Integral of $$$14 - 4 r^{2}$$$

Find $$$\int \left(14 - 4 r^{2}\right)\, dr$$$.

Integrate term by term:

$${\color{red}{\int{\left(14 - 4 r^{2}\right)d r}}} = {\color{red}{\left(\int{14 d r} - \int{4 r^{2} d r}\right)}}$$

Apply the constant rule $$$\int c\, dr = c r$$$ with $$$c=14$$$:

$$- \int{4 r^{2} d r} + {\color{red}{\int{14 d r}}} = - \int{4 r^{2} d r} + {\color{red}{\left(14 r\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ with $$$c=4$$$ and $$$f{\left(r \right)} = r^{2}$$$:

$$14 r - {\color{red}{\int{4 r^{2} d r}}} = 14 r - {\color{red}{\left(4 \int{r^{2} d r}\right)}}$$

Apply the power rule $$$\int r^{n}\, dr = \frac{r^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$14 r - 4 {\color{red}{\int{r^{2} d r}}}=14 r - 4 {\color{red}{\frac{r^{1 + 2}}{1 + 2}}}=14 r - 4 {\color{red}{\left(\frac{r^{3}}{3}\right)}}$$

Therefore,

$$\int{\left(14 - 4 r^{2}\right)d r} = - \frac{4 r^{3}}{3} + 14 r$$

Add the constant of integration:

$$\int{\left(14 - 4 r^{2}\right)d r} = - \frac{4 r^{3}}{3} + 14 r+C$$

$$$\int \left(14 - 4 r^{2}\right)\, dr = \left(- \frac{4 r^{3}}{3} + 14 r\right) + C$$$A