Integral de $$$e^{2 t}$$$

Encontre $$$\int e^{2 t}\, dt$$$.

Seja $$$u=2 t$$$.

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

A integral pode ser reescrita como

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

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 $$$f{\left(u \right)} = e^{u}$$$:

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

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 $$$u=2 t$$$:

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

Portanto,

$$\int{e^{2 t} d t} = \frac{e^{2 t}}{2}$$

Adicione a constante de integração:

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

$$$\int e^{2 t}\, dt = \frac{e^{2 t}}{2} + C$$$A