Integraal van $$$e^{- t \left(a + s\right)}$$$ met betrekking tot $$$t$$$

Bepaal $$$\int e^{- t \left(a + s\right)}\, dt$$$.

Zij $$$u=- t \left(a + s\right)$$$.

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

Dus,

$${\color{red}{\int{e^{- t \left(a + s\right)} d t}}} = {\color{red}{\int{\frac{e^{u}}{- a - s} d u}}}$$

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

$${\color{red}{\int{\frac{e^{u}}{- a - s} d u}}} = {\color{red}{\frac{\int{e^{u} d u}}{- a - s}}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{- a - s} = \frac{{\color{red}{e^{u}}}}{- a - s}$$

We herinneren eraan dat $$$u=- t \left(a + s\right)$$$:

$$\frac{e^{{\color{red}{u}}}}{- a - s} = \frac{e^{{\color{red}{\left(- t \left(a + s\right)\right)}}}}{- a - s}$$

Dus,

$$\int{e^{- t \left(a + s\right)} d t} = \frac{e^{- t \left(a + s\right)}}{- a - s}$$

Vereenvoudig:

$$\int{e^{- t \left(a + s\right)} d t} = - \frac{e^{- t \left(a + s\right)}}{a + s}$$

Voeg de integratieconstante toe:

$$\int{e^{- t \left(a + s\right)} d t} = - \frac{e^{- t \left(a + s\right)}}{a + s}+C$$

$$$\int e^{- t \left(a + s\right)}\, dt = - \frac{e^{- t \left(a + s\right)}}{a + s} + C$$$A