NUMERICAL METHOD'S

NUMERICAL METHODS
SARANYA A
ASST PROF OF MATHEMATICS
SAC WOMEN’S COLLEGE,COLLEGE.

Numerical Integration Methods
• We will present three numerical integration methods:
(1) The trapezoidal rule,
(2) The Simpson’s one-third rule, and
(3) The Gaussian quadrature
The first two methods require given functionsin the integrand of the definite
integrals. The last method, the “Gaussian quadrature” will automatically define
optimal locations on which the integrations are performed. It is a popular numerical
integration method in advanced numerical analyses, such as the finite element
method presented in the subsequent .

The trapezoidal rule for numerical integration
• The value of a definite integral of a function y(x) is equal to the area under the
curve produced by this function between the upper and lower limits variables of the
function.
• Let y = f(x) be continuous on [a, b]. We divide the interval [a, b] into n equal
subintervals, each of width, h = (b - a)/n,
• such that a = x0 < x1 < x2 < ⋯ < xn = b
• Area = h2[y0+2(y1+y2+y3+.....+yn−1)+yn]
• where,
• y0, y1,y2…. are the values of function at x = 1, 2, 3….. respectively.

x 0 0.5 1 1.5
y 5 6 9 11
Example 1: Find the area under the curve using trapezoidal rule formula which passes through the following
points:
Solution:
Given: y00 = 5
y1y1= 6
y2y2= 9
y3y3= 11
h = (0.5 - 0) = (1 - 0.5) = (1.5 - 1) = 0.5
Using Trapezoidal Rule Formula,
Area = h2[y0+yn+2(y1+y2+y3+.....+yn−1)]h2[y0+yn+2(y1+y2+y3+.....+yn−1)]
= 0.52[5+11+2(6+9)]0.52[5+11+2(6+9)]
= 0.25[16+30]
= 0.25[46]
= 11.5
Therefore, the area under the curve is 11.5 sq units.

• Simpson’s rule:
• Simpson’s rule is one of the numerical methods which is used to evaluate the
definite integral. Usually, to find the definite integral, we use the fundamental theorem of
calculus, where we have to apply the antiderivative techniques of integration.
• However, sometimes, it isn’t easy to find the antiderivative of an integral, like in
Scientific Experiments, where the function has to be determined from the observed
readings.
• Therefore, numerical methods are used to approximate the integral in such conditions.
Other numerical methods used are trapezoidal rule, midpoint rule, left or right
approximation using Riemann sums. Here, we will discuss Simpson’s rule formula, 1/3 rule,
3/8 rule and examples.

∫a
b f(x) dx = h/3 [(y0 + yn) + 4(y1 + y3 + y5 + …. + yn-1) + 2(y2 + y4 + y6 +
….. + yn-2)]
Where n is the even number, △x = (b – a)/n and xi = a + i△x
• If we have f(x) = y, which is equally spaced between [a, b] and if a = x0, x1 = x0 + h,
x2 = x0 + 2h …., xn = x0 + nh, where h is the difference between the terms.
• we can say that y0 = f(x0), y1 = f(x1), y2 = f(x2),……,yn = f(xn) are the analogous
values of y with each value of x.

Simpson’s ⅓ Rule for Integration
• We can get a quick approximation for definite integrals when we divide a small
interval [a, b] into two parts. Therefore, after dividing the interval, we get;
• x0= a, x1= a + b, x2 = b
• Hence, we can write the approximation as;
• ∫a
b f(x) dx ≈ S2 = h/3[f(x0) + 4f(x1) + f(x2)]
• S2 = h/3 [f(a) + 4 f((a+b)/2) + f(b)]
• Where h = (b – a)/2
• This is the Simpson’s ⅓ rule for integration.

Simpson’s 3/8 Rule
• Another method of numerical integration is called “Simpson’s 3/8 rule”. It is
completely based on the cubic interpolation rather than the quadratic
interpolation. Simpson’s 3/8 or three-eight rule is given by:
• ∫a
b f(x) dx = 3h/8 [(y0 + yn) + 3(y1 + y2 + y4 + y5 + …. + yn-1) + 2(y3 + y6 +
y9 + ….. + yn-3)]

• This rule is more accurate than the standard method, as it uses one more functional
value. For 3/8 rule, the composite Simpson’s 3/8 rule also exists which is similar to
the generalized form.
• The 3/8 rule is known as Simpson’s second rule of integration.
• Simpson’s Rule Error
• Although in Simpson’s rule method we get a more accurate approximation for
definite integral, still the error occurs which is defined when n = 2;
• -(1/90)[(b-a)/2]5 f(4) (ξ)
• Where ξ is some number between a and b.

Example: Evaluate ∫0
1exdx, by Simpson’s ⅓ rule.
• Solution:
• Let us divide the range [0, 1] into six equal parts by taking h = 1/6.
• If x0 = 0 then y0 = e0 = 1.
• If x1 = x0 + h = ⅙, then y1 = e1/6 = 1.1813
• If x2 = x0 + 2h = 2/6 = 1/3 then, y2 = e1/3 = 1.3956
• If x3 = x0 + 3h = 3/6 = ½ then y3 = e1/2= 1.6487
• If x4 = x0 + 4h = 4/6 ⅔ then y4 = e2/3 = 1.9477
• If x5 = x0 + 5h = ⅚ then y5 = e5/6 = 2.3009
• If x6 = x0 + 6h = 6/6 = 1 then y6 = e1 = 2.7182
• We know by Simpson’s ⅓ rule;
• ∫a
b f(x) dx = h/3 [(y0 + yn) + 4(y1 + y3 + y5 + …. + yn-1) + 2(y2 + y4 + y6 + ….. + yn-2)]
• Therefore,
• ∫0
1exdx = (1/18) [(1 + 2.7182) + 4(1.1813 + 1.6487 + 2.3009) + 2(1.39561 + 1.9477)]
• = (1/18)[3.7182 + 20.5236 + 6.68662]
• = 1.7182 (approx.)

NUMERICAL METHOD'S

More Related Content

What's hot (20)

Similar to NUMERICAL METHOD'S (20)

Recently uploaded (20)

NUMERICAL METHOD'S