Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

Why is numerical integration necessary in the first place?

Where do these formulas come from (especially Simpson’s Rule)?

How can I do all of these formulas quickly?

Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?

Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?

Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In the previous post, we showed that the midpoint approximation of has error
In this post, we consider the global error when integrating on the interval instead of a subinterval . The logic for determining the global error is much the same as what we used earlier for the left-endpoint rule.
The total error when approximating will be the sum of the errors for the integrals over , , through . Therefore, the total error will be

.

So that this formula doesn’t appear completely mystical, this actually matches the numerical observations that we made earlier. The figure below shows the left-endpoint approximations to for different numbers of subintervals. If we take and , then the error should be approximately equal to

,

which, as expected, is close to the actual error of .
Let , so that the error becomes

,

where is the average of the . Clearly, this average is somewhere between the smallest and the largest of the . Since is a continuous function, that means that there must be some value of between and — and therefore between and — so that by the Intermediate Value Theorem. We conclude that the error can be written as

,

Finally, since is the length of one subinterval, we see that is the total length of the interval . Therefore,

,

where the constant is determined by , , and . In other words, for the special case , we have established that the error from the Midpoint Rule is approximately quadratic in — without resorting to the generalized mean-value theorem and confirming the numerical observations we made earlier.

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.
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