Numerical Methods/Numerical Integration

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Often, we need to find the integral of a function that may be difficult to integrate analytically (ie, as a definite integral) or impossible (the function only existing as a table of values).

Some methods of approximating said integral are listed below.

Trapezoidal Rule[edit]

Consider some function, possibly unknown, f(x), with known values over the interval [a,b] at n+1 evenly spaced points xi of spacing h={(b-a) \over n}, x_0=a and x_n=b.

Further, denote the function value at the ith mesh point as f(x_i).

Using the notion of integration as "finding the area under the function curve", we can denote the integral over the ith segment of the interval, from x_{i-1} to x_i as:

\int_{x_{i-1}}^{x_i} f(x)\, dx = (1)

Since we may not know the antiderivative of f(x), we must approximate it. Such approximation in the Trapezoidal Rule, unsurprisingly, involves approximating (1) with a trapezoid of width h, left height f(x_{i-1}), right height f(x_i). Thus,

(1) \simeq {1 \over 2}h(f(x_{i-1}) + f(x_i)) = (2)

(2) gives us an approximation to the area under one interval of the curve, and must be repeated to cover the entire interval.

For the case where n = 2,

\int_{x_a}^{x_b} f(x)\, dx \simeq {1 \over 2}h(f(x_0) + f(x_1)) + {1 \over 2}h(f(x_{1}) + f(x_2)) = (3)


Collecting like terms on the right hand side of (3) gives us:

 {1 \over 2} h (f(x_0) + f(x_1) + f(x_1) + f(x_2))

or

 {1 \over 2} h (f(x_0) + 2f(x_1) + f(x_2))

Now, substituting in for h and cleaning up,

 {(b-a) \over 2\cdot2} (f(x_0) + 2f(x_1) + f(x_2))


To motivate the general version of the trapezoidal rule, now consider n = 4,

\int_{x_a}^{x_b} f(x)\, dx \simeq {1 \over 2}h(f(x_0) + f(x_1)) + {1 \over 2}h(f(x_1) + f(x_2)) + {1 \over 2}h(f(x_2) + f(x_3)) + {1 \over 2}h(f(x_3) + f(x_4))

Following a similar process as for the case when n=2, we obtain

 {(b-a) \over 2\cdot4} (f(x_0) + 2(f(x_1) + f(x_2) + f(x_3)) + f(x_4))

Proceeding to the general case where n = N,

 \int_{x_a}^{x_b} f(x)\, dx \simeq {(b-a) \over 2 \cdot n} (f(x_0) + 2(\sum_{k=1}^Nf(x_k)) + f(x_n))

Trapizoidruleexample.jpg This is an example of what the trapezoidal rule would represent graphicly, here  y = -x^2 + 5 .


Example[edit]

Approximate  \int_{0}^{1} x^3\, dx to within 5%.

First, since the function can be exactly integrated, let us do so, to provide a check on our answer.

 \int_{0}^{1} x^3\, dx = \left [{x^4 \over 4}\right ]_0^1 = {1 \over 4} = 0.25 = (4)

We will start with an interval size of 1, only considering the end points.

 f(0) = 0

 f(1) = 1

(4)  \simeq {(1-0) \over (2\cdot1)} (f(0) + f(1)) = {1 \over 2.1} (0 + 1) = {1 \over 2} = 0.5

Relative error =  \left |{ (0.5-0.25) \over 0.25}  \right | = 1

Hmm, a little high for our purposes. So, we halve the interval size to 0.5 and add to the list

 f(0.5) = 0.125

(4)  \simeq {(1-0) \over (2\cdot2)} (f(0) + 2f(0.5) +  f(1)) = {1 \over 2\cdot2} (0 + 2(0.125) + 1) = {1.25 \over 4} = 0.3125

Relative error =  \left |{ (0.3125-0.25) \over 0.25}  \right | = 0.25

Still above 0.01, but vastly improved from the initial step. We continue in the same fashion, calculating f(0.25) and f(0.75), rounding off to four decimal places.

 f(0.25) = 0.0156

 f(0.75) = 0.4219

(4)  \simeq {(1-0) \over (2\cdot4)} ( 0 + 2(0.0156 + 0.125 + 0.4219) + 1) = {1 \over 8} (2.2150) = 0.2656

Relative error =  \left |{ (0.2656-0.25) \over 0.25}  \right | = 0.0624

We are well on our way. Continuing, with interval size 0.125 and rounding as before,

 f(0.125) = 0.0020

 f(0.375) = 0.0527

 f(0.625) = 0.2441

 f(0.875) = 0.6699


(4)  \simeq {(1-0) \over (2\cdot8)} (0 + 2(0.0020 + 0.0156 + 0.0527 + 0.0125 + 0.2441 + 0.4219 + 0.6699) + 1) = {1 \over 16} (4.0624) = 0.2539

Relative error =  \left |{ (0.2539-0.25) \over 0.25}  \right | = 0.0156

Since our relative error is less than 5%, we stop.

