Evaluate the double integral of x times e to the power minus y squared over the region where √y/3 ≤ x ≤ √y/2 and 0 ≤ y < ∞.

Step-by-step solution

1. 1

   Integrate in $x$ first — the order is already the convenient one. The region is bounded on the left and right by curves given as $x$ in terms of $y$, so $dx$ belongs inside. And $e^{-y^{2}}$ is a constant for that inner integration, which matters because $e^{-y^{2}}$ has no elementary antiderivative in $y$; postponing it is essential.

   $\int_{\sqrt y/3}^{\sqrt y/2} x\,dx=\left[\frac{x^{2}}{2}\right]_{\sqrt y/3}^{\sqrt y/2}.$

2. 2

   Evaluate the bracket. Squaring the limits removes the square roots: $(\sqrt y/2)^{2}=y/4$ and $(\sqrt y/3)^{2}=y/9$. So

   $\frac12\left(\frac{y}{4}-\frac{y}{9}\right)=\frac12\cdot\frac{9y-4y}{36}=\frac{5y}{72}.$

   The strip therefore weights each $y$ by $5y/72$, a factor that is linear in $y$ — exactly the factor that will make the outer integral elementary.

3. 3

   Reduce to a single integral.

   $\int_{0}^{\infty}\frac{5y}{72}e^{-y^{2}}\,dy=\frac{5}{72}\int_{0}^{\infty}y\,e^{-y^{2}}\,dy.$

4. 4

   Substitute $u=y^{2}$. Then $du=2y\,dy$, so $y\,dy=\tfrac12\,du$, and the limits $y=0,\infty$ become $u=0,\infty$:

   $\int_{0}^{\infty}y\,e^{-y^{2}}\,dy=\frac12\int_{0}^{\infty}e^{-u}\,du=\frac12\bigl[-e^{-u}\bigr]_{0}^{\infty}=\frac12.$

   This is the reason the improper upper limit causes no trouble: the extra factor of $y$ converts a Gaussian into a plain exponential.

5. 5

   Multiply the two factors and check numerically.

   $\frac{5}{72}\cdot\frac12=\frac{5}{144}.$

   Adaptive quadrature applied to $\int_{0}^{\infty}\tfrac{5y}{72}e^{-y^{2}}dy$ returns $0.0347222222$, and $5/144=0.034722\overline{2}$ — agreement to ten digits.