Integral de $$$- r^{2} + 2 z^{2}$$$ con respecto a $$$r$$$

Halla $$$\int \left(- r^{2} + 2 z^{2}\right)\, dr$$$.

Integra término a término:

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

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

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

Aplica la regla de la constante $$$\int c\, dr = c r$$$ con $$$c=2 z^{2}$$$:

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

Por lo tanto,

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

Simplificar:

$$\int{\left(- r^{2} + 2 z^{2}\right)d r} = \frac{r \left(- r^{2} + 6 z^{2}\right)}{3}$$

Añade la constante de integración:

$$\int{\left(- r^{2} + 2 z^{2}\right)d r} = \frac{r \left(- r^{2} + 6 z^{2}\right)}{3}+C$$

$$$\int \left(- r^{2} + 2 z^{2}\right)\, dr = \frac{r \left(- r^{2} + 6 z^{2}\right)}{3} + C$$$A