Integraal van $$$2 x^{3} - 8 x - 1$$$

Bepaal $$$\int \left(2 x^{3} - 8 x - 1\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(2 x^{3} - 8 x - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} - \int{8 x d x} + \int{2 x^{3} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$- \int{8 x d x} + \int{2 x^{3} d x} - {\color{red}{\int{1 d x}}} = - \int{8 x d x} + \int{2 x^{3} d x} - {\color{red}{x}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=8$$$ en $$$f{\left(x \right)} = x$$$:

$$- x + \int{2 x^{3} d x} - {\color{red}{\int{8 x d x}}} = - x + \int{2 x^{3} d x} - {\color{red}{\left(8 \int{x d x}\right)}}$$

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

$$- x + \int{2 x^{3} d x} - 8 {\color{red}{\int{x d x}}}=- x + \int{2 x^{3} d x} - 8 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- x + \int{2 x^{3} d x} - 8 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = x^{3}$$$:

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

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

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

Dus,

$$\int{\left(2 x^{3} - 8 x - 1\right)d x} = \frac{x^{4}}{2} - 4 x^{2} - x$$

Vereenvoudig:

$$\int{\left(2 x^{3} - 8 x - 1\right)d x} = \frac{x \left(x^{3} - 8 x - 2\right)}{2}$$

Voeg de integratieconstante toe:

$$\int{\left(2 x^{3} - 8 x - 1\right)d x} = \frac{x \left(x^{3} - 8 x - 2\right)}{2}+C$$

$$$\int \left(2 x^{3} - 8 x - 1\right)\, dx = \frac{x \left(x^{3} - 8 x - 2\right)}{2} + C$$$A