Integraal van $$$9 \sqrt{2} t^{16}$$$

Bepaal $$$\int 9 \sqrt{2} t^{16}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=9 \sqrt{2}$$$ en $$$f{\left(t \right)} = t^{16}$$$:

$${\color{red}{\int{9 \sqrt{2} t^{16} d t}}} = {\color{red}{\left(9 \sqrt{2} \int{t^{16} d t}\right)}}$$

Pas de machtsregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=16$$$:

$$9 \sqrt{2} {\color{red}{\int{t^{16} d t}}}=9 \sqrt{2} {\color{red}{\frac{t^{1 + 16}}{1 + 16}}}=9 \sqrt{2} {\color{red}{\left(\frac{t^{17}}{17}\right)}}$$

Dus,

$$\int{9 \sqrt{2} t^{16} d t} = \frac{9 \sqrt{2} t^{17}}{17}$$

Voeg de integratieconstante toe:

$$\int{9 \sqrt{2} t^{16} d t} = \frac{9 \sqrt{2} t^{17}}{17}+C$$

$$$\int 9 \sqrt{2} t^{16}\, dt = \frac{9 \sqrt{2} t^{17}}{17} + C$$$A