<s>[INST] Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$  \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$$  = 2\sqrt{0}-2\sqrt{3} = -2\sqrt{3}$$
hi sorry i meant 3-x
 [/INST] $u=3-x$, $dx=-du$, and as $x$ goes from $0$ to $3$, then $u$ goes from $3$ to $0$, so
$$
\int_0^3 \frac{dx}{\sqrt{3-x}} = \int_3^0\frac{-du}{\sqrt{u}} = \int_0^3\frac{du}{\sqrt{u}} = \left. 2\sqrt{u} \vphantom{\frac 1 1}\ \right|_0^3 = 2\sqrt{3}.
$$
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