Integral of $$$\frac{t}{e^{4}}$$$

Find $$$\int \frac{t}{e^{4}}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=e^{-4}$$$ and $$$f{\left(t \right)} = t$$$:

$${\color{red}{\int{\frac{t}{e^{4}} d t}}} = {\color{red}{\frac{\int{t d t}}{e^{4}}}}$$

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{{\color{red}{\int{t d t}}}}{e^{4}}=\frac{{\color{red}{\frac{t^{1 + 1}}{1 + 1}}}}{e^{4}}=\frac{{\color{red}{\left(\frac{t^{2}}{2}\right)}}}{e^{4}}$$

Therefore,

$$\int{\frac{t}{e^{4}} d t} = \frac{t^{2}}{2 e^{4}}$$

Add the constant of integration:

$$\int{\frac{t}{e^{4}} d t} = \frac{t^{2}}{2 e^{4}}+C$$

$$$\int \frac{t}{e^{4}}\, dt = \frac{t^{2}}{2 e^{4}} + C$$$A