# Numerical Methods/Numerical Integration

<< Numerical Methods

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 $h = \frac{{b - a}}{n} = \frac{{1.2 - 0}}{6} = 0.2$

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

## 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.