Integraal van $$$\frac{3 x^{3}}{4} - x$$$

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

Integreer termgewijs:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{3}{4}$$$ en $$$f{\left(x \right)} = x^{3}$$$:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=3$$$:

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

Dus,

$$\int{\left(\frac{3 x^{3}}{4} - x\right)d x} = \frac{3 x^{4}}{16} - \frac{x^{2}}{2}$$

Vereenvoudig:

$$\int{\left(\frac{3 x^{3}}{4} - x\right)d x} = \frac{x^{2} \left(3 x^{2} - 8\right)}{16}$$

Voeg de integratieconstante toe:

$$\int{\left(\frac{3 x^{3}}{4} - x\right)d x} = \frac{x^{2} \left(3 x^{2} - 8\right)}{16}+C$$

$$$\int \left(\frac{3 x^{3}}{4} - x\right)\, dx = \frac{x^{2} \left(3 x^{2} - 8\right)}{16} + C$$$A