Simpson's rule

Question

Use Simpson's rule to estimate using 4 intervals:

 

∫(from 1 to 3) e^x^2 dx

Answer 1

Reading this as ∫₁³ e^x2 dx — the classic example of an integral with no elementary antiderivative, which is exactly why you are being asked to approximate it.

Step 1: find h and the nodes

With a = 1, b = 3, and n = 4 subintervals, h = (b − a)/n = (3 − 1)/4 = 0.5.

So the nodes are x₀ = 1, x₁ = 1.5, x₂ = 2, x₃ = 2.5, x₄ = 3.

Step 2: evaluate the function at each node

f(x₀) = e¹ ≈ 2.718282

f(x₁) = e^2.25 ≈ 9.487736

f(x₂) = e⁴ ≈ 54.598150

f(x₃) = e^6.25 ≈ 518.012825

f(x₄) = e⁹ ≈ 8103.083928

Notice how violently these grow — that will matter at the end.

Step 3: apply the rule

Simpson’s rule is

(h/3)[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + f(x₄)]

The coefficient pattern is 1, 4, 2, 4, …, 4, 1: endpoints get 1, odd-indexed nodes get 4, even-indexed interior nodes get 2. A quick check — the coefficients here sum to 1 + 4 + 2 + 4 + 1 = 12, and (h/3)(12) = (0.5/3)(12) = 2, which is exactly the width of the interval. If that check fails, you have miscounted the pattern.

Filling it in:

2.718282 + 4(9.487736) + 2(54.598150) + 4(518.012825) + 8103.083928

= 2.718282 + 37.950943 + 109.196300 + 2072.051298 + 8103.083928

= 10325.00075

Then (0.5/3)(10325.00075) = ≈ 1720.83

One honest caveat

That is the answer to the question as asked, but it is not a good estimate of the integral. The true value is about 1443.08, so four subintervals overshoot by roughly 19%.

The reason is in Simpson’s error term, which is proportional to the fourth derivative of f. For e^x2 that fourth derivative is enormous near x = 3, and Simpson’s rule works by fitting parabolas through consecutive triples of points — a parabola simply cannot keep up with something growing that fast over an interval half a unit wide.

Watch it settle down as you refine: n = 8 gives about 1475.43, n = 16 gives about 1445.62, closing in on 1443.08. The error drops by roughly a factor of 16 each time you double n, which is the hallmark of a fourth-order method.

So the takeaway is not that Simpson’s rule is bad, but that reporting an approximation without a feel for its error is the real mistake. When a function is exploding across your interval, n = 4 is a demonstration, not a measurement.