Answer
$$\int_1^2x\sqrt{x-1}dx=\frac{16}{15}$$
Work Step by Step
To evaluate the integral $$\int_1^2x\sqrt{x-1}dx$$
we will use substitution $x-1=t$ which gives us $dx=dt$ and $x=t+1$. New integration bounds are: for $x=1$ we have $t=0$ and for $x=2$ we have $t=1$. Putting this into the integral we get:
$$\int_1^2x\sqrt{x-1}dx=\int_0^1(t+1)\sqrt t dt=\int_0^1t^{3/2}dt+\int_0^1t^{1/2}dt=
\left.\frac{t^{5/2}}{\frac{5}{2}}\right|_0^1+\left.\frac{t^{3/2}}{\frac{3}{2}}\right|_0^1=
\frac{2}{5}(1^{5/2}-0^{5/2})+\frac{2}{3}(1^{3/2}-0^{3/2})=\frac{2}{5}\cdot1+\frac{2}{3}\cdot1=
\frac{16}{15}$$
