Integral of $$$x \sqrt{4 - x}$$$

Find $$$\int x \sqrt{4 - x}\, dx$$$.

Let $$$u=4 - x$$$.

Then $$$du=\left(4 - x\right)^{\prime }dx = - dx$$$ (steps can be seen »), and we have that $$$dx = - du$$$.

The integral becomes

$${\color{red}{\int{x \sqrt{4 - x} d x}}} = {\color{red}{\int{\sqrt{u} \left(u - 4\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{\sqrt{u} \left(u - 4\right) d u}}} = {\color{red}{\int{\left(u^{\frac{3}{2}} - 4 \sqrt{u}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{\frac{3}{2}} - 4 \sqrt{u}\right)d u}}} = {\color{red}{\left(- \int{4 \sqrt{u} d u} + \int{u^{\frac{3}{2}} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{3}{2}$$$:

$$- \int{4 \sqrt{u} d u} + {\color{red}{\int{u^{\frac{3}{2}} d u}}}=- \int{4 \sqrt{u} d u} + {\color{red}{\frac{u^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=- \int{4 \sqrt{u} d u} + {\color{red}{\left(\frac{2 u^{\frac{5}{2}}}{5}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=4$$$ and $$$f{\left(u \right)} = \sqrt{u}$$$:

$$\frac{2 u^{\frac{5}{2}}}{5} - {\color{red}{\int{4 \sqrt{u} d u}}} = \frac{2 u^{\frac{5}{2}}}{5} - {\color{red}{\left(4 \int{\sqrt{u} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\frac{2 u^{\frac{5}{2}}}{5} - 4 {\color{red}{\int{\sqrt{u} d u}}}=\frac{2 u^{\frac{5}{2}}}{5} - 4 {\color{red}{\int{u^{\frac{1}{2}} d u}}}=\frac{2 u^{\frac{5}{2}}}{5} - 4 {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\frac{2 u^{\frac{5}{2}}}{5} - 4 {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Recall that $$$u=4 - x$$$:

$$- \frac{8 {\color{red}{u}}^{\frac{3}{2}}}{3} + \frac{2 {\color{red}{u}}^{\frac{5}{2}}}{5} = - \frac{8 {\color{red}{\left(4 - x\right)}}^{\frac{3}{2}}}{3} + \frac{2 {\color{red}{\left(4 - x\right)}}^{\frac{5}{2}}}{5}$$

Therefore,

$$\int{x \sqrt{4 - x} d x} = \frac{2 \left(4 - x\right)^{\frac{5}{2}}}{5} - \frac{8 \left(4 - x\right)^{\frac{3}{2}}}{3}$$

Simplify:

$$\int{x \sqrt{4 - x} d x} = \frac{2 \left(4 - x\right)^{\frac{3}{2}} \left(- 3 x - 8\right)}{15}$$

Add the constant of integration:

$$\int{x \sqrt{4 - x} d x} = \frac{2 \left(4 - x\right)^{\frac{3}{2}} \left(- 3 x - 8\right)}{15}+C$$

$$$\int x \sqrt{4 - x}\, dx = \frac{2 \left(4 - x\right)^{\frac{3}{2}} \left(- 3 x - 8\right)}{15} + C$$$A