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

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

Sea $$$u=2 t$$$.

Entonces $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = \frac{du}{2}$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

Recordemos que $$$u=2 t$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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