I understand how to do the computation here:
$$\int_{0}^{\infty}xe^{-x^2}dx = \frac{1}{2}\int_{-\infty}^{0}e^{u}du = \frac{1}{2}$$
But I don't have any intuition for why the right-hand area under the curve $y=xe^{-x^2}$ would happen to be exactly half the left-hand area under the curve $y=e^x$. Or more generally, why this relationship holds:
$$\int_a^b xe^{-x^2}dx = \frac12\int_{-b^2}^{-a^2} e^xdx$$
Just to be clear, I'm not at all confused about the computation itself. I get that $u=-x^2$ and then you just substitute all the right things in. What I'm looking for is something like a geometric or behavioral intuition for why these two functions ($x \mapsto xe^{-x^2}$ and $x \mapsto e^{x}$) have such closely related areas under the curve.
This is just integration by substitution,
Let $u=-x^2$, then $du=-2x dx$ and $\frac{-1}{2}du=x dx$
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$$ \int^b_a xe^{-x^2}dx = -\frac{1}{2}\int e^{u} du $$
Now, $x=b \iff u=-b^2$ and $x=a \iff u=-a^2$, so the integral becomes
$$\begin{align}\int^b_a xe^{-x^2}dx &= -\frac{1}{2}\int^{-b^2}_{-a^2} e^{u} du\\ &= \frac{1}{2}\int_{-b^2}^{-a^2} e^{u}du \end{align}$$
To get the intuition, consider,
$$\int_a^b xe^{-x^2}dx = \int_a^b \frac{d}{dx}\left(\frac{-1}{2}e^{-x^2} \right) dx= \frac{-1}{2}e^{-x^2}|_a^b=\frac{-1}{2}(e^{-b^2}-e^{-a^2})$$
and $$\int_c^d e^xdx= e^x|_c^d=e^d-e^c$$