Integraal van $$$\frac{1}{5 y^{4}}$$$

Bepaal $$$\int \frac{1}{5 y^{4}}\, dy$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ toe met $$$c=\frac{1}{5}$$$ en $$$f{\left(y \right)} = \frac{1}{y^{4}}$$$:

$${\color{red}{\int{\frac{1}{5 y^{4}} d y}}} = {\color{red}{\left(\frac{\int{\frac{1}{y^{4}} d y}}{5}\right)}}$$

Pas de machtsregel $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-4$$$:

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

Dus,

$$\int{\frac{1}{5 y^{4}} d y} = - \frac{1}{15 y^{3}}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{5 y^{4}} d y} = - \frac{1}{15 y^{3}}+C$$

$$$\int \frac{1}{5 y^{4}}\, dy = - \frac{1}{15 y^{3}} + C$$$A