Integraal van $$$x \sqrt{4 - x}$$$

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

Zij $$$u=4 - x$$$.

Dan $$$du=\left(4 - x\right)^{\prime }dx = - dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = - du$$$.

Dus,

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

Expand the expression:

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

Integreer termgewijs:

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=\frac{3}{2}$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=4$$$ en $$$f{\left(u \right)} = \sqrt{u}$$$:

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=\frac{1}{2}$$$:

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

We herinneren eraan dat $$$u=4 - x$$$:

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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