Integralen av $$$- 2 x^{2} + 16 x$$$

Bestäm $$$\int \left(- 2 x^{2} + 16 x\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(- 2 x^{2} + 16 x\right)d x}}} = {\color{red}{\left(\int{16 x d x} - \int{2 x^{2} d x}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=2$$$ och $$$f{\left(x \right)} = x^{2}$$$:

$$\int{16 x d x} - {\color{red}{\int{2 x^{2} d x}}} = \int{16 x d x} - {\color{red}{\left(2 \int{x^{2} d x}\right)}}$$

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

$$\int{16 x d x} - 2 {\color{red}{\int{x^{2} d x}}}=\int{16 x d x} - 2 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\int{16 x d x} - 2 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=16$$$ och $$$f{\left(x \right)} = x$$$:

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

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

$$- \frac{2 x^{3}}{3} + 16 {\color{red}{\int{x d x}}}=- \frac{2 x^{3}}{3} + 16 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \frac{2 x^{3}}{3} + 16 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Alltså,

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

Förenkla:

$$\int{\left(- 2 x^{2} + 16 x\right)d x} = \frac{2 x^{2} \left(12 - x\right)}{3}$$

Lägg till integrationskonstanten:

$$\int{\left(- 2 x^{2} + 16 x\right)d x} = \frac{2 x^{2} \left(12 - x\right)}{3}+C$$

$$$\int \left(- 2 x^{2} + 16 x\right)\, dx = \frac{2 x^{2} \left(12 - x\right)}{3} + C$$$A