How does $u$-substitution work: $\int \frac{\sqrt y e^{-y/2}}{2} \mathrm{d}y$

Question

I have been banging my head on the table for 3 hours trying to understand the final step here. Where did $-2^{(3/2)}$ come from? Why is there a $-u$ inside the square root?

Answer 1

From $u = -\frac{y}{2}$ we get $y = -2u$. They then pulled the $2$ out of the square root leaving just $-u$ under the square root. Then when the $\sqrt{2}$ is multiplied with the $-2$ from $dy = -2du$ gives $-2^{\frac{3}{2}}$, which they then pulled out of the integral.

Answer 2

$$\int\sqrt y \,e^{-\frac y2}\,\mathrm dy\\= \int\sqrt {-2u}\, e^u\,(-2\,\mathrm du)\\= -\sqrt2\,2\int\sqrt {-u}\, e^u\,\mathrm du\\= -2^{\frac32}\int\sqrt {-u}\, e^u\,\mathrm du.$$