Integraal van $$$i_{n} k_{n} x^{3}$$$ met betrekking tot $$$x$$$

Bepaal $$$\int i_{n} k_{n} x^{3}\, dx$$$.

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

$${\color{red}{\int{i_{n} k_{n} x^{3} d x}}} = {\color{red}{i_{n} k_{n} \int{x^{3} d x}}}$$

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

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

Dus,

$$\int{i_{n} k_{n} x^{3} d x} = \frac{i_{n} k_{n} x^{4}}{4}$$

Voeg de integratieconstante toe:

$$\int{i_{n} k_{n} x^{3} d x} = \frac{i_{n} k_{n} x^{4}}{4}+C$$

$$$\int i_{n} k_{n} x^{3}\, dx = \frac{i_{n} k_{n} x^{4}}{4} + C$$$A