Integraal van $$$a \epsilon \sigma t^{4}$$$ met betrekking tot $$$t$$$

Bepaal $$$\int a \epsilon \sigma t^{4}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=a \epsilon \sigma$$$ en $$$f{\left(t \right)} = t^{4}$$$:

$${\color{red}{\int{a \epsilon \sigma t^{4} d t}}} = {\color{red}{a \epsilon \sigma \int{t^{4} d t}}}$$

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

$$a \epsilon \sigma {\color{red}{\int{t^{4} d t}}}=a \epsilon \sigma {\color{red}{\frac{t^{1 + 4}}{1 + 4}}}=a \epsilon \sigma {\color{red}{\left(\frac{t^{5}}{5}\right)}}$$

Dus,

$$\int{a \epsilon \sigma t^{4} d t} = \frac{a \epsilon \sigma t^{5}}{5}$$

Voeg de integratieconstante toe:

$$\int{a \epsilon \sigma t^{4} d t} = \frac{a \epsilon \sigma t^{5}}{5}+C$$

$$$\int a \epsilon \sigma t^{4}\, dt = \frac{a \epsilon \sigma t^{5}}{5} + C$$$A