Does Simpson’s rule converge to the integral of e to the x squared?

Expert intro: The notation changes a little, but the mathematical question is the same: as the Simpson partition gets finer, where does the approximation land?

Detailed walkthrough

Let

$I=\int_0^1 e^{x^2}\,dx$

and let $I_n$ be the Simpson’s rule $\mathrm{Simp}(2n)$ approximation.

Step 1: Check the function

The integrand

$f(x)=e^{x^2}$

is continuous, smooth, and very well-behaved on $[0,1]$.

Step 2: Use the convergence of Simpson’s rule

Simpson’s rule is a convergent quadrature method for smooth functions. As $n\to\infty$, the subinterval width shrinks to $0$, so the approximation error goes to $0$.

Therefore,

$\lim_{n\to\infty} I_n=\int_0^1 e^{x^2}\,dx.$

Step 3: Answer the question

Yes, the sequence $(I_n)$ converges, and its limit is

$\boxed{\int_0^1 e^{x^2}\,dx}.$

💡 Pitfall guide

One thing students sometimes miss: the integral does not need an elementary antiderivative for Simpson’s rule to converge. Numerical convergence is about the function’s regularity, not about whether you can write the integral in closed form.

🔄 Real-world variant

If the integrand were replaced by a polynomial of degree at most 3, Simpson’s rule would actually be exact for every partition size. For $e^{x^2}$, it is not exact, but it still converges to the same integral as the mesh refines.

🔍 Related terms

Simpson’s rule, quadrature error, continuous function