Integraal van $$$- t^{2} + 2 z^{20}$$$ met betrekking tot $$$t$$$

Bepaal $$$\int \left(- t^{2} + 2 z^{20}\right)\, dt$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(- t^{2} + 2 z^{20}\right)d t}}} = {\color{red}{\left(- \int{t^{2} d t} + \int{2 z^{20} d t}\right)}}$$

Pas de machtsregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

$$\int{2 z^{20} d t} - {\color{red}{\int{t^{2} d t}}}=\int{2 z^{20} d t} - {\color{red}{\frac{t^{1 + 2}}{1 + 2}}}=\int{2 z^{20} d t} - {\color{red}{\left(\frac{t^{3}}{3}\right)}}$$

Pas de constantenregel $$$\int c\, dt = c t$$$ toe met $$$c=2 z^{20}$$$:

$$- \frac{t^{3}}{3} + {\color{red}{\int{2 z^{20} d t}}} = - \frac{t^{3}}{3} + {\color{red}{\left(2 t z^{20}\right)}}$$

Dus,

$$\int{\left(- t^{2} + 2 z^{20}\right)d t} = - \frac{t^{3}}{3} + 2 t z^{20}$$

Vereenvoudig:

$$\int{\left(- t^{2} + 2 z^{20}\right)d t} = \frac{t \left(- t^{2} + 6 z^{20}\right)}{3}$$

Voeg de integratieconstante toe:

$$\int{\left(- t^{2} + 2 z^{20}\right)d t} = \frac{t \left(- t^{2} + 6 z^{20}\right)}{3}+C$$

$$$\int \left(- t^{2} + 2 z^{20}\right)\, dt = \frac{t \left(- t^{2} + 6 z^{20}\right)}{3} + C$$$A