xn] the error in the quadrature formula is given by
E =
xn
x0
f(x) =
xn
x0
ypdx

Error
Error in Trapezoidal rule:
|error| ≤ (b − a)
h2
12
|f (M)|
where f (M) = max |f 0(x)|, |f 1(x)|, ..., |f n−1(x)|
∴ error is of order h2
total error =
dh3
12
y 0 + y 1 + ... + y n−1

Error
Error in Simpson’s
1
3
rd rule:
h4
180
|f4
(M)|
where f4(M) = max |y4
0|, |y4
2|, ..., |y4
n−2|
total error =
h5
90
y4
0 + y4
2 + ... + y4
n−2

Error
Error in Simpson’s
3
8
th rule:
h4
80
|f4
(M)|
where f4(M) = max |y4
0|, |y4
3|, ..., |y4
n−3|
total error =
3h5
80
y4
0 + y4
3 + ... + y4
n−3

Numerical Methods 3

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