Integral de $$$\frac{25 t}{e^{\frac{1}{20}}}$$$

Halla $$$\int \frac{25 t}{e^{\frac{1}{20}}}\, dt$$$.

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{25}{e^{\frac{1}{20}}}$$$ y $$$f{\left(t \right)} = t$$$:

$${\color{red}{\int{\frac{25 t}{e^{\frac{1}{20}}} d t}}} = {\color{red}{\left(\frac{25 \int{t d t}}{e^{\frac{1}{20}}}\right)}}$$

Aplica la regla de la potencia $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

$$\frac{25 {\color{red}{\int{t d t}}}}{e^{\frac{1}{20}}}=\frac{25 {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}}{e^{\frac{1}{20}}}=\frac{25 {\color{red}{\left(\frac{t^{2}}{2}\right)}}}{e^{\frac{1}{20}}}$$

Por lo tanto,

$$\int{\frac{25 t}{e^{\frac{1}{20}}} d t} = \frac{25 t^{2}}{2 e^{\frac{1}{20}}}$$

Añade la constante de integración:

$$\int{\frac{25 t}{e^{\frac{1}{20}}} d t} = \frac{25 t^{2}}{2 e^{\frac{1}{20}}}+C$$

$$$\int \frac{25 t}{e^{\frac{1}{20}}}\, dt = \frac{25 t^{2}}{2 e^{\frac{1}{20}}} + C$$$A