Answer
$\int_{1}^{2}x~\sqrt{x-1}~dx = \frac{16}{15}$
Work Step by Step
$\int_{1}^{2}x~\sqrt{x-1}~dx$
Let $u = \sqrt{x-1}$
$\frac{du}{dx} = \frac{1}{2 \sqrt{x-1}}$
$dx = 2 \sqrt{x-1}~du = 2u~du$
Also:
$u = \sqrt{x-1}$
$u^2 = x-1$
$x = u^2+1$
When $x = 1$, then $u =0$
When $x = 2$, then $u = 1$
$\int_{0}^{1} (u^2+1)~(u)~(2u)~du$
$=\int_{0}^{1} (2u^4+2u^2)~du$
$= (\frac{2}{5}u^5+\frac{2}{3}u^3)~\vert_{0}^{1}$
$= [\frac{2}{5}(1)^5+\frac{2}{3}(1)^3]-[\frac{2}{5}(0)^5+\frac{2}{3}(0)^3]$
$= (\frac{2}{5}+\frac{2}{3})-(0)$
$= \frac{16}{15}$
