Integraal van $$$e^{5 t}$$$

Bepaal $$$\int e^{5 t}\, dt$$$.

Zij $$$u=5 t$$$.

Dan $$$du=\left(5 t\right)^{\prime }dt = 5 dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = \frac{du}{5}$$$.

Dus,

$${\color{red}{\int{e^{5 t} d t}}} = {\color{red}{\int{\frac{e^{u}}{5} d u}}}$$

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

$${\color{red}{\int{\frac{e^{u}}{5} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{5}\right)}}$$

De integraal van de exponentiële functie is $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{{\color{red}{\int{e^{u} d u}}}}{5} = \frac{{\color{red}{e^{u}}}}{5}$$

We herinneren eraan dat $$$u=5 t$$$:

$$\frac{e^{{\color{red}{u}}}}{5} = \frac{e^{{\color{red}{\left(5 t\right)}}}}{5}$$

Dus,

$$\int{e^{5 t} d t} = \frac{e^{5 t}}{5}$$

Voeg de integratieconstante toe:

$$\int{e^{5 t} d t} = \frac{e^{5 t}}{5}+C$$

$$$\int e^{5 t}\, dt = \frac{e^{5 t}}{5} + C$$$A