Integraal van $$$6 \sqrt{2} \sqrt{x^{3}}$$$

Bepaal $$$\int 6 \sqrt{2} \sqrt{x^{3}}\, dx$$$.

De invoer is herschreven: $$$\int{6 \sqrt{2} \sqrt{x^{3}} d x}=\int{6 \sqrt{2} x^{\frac{3}{2}} d x}$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=6 \sqrt{2}$$$ en $$$f{\left(x \right)} = x^{\frac{3}{2}}$$$:

$${\color{red}{\int{6 \sqrt{2} x^{\frac{3}{2}} d x}}} = {\color{red}{\left(6 \sqrt{2} \int{x^{\frac{3}{2}} d x}\right)}}$$

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

$$6 \sqrt{2} {\color{red}{\int{x^{\frac{3}{2}} d x}}}=6 \sqrt{2} {\color{red}{\frac{x^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=6 \sqrt{2} {\color{red}{\left(\frac{2 x^{\frac{5}{2}}}{5}\right)}}$$

Dus,

$$\int{6 \sqrt{2} x^{\frac{3}{2}} d x} = \frac{12 \sqrt{2} x^{\frac{5}{2}}}{5}$$

Voeg de integratieconstante toe:

$$\int{6 \sqrt{2} x^{\frac{3}{2}} d x} = \frac{12 \sqrt{2} x^{\frac{5}{2}}}{5}+C$$

$$$\int 6 \sqrt{2} \sqrt{x^{3}}\, dx = \frac{12 \sqrt{2} x^{\frac{5}{2}}}{5} + C$$$A