<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] The integral is undefined at 3.  We must use an improper integral. 
$$
\int_0^3 \frac{1}{\sqrt{3-x}} dx = \lim_{x \to 3^-}\int_0^b \frac{1}{\sqrt{3-x}} dx
$$
using a u substitution where u = 3 - x, then du = -dx which gives the following:
$$
\int \frac{-1}{\sqrt{u}}du = -2\sqrt{u}
$$
Putting this back in terms of x gives us 
$$ -2\sqrt{3-x}
$$
Thus we get
$$
-2\lim_{x \to 3^-}[\sqrt{3-b} - \sqrt{3-0}] = (-2)(-\sqrt{3}) = 2\sqrt{3}
$$
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