Simpson's rule

Question

Estimate the number of intervals necessary to approximate the following definite integral with an error less than 0.001 using Simpson's rule.

 

∫(from 1 to 5) e^x^3 dx

Answer 1

Reading this as ∫₁⁵ e^x3 dx. The tool you want is Simpson’s error bound:

|E_S| ≤ (b − a)⁵ · M / (180 n⁴), where M is the maximum of |f⁽⁴⁾(x)| on [a, b].

So the work splits into two parts: find M, then solve for n. Most of the effort is in M.

Step 1: the fourth derivative

With f(x) = e^x3, every derivative keeps the e^x3 factor and picks up polynomial pieces from the chain rule. Working up:

f′(x) = 3x² e^x3

f″(x) = (6x + 9x⁴) e^x3

f′″(x) = (6 + 54x³ + 27x⁶) e^x3

f⁽⁴⁾(x) = (180x² + 324x⁵ + 81x⁸) e^x3

Step 2: maximize it on [1, 5]

Both factors are positive and strictly increasing on this interval, so no calculus is needed — the max sits at the right endpoint, x = 5.

M = f⁽⁴⁾(5) = (180·25 + 324·3125 + 81·390625) e¹²⁵ = 32,657,625 · e¹²⁵ ≈ 6.32 × 10⁶¹

Step 3: solve for n

With b − a = 4 and the tolerance 0.001:

4⁵ · M / (180 n⁴) ≤ 0.001

n⁴ ≥ 1024 · M / 0.18 ≈ 3.60 × 10⁶⁵

n ≥ (3.60 × 10⁶⁵)^1/4 ≈ 2.4 × 10¹⁶

Rounding up to an even integer (Simpson’s rule requires an even n), that is roughly 25 quadrillion subintervals.

Why the answer is absurd, and why that is the lesson

That number is not a mistake. It is what the bound honestly demands, and it is worth understanding why it comes out so large.

The bound is a worst-case estimate. It takes the single largest value of |f⁽⁴⁾| anywhere on the interval and pretends the function is that badly behaved everywhere. For e^x3 that is wildly pessimistic: at x = 1 the fourth derivative is about 1,590, while at x = 5 it is about 6 × 10⁶¹. Those differ by roughly 58 orders of magnitude. Almost the entire interval is easy, and the bound is set by a sliver near x = 5.

Two things follow, and they are the real takeaway:

First, if a problem asks you to “estimate the number of intervals,” it is asking you to apply the bound, not to predict what would actually work. The true error at a modest n is far smaller than the bound allows.

Second, this is exactly why adaptive quadrature exists. Splitting [1, 5] and spending nearly all your subintervals near x = 5, where the function is genuinely difficult, reaches the same accuracy with a workable number of evaluations. A uniform grid is forced to treat the whole interval as badly as its worst point — and when a function grows like e^x3, that is a very expensive assumption.