Both methods start by dividing the interval into subintervals of equal length by choosing a partition

satisfying

where

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

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.

>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(int(f(x),x=0..2));The command for using the trapezoidal rule is

>trapezoid(f(x),x=0..2,10);Putting an

>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));

- For the function
, use the trapezoidal rule formula to approximate the area under over the interval ising . Verify your answer using the
`trapezoid`command in Maple. Repeat this exercise with simpson's rule. - Plot the function
between and .
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?
- There is an error term associated with the trapezoidal rule that can be used to estimate the error. More precisely, we have

where

for some value between and that maximizes the second derivative in absolute value. Solving the error formula for guarantees a number of subintervals such that the error term is less than some desired tolerance . This gives:

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

over the interval , find the value of using the error estimate and see if the actual number of subintervals required to satisfy a tolerance of 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 is

where is the maximum of on the interval . Again, find the value of 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.

2007-02-19