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
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,

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

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

Example
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
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
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
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 $$h = \frac{n} = \frac{6} = 0.2$$

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

Simpson's 3/8
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.

References and further reading

 * Eric W. Weisstein. "Trapezoidal Rule." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/TrapezoidalRule.html
 * Davis, P. J., & Rabinowitz, P. (2007). Methods of numerical integration. Courier Corporation.

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