Does Simpson’s rule converge for $int_0^1 e^{x^2},dx$?

Key takeaway: This is a classic numerical integration question: once you know what Simpson’s rule is doing on a finer and finer partition, the limit becomes much easier to identify than the notation suggests.

We are approximating

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

by Simpson’s rule with $2n$ subintervals, and we call that approximation $I_n$.

Step 1: Recall the key property of Simpson’s rule

For a sufficiently smooth function, Simpson’s rule is a convergent quadrature method. Here,

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

is smooth on $[0,1]$, so the Simpson approximation converges as the mesh size goes to $0$.

Step 2: Identify the target of the convergence

Since each $I_n$ is just a numerical approximation to the same integral, the limit must be the integral itself:

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

Step 3: State the conclusion

So the sequence $(I_n)$ does converge, and its limit is

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

There is no elementary closed form for this integral, so the best exact answer is the integral itself.

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Pitfalls the pros know 👇 A common mistake is trying to force a closed-form antiderivative for $e^{x^2}$. That is not needed here. The point is convergence of the numerical method, not symbolic integration. Another trap is confusing Simpson’s rule with an exact formula for every $n$; it is only an approximation, though a very accurate one for smooth functions.

What if the problem changes? If the integrand were less regular, convergence would depend on the smoothness assumptions behind Simpson’s rule. For example, if $f$ had a singularity or a jump on $[0,1]$, the same conclusion would not be automatic. But for $e^{x^2}$, smoothness is more than enough, so the Simpson sequence still converges to the exact integral.

Tags: Simpson’s rule, numerical integration, quadrature convergence