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Integrate 1/sqrt(3-x) from 2 to 3
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\[\int\limits_{2}^{3}\frac{1}{\sqrt{3-x}} dx\]
let u=3-x => du=-1 dx
so if du=-1 dx then -du=1 dx
\[\int\limits_{3-2}^{3-3}\frac{-du}{\sqrt{u}}\]
\[-\int\limits_{1}^{0}u^\frac{-1}{2} du=-2u^\frac{1}{2}|_1^0=-2(0^\frac{1}{2}-1^\frac{1}{2})=-2(-1)=2\]
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