Integral of $$$\frac{2 \epsilon r^{2}}{5}$$$ with respect to $$$\epsilon$$$

Find $$$\int \frac{2 \epsilon r^{2}}{5}\, d\epsilon$$$.

Apply the constant multiple rule $$$\int c f{\left(\epsilon \right)}\, d\epsilon = c \int f{\left(\epsilon \right)}\, d\epsilon$$$ with $$$c=\frac{2 r^{2}}{5}$$$ and $$$f{\left(\epsilon \right)} = \epsilon$$$:

$${\color{red}{\int{\frac{2 \epsilon r^{2}}{5} d \epsilon}}} = {\color{red}{\left(\frac{2 r^{2} \int{\epsilon d \epsilon}}{5}\right)}}$$

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

$$\frac{2 r^{2} {\color{red}{\int{\epsilon d \epsilon}}}}{5}=\frac{2 r^{2} {\color{red}{\frac{\epsilon^{1 + 1}}{1 + 1}}}}{5}=\frac{2 r^{2} {\color{red}{\left(\frac{\epsilon^{2}}{2}\right)}}}{5}$$

Therefore,

$$\int{\frac{2 \epsilon r^{2}}{5} d \epsilon} = \frac{\epsilon^{2} r^{2}}{5}$$

Add the constant of integration:

$$\int{\frac{2 \epsilon r^{2}}{5} d \epsilon} = \frac{\epsilon^{2} r^{2}}{5}+C$$

$$$\int \frac{2 \epsilon r^{2}}{5}\, d\epsilon = \frac{\epsilon^{2} r^{2}}{5} + C$$$A