Answer
$$\int x\sqrt{4-x}dx=\frac{2(4-x)^{5/2}}{5}-\frac{8u(4-x)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int x\sqrt{4-x}dx$$
We set $u=4-x$, which means $x=4-u$
Then $$du=-dx$$ $$dx=-du$$
Therefore, $$A=-\int(4-u)\sqrt udu=-\int(4-u)u^{1/2}du$$ $$A=-\int\Big(4u^{1/2}-u^{3/2}\Big)du$$ $$A=-\Big(\frac{4u^{3/2}}{\frac{3}{2}}-\frac{u^{5/2}}{\frac{5}{2}}\Big)+C$$ $$A=\frac{2u^{5/2}}{5}-\frac{8u^{3/2}}{3}+C$$ $$A=\frac{2(4-x)^{5/2}}{5}-\frac{8u(4-x)^{3/2}}{3}+C$$
