Integraal van $$$- \frac{x^{5}}{3} + x^{3}$$$

Bepaal $$$\int \left(- \frac{x^{5}}{3} + x^{3}\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(- \frac{x^{5}}{3} + x^{3}\right)d x}}} = {\color{red}{\left(\int{x^{3} d x} - \int{\frac{x^{5}}{3} d x}\right)}}$$

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

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(x \right)} = x^{5}$$$:

$$\frac{x^{4}}{4} - {\color{red}{\int{\frac{x^{5}}{3} d x}}} = \frac{x^{4}}{4} - {\color{red}{\left(\frac{\int{x^{5} d x}}{3}\right)}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=5$$$:

$$\frac{x^{4}}{4} - \frac{{\color{red}{\int{x^{5} d x}}}}{3}=\frac{x^{4}}{4} - \frac{{\color{red}{\frac{x^{1 + 5}}{1 + 5}}}}{3}=\frac{x^{4}}{4} - \frac{{\color{red}{\left(\frac{x^{6}}{6}\right)}}}{3}$$

Dus,

$$\int{\left(- \frac{x^{5}}{3} + x^{3}\right)d x} = - \frac{x^{6}}{18} + \frac{x^{4}}{4}$$

Voeg de integratieconstante toe:

$$\int{\left(- \frac{x^{5}}{3} + x^{3}\right)d x} = - \frac{x^{6}}{18} + \frac{x^{4}}{4}+C$$

$$$\int \left(- \frac{x^{5}}{3} + x^{3}\right)\, dx = \left(- \frac{x^{6}}{18} + \frac{x^{4}}{4}\right) + C$$$A