Integralen av $$$- 4 x^{3} + x^{2}$$$

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

Integrera termvis:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=2$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = x^{3}$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=3$$$:

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

$$$\int \left(- 4 x^{3} + x^{2}\right)\, dx = x^{3} \left(\frac{1}{3} - x\right) + C$$$A