[tex]\displaystyle\int_{x=9}^{x=10}x\sqrt{x-9}\,\mathrm dx[/tex]
Let [tex]y=x-9[/tex], so that [tex]x=y+9[/tex] and [tex]\mathrm dx=\mathrm dy[/tex]. The integral is then equivalent to
[tex]\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[/tex]
