Integral de $$$9 e^{\frac{t}{2}}$$$

Halla $$$\int 9 e^{\frac{t}{2}}\, dt$$$.

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=9$$$ y $$$f{\left(t \right)} = e^{\frac{t}{2}}$$$:

$${\color{red}{\int{9 e^{\frac{t}{2}} d t}}} = {\color{red}{\left(9 \int{e^{\frac{t}{2}} d t}\right)}}$$

Sea $$$u=\frac{t}{2}$$$.

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

Por lo tanto,

$$9 {\color{red}{\int{e^{\frac{t}{2}} d t}}} = 9 {\color{red}{\int{2 e^{u} d u}}}$$

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

$$9 {\color{red}{\int{2 e^{u} d u}}} = 9 {\color{red}{\left(2 \int{e^{u} d u}\right)}}$$

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

$$18 {\color{red}{\int{e^{u} d u}}} = 18 {\color{red}{e^{u}}}$$

Recordemos que $$$u=\frac{t}{2}$$$:

$$18 e^{{\color{red}{u}}} = 18 e^{{\color{red}{\left(\frac{t}{2}\right)}}}$$

Por lo tanto,

$$\int{9 e^{\frac{t}{2}} d t} = 18 e^{\frac{t}{2}}$$

Añade la constante de integración:

$$\int{9 e^{\frac{t}{2}} d t} = 18 e^{\frac{t}{2}}+C$$

$$$\int 9 e^{\frac{t}{2}}\, dt = 18 e^{\frac{t}{2}} + C$$$A