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

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=4$$$:

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

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

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

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

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

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

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

Alltså,

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

Förenkla:

$$\int{\left(- x^{2} - 3 x + 4\right)d x} = \frac{x \left(- 2 x^{2} - 9 x + 24\right)}{6}$$

Lägg till integrationskonstanten:

$$\int{\left(- x^{2} - 3 x + 4\right)d x} = \frac{x \left(- 2 x^{2} - 9 x + 24\right)}{6}+C$$

$$$\int \left(- x^{2} - 3 x + 4\right)\, dx = \frac{x \left(- 2 x^{2} - 9 x + 24\right)}{6} + C$$$A