Question

Evaluate the definite integral

Answer 1

$\displaystyle\int_{x=9}^{x=10}x\sqrt{x-9}\,\mathrm dx$

Let $y=x-9$, so that $x=y+9$ and $\mathrm dx=\mathrm dy$. The integral is then equivalent to

$\displaystyle\int_{y=0}^{y=1}(y+9)\sqrt y\,\mathrm dy=\int_0^1(y^{3/2}+9y^{1/2})\,\mathrm dy=\dfrac{32}5$