Integral of $$$- \frac{3 x^{2}}{4}$$$

Find $$$\int \left(- \frac{3 x^{2}}{4}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{3}{4}$$$ and $$$f{\left(x \right)} = x^{2}$$$:

$${\color{red}{\int{\left(- \frac{3 x^{2}}{4}\right)d x}}} = {\color{red}{\left(- \frac{3 \int{x^{2} d x}}{4}\right)}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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