Integral de $$$\frac{48 t^{2}}{e^{5}}$$$

Halla $$$\int \frac{48 t^{2}}{e^{5}}\, dt$$$.

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{48}{e^{5}}$$$ y $$$f{\left(t \right)} = t^{2}$$$:

$${\color{red}{\int{\frac{48 t^{2}}{e^{5}} d t}}} = {\color{red}{\left(\frac{48 \int{t^{2} d t}}{e^{5}}\right)}}$$

Aplica la regla de la potencia $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$\frac{48 {\color{red}{\int{t^{2} d t}}}}{e^{5}}=\frac{48 {\color{red}{\frac{t^{1 + 2}}{1 + 2}}}}{e^{5}}=\frac{48 {\color{red}{\left(\frac{t^{3}}{3}\right)}}}{e^{5}}$$

Por lo tanto,

$$\int{\frac{48 t^{2}}{e^{5}} d t} = \frac{16 t^{3}}{e^{5}}$$

Añade la constante de integración:

$$\int{\frac{48 t^{2}}{e^{5}} d t} = \frac{16 t^{3}}{e^{5}}+C$$

$$$\int \frac{48 t^{2}}{e^{5}}\, dt = \frac{16 t^{3}}{e^{5}} + C$$$A