Integral de $$$- \frac{\pi d \theta}{8}$$$ em relação a $$$d$$$

Encontre $$$\int \left(- \frac{\pi d \theta}{8}\right)\, dd$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(d \right)}\, dd = c \int f{\left(d \right)}\, dd$$$ usando $$$c=- \frac{\pi \theta}{8}$$$ e $$$f{\left(d \right)} = d$$$:

$${\color{red}{\int{\left(- \frac{\pi d \theta}{8}\right)d d}}} = {\color{red}{\left(- \frac{\pi \theta \int{d d d}}{8}\right)}}$$

Aplique a regra da potência $$$\int d^{n}\, dd = \frac{d^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

$$- \frac{\pi \theta {\color{red}{\int{d d d}}}}{8}=- \frac{\pi \theta {\color{red}{\frac{d^{1 + 1}}{1 + 1}}}}{8}=- \frac{\pi \theta {\color{red}{\left(\frac{d^{2}}{2}\right)}}}{8}$$

Portanto,

$$\int{\left(- \frac{\pi d \theta}{8}\right)d d} = - \frac{\pi d^{2} \theta}{16}$$

Adicione a constante de integração:

$$\int{\left(- \frac{\pi d \theta}{8}\right)d d} = - \frac{\pi d^{2} \theta}{16}+C$$

$$$\int \left(- \frac{\pi d \theta}{8}\right)\, dd = - \frac{\pi d^{2} \theta}{16} + C$$$A