Integral de $$$e^{- 2 n}$$$

Encontre $$$\int e^{- 2 n}\, dn$$$.

Seja $$$u=- 2 n$$$.

Então $$$du=\left(- 2 n\right)^{\prime }dn = - 2 dn$$$ (veja os passos »), e obtemos $$$dn = - \frac{du}{2}$$$.

A integral pode ser reescrita como

$${\color{red}{\int{e^{- 2 n} d n}}} = {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{2}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{2}\right)}}$$

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{2} = - \frac{{\color{red}{e^{u}}}}{2}$$

Recorde que $$$u=- 2 n$$$:

$$- \frac{e^{{\color{red}{u}}}}{2} = - \frac{e^{{\color{red}{\left(- 2 n\right)}}}}{2}$$

Portanto,

$$\int{e^{- 2 n} d n} = - \frac{e^{- 2 n}}{2}$$

Adicione a constante de integração:

$$\int{e^{- 2 n} d n} = - \frac{e^{- 2 n}}{2}+C$$

$$$\int e^{- 2 n}\, dn = - \frac{e^{- 2 n}}{2} + C$$$A