Integral de $$$e^{- t \left(a + s\right)}$$$ con respecto a $$$t$$$

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

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

Entonces $$$du=\left(- t \left(a + s\right)\right)^{\prime }dt = - (a + s) dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = - \frac{du}{a + s}$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{- a - s}$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

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

Recordemos que $$$u=- t \left(a + s\right)$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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