Error Analysis[edit]

Let y=f(x) be continuous,well-behaved and have continuous derivatives in [x0,xn]. We expand y in a Taylor series about x=x0,thus-
\int_{x_0}^{x_1}y\, dx=\int_{x_0}^{x_1}[y_0+(x-x_0)y'_0+(x-x_0)^2y''_0/2!+......]\,dx

Simpson's Rule[edit]

Consider some function y=f(x) possibily unknown with known values over the interval [a,b] at n+1 evently spaced points then it defined as

\int_{x_0}^{x_n} f(x)\, dx \simeq {1 \over 3}h \bigg\{f(x_0) + f(x_n) + 2\Big(f(x_2) + f(x_4) + ... + f(x_{n-2})\Big) + 4\Big(f(x_1) + f(x_3) + ... + f(x_{n-1})\Big)\bigg\}

where h={(b-a) \over n} and x_0=a and x_n=b.

Example[edit]

Evaluate 
\int\limits_0^{1.2} {x\left( {8 - x^3 } \right)^{\frac{1}
{2}} dx} by taking n = 6 (n must be even)

Solution: Here f(x) = x\left( {8 - x^3 } \right)^{\frac{1}{2}}

Since a = 0 & b = 1.2 so h = \frac{{b - a}}{n} = \frac{{1.2 - 0}}{6} = 0.2

Now when a = x_0 = 0 then  f(x_0) = 0

And since x_n  = x_{n - 1}  + h, therefore for x_1 = 0.2 , x_2 = 0.4 , x_3 = 0.6 , x_4 = 0.8 , x_5 = 1 , x_6 = b = 1.2 the corresponding values are f(x_1) = 0.7784 , f(x_2) = 1.58721 , f(x_3) = 1.6740 , f(x_4) = 2.1891 , f(x_5) = 2.6458 , f(x_6) = 3.0053

Incomplete ... Completed soon

Error Analysis[edit]

Simpson's 3/8[edit]

The numerical integration technique known as "Simpson's 3/8 rule" is credited to the mathematician Thomas Simpson (1710-1761) of Leicestershire, England. His also worked in the areas of numerical interpolation and probability theory.


Theorem (Simpson's 3/8 Rule) Consider over , where , , and . Simpson's 3/8 rule is

   .   

This is an numerical approximation to the integral of over and we have the expression

   .  

The remainder term for Simpson's 3/8 rule is , where lies somewhere between , and have the equality

   .


Proof Simpson's 3/8 Rule Simpson's 3/8 Rule


Composite Simpson's 3/8 Rule

   Our next method of finding the area under a curve  is by approximating that curve with a series of cubic segments that lie above the intervals  .  When several cubics are used, we call it the composite Simpson's 3/8 rule.  


Theorem (Composite Simpson's 3/8 Rule) Consider over . Suppose that the interval is subdivided into subintervals of equal width by using the equally spaced sample points for . The composite Simpson's 3/8 rule for subintervals is

   .  

This is an numerical approximation to the integral of over and we write

   .  


Proof Simpson's 3/8 Rule Simpson's 3/8 Rule


Remainder term for the Composite Simpson's 3/8 Rule

Corollary (Simpson's 3/8 Rule: Remainder term) Suppose that is subdivided into subintervals of width . The composite Simpson's 3/8 rule

   .  

is an numerical approximation to the integral, and

   .  

Furthermore, if , then there exists a value with so that the error term has the form

   .  

This is expressed using the "big " notation .


Remark. When the step size is reduced by a factor of the remainder term should be reduced by approximately .


Algorithm Composite Simpson's 3/8 Rule. To approximate the integral

   ,  


by sampling at the equally spaced sample points for , where . Notice that and .


Animations (Simpson's 3/8 Rule Simpson's 3/8 Rule). Internet hyperlinks to animations.


Computer Programs Simpson's 3/8 Rule Simpson's 3/8 Rule


Mathematica Subroutine (Simpson's 3/8 Rule). Object oriented programming.


Example 1. Numerically approximate the integral by using Simpson's 3/8 rule with m = 1, 2, 4. Solution 1.


Example 2. Numerically approximate the integral by using Simpson's 3/8 rule with m = 10, 20, 40, 80, and 160. Solution 2.


Example 3. Find the analytic value of the integral (i.e. find the "true value"). Solution 3.


Example 4. Use the "true value" in example 3 and find the error for the Simpson' 3/8 rule approximations in example 2. Solution 4.


Example 5. When the step size is reduced by a factor of the error term should be reduced by approximately . Explore this phenomenon. Solution 5.


Example 6. Numerically approximate the integral by using Simpson's 3/8 rule with m = 1, 2, 4. Solution 6.


Example 7. Numerically approximate the integral by using Simpson's 3/8 rule with m = 10, 20, 40, 80, and 160. Solution 7.


Example 8. Find the analytic value of the integral (i.e. find the "true value"). Solution 8.


Example 9. Use the "true value" in example 8 and find the error for the Simpson's 3/8 rule approximations in example 7. Solution 9.


Example 10. When the step size is reduced by a factor of the error term should be reduced by approximately . Explore this phenomenon. Solution 10.


Various Scenarios and Animations for Simpson's 3/8 Rule.

Example 11. Let over . Use Simpson's 3/8 rule to approximate the value of the integral. Solution 11.


Animations (Simpson's 3/8 Rule Simpson's 3/8 Rule). Internet hyperlinks to animations.


Research Experience for Undergraduates

Simpson's Rule for Numerical Integration Simpson's Rule for Numerical Integration Internet hyperlinks to web sites and a bibliography of articles.

Headline text[edit]

Example[edit]

Error Analysis[edit]

References[edit]

Eric W. Weisstein. "Trapezoidal Rule." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/TrapezoidalRule.html



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