Thoughts on Numerical Integration (Part 17): Midpoint rule and global rate of convergence

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 \displaystyle \int_{x_i}^{x_i+h} x^k \, dx  has error

= \displaystyle \frac{k(k-1)}{24} x_i^{k-2} h^3 + O(h^4)

In this post, we consider the global error when integrating on the interval [a,b] instead of a subinterval [x_i,x_i+h]. 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 \displaystyle \int_a^b x^k \, dx = \int_{x_0}^{x_n} x^k \, dx will be the sum of the errors for the integrals over [x_0,x_1], [x_1,x_2], through [x_{n-1},x_n]. Therefore, the total error will be

E \approx \displaystyle \frac{k(k-1)}{24} \left(x_0^{k-2} + x_1^{k-2} + \dots + x_{n-1}^{k-2} \right) h^3.

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 \displaystyle \int_1^2 x^9 \, dx for different numbers of subintervals. If we take n = 100 and h = 0.01, then the error should be approximately equal to

\displaystyle \frac{9 \times 8}{24} \left(1^7 + 1.01^7 + \dots + 1.99^7 \right) (0.01)^3 \approx 0.0093731,

which, as expected, is close to the actual error of 102.3 - 102.2904379 \approx 0.00956211.
Let y_i = x_i^{k-2}, so that the error becomes

E \approx \displaystyle \frac{k(k-1)}{24} \left(y_0 + y_1 + \dots + y_{n-1} \right) h^3 + O(h^4) = \displaystyle \frac{k(k-1)}{24} \overline{y} n h^3,

where \overline{y} = (y_0 + y_1 + \dots + y_{n-1})/n is the average of the y_i. Clearly, this average is somewhere between the smallest and the largest of the y_i. Since y = x^{k-2} is a continuous function, that means that there must be some value of x_* between x_0 and x_{k-1} — and therefore between a and b — so that x_*^{k-2} = \overline{y} by the Intermediate Value Theorem. We conclude that the error can be written as

E \approx \displaystyle \frac{k(k-1)}{24} x_*^{k-2} nh^3,

Finally, since h is the length of one subinterval, we see that nh = b-a is the total length of the interval [a,b]. Therefore,

E \approx \displaystyle \frac{k(k-1)}{24} x_*^{k-2} (b-a)h^2 \equiv ch^2,

where the constant c is determined by a, b, and k. In other words, for the special case f(x) = x^k, we have established that the error from the Midpoint Rule is approximately quadratic in h — without resorting to the generalized mean-value theorem and confirming the numerical observations we made earlier.

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