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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?

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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.

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

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    $\begingroup$ P.S. Even when just reading mathematics, generous use of rough paper (instead of analysing all in your head) is invaluable: it extends your mental space and allows room to unpack problems to make them tractable. $\endgroup$
    – ryang
    Feb 3, 2022 at 21:17

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