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Subsections


Numerical Integration

Purpose

The purpose of this lab is to give you some experience with using the trapezoidal rule and Simpson's rule to approximate integrals.

Background

The trapezoidal rule and Simpson's rule are used for approximating area under a curve or the definite integral

\begin{displaymath}\int_{a}^{b} f(x) \, dx \end{displaymath}

Both methods start by dividing the interval $[a,b]$ into $n$ subintervals of equal length by choosing a partition

\begin{displaymath}a = x_0 < x_1 < x_2 < \ldots < x_n = b \end{displaymath}

satisfying

\begin{displaymath}x_i = a + i h, \mbox{ for $i=0, \ldots, n$} \end{displaymath}

where

\begin{displaymath}h = \frac{b-a}{n} \end{displaymath}

is the length of each subinterval. For the trapezoidal rule, the integral over each subinterval is approximated by the area of a trapezoid. This gives the following approximation to the integral

\begin{displaymath}T_n = \frac{h}{2} \left( f(x_0) + 2 f(x_1) + 2 f(x_2) + \ldots + 2
f(x_{n-1}) + f(x_n) \right) \end{displaymath}

For Simpson's rule, the function is approximated by a parabola over pairs of subintervals. When the areas under the parabolas are computed and summed up, the result is the following approximation.

\begin{displaymath}S_n = \frac{h}{3} \left( f(x_0) + 4 f(x_1) + 2 f(x_2) + \ldots + 4
f(x_{n-1}) + f(x_n) \right) \end{displaymath}

Maple commands

The commands for the trapezoidal rule and Simpson's rule are in the student package.
>with(student);
The following example will use the function
>f:=x->x^2*exp(x);
This computes the integral of the function from 0 to 2.
>int(f(x),x=0..2);
Using the evalf command provides a decimal approximation.
>evalf(int(f(x),x=0..2));
The command for using the trapezoidal rule is trapezoid. The syntax is very similar to that of the int command. the last argument specifies the number of subintervals to use. In the command below, the number of subintervals is set to 10, but you should experiment with increasing or decreasing this number. Note that Maple writes out the sum and doesn't evaluate it to a number.
>trapezoid(f(x),x=0..2,10);
Putting an evalf command on the outside computes the trapezoidal approximation.
>evalf(trapezoid(f(x),x=0..2,10));
The command for Simpson's rule is very similar.
>simpson(f(x),x=0..2,10);
>evalf(simpson(f(x),x=0..2,10));

Exercises

  1. For the function $f(x)=x \sin(x)+4$, use the trapezoidal rule formula to approximate the area under $f(x)$ over the interval $[0,4]$ ising $n=8$. Verify your answer using the trapezoid command in Maple. Repeat this exercise with simpson's rule.

  2. Plot the function $f(x)= \sin (x)$ between $0$ and $\pi$. Using the trapezoid rule find the minimum number of subintervals necessary to approximate the area to within 0.001. Then do the same with Simpson's rule. Which rule is more accurate and why?

  3. There is an error term associated with the trapezoidal rule that can be used to estimate the error. More precisely, we have

    \begin{displaymath}\int_{a}^{b} f(x) \, dx = T_n + E_n \end{displaymath}

    where

    \begin{displaymath}E_n = - \frac{(b-a)^3 f''(c)}{12n^2} \end{displaymath}

    for some value $c$ between $a$ and $b$ that maximizes the second derivative in absolute value. Solving the error formula for $n$ guarantees a number of subintervals such that the error term $\vert E_n\vert$ is less than some desired tolerance $\varepsilon$. This gives:

    \begin{displaymath}n > \sqrt{\frac{(b-a)^3 M}{12 \varepsilon}} \end{displaymath}

    The way to think about this result is that it gives a value for $n$ which guarantees that the error of the trapezoidal rule is less than the tolerance $\varepsilon$. It is generally a very conservative result. For the function

    \begin{displaymath}f(x)=\frac{x}{1+x^3} \end{displaymath}

    over the interval $[0,5]$, find the value of $n$ using the error estimate and see if the actual number of subintervals required to satisfy a tolerance of $\varepsilon = 0.001$ is smaller than the number given by the error estimate.

    Similarly, the number of subintervals for the simpson rule approximation to guarantee an error smaller than $\varepsilon$ is

    \begin{displaymath}n > \sqrt[4]{\frac{(b-a)^5 M}{180 \varepsilon}} \end{displaymath}

    where $M$ is the maximum of $\mid f^{(4)}(x) \mid$ on the interval $[a,b]$. Again, find the value of $n$ guaranteed by the error estimate to satisfy the same tolerance and see if the actual number of subintervals needed for simpson's rule to satisfy the tolerance is smaller.

next up previous
Next: About this document ... Up: lab_template Previous: lab_template
Dina Solitro
2007-02-19