Integralen av $$$- y^{2} + y + 12$$$

Bestäm $$$\int \left(- y^{2} + y + 12\right)\, dy$$$.

Integrera termvis:

$${\color{red}{\int{\left(- y^{2} + y + 12\right)d y}}} = {\color{red}{\left(\int{12 d y} + \int{y d y} - \int{y^{2} d y}\right)}}$$

Tillämpa konstantregeln $$$\int c\, dy = c y$$$ med $$$c=12$$$:

$$\int{y d y} - \int{y^{2} d y} + {\color{red}{\int{12 d y}}} = \int{y d y} - \int{y^{2} d y} + {\color{red}{\left(12 y\right)}}$$

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

$$12 y - \int{y^{2} d y} + {\color{red}{\int{y d y}}}=12 y - \int{y^{2} d y} + {\color{red}{\frac{y^{1 + 1}}{1 + 1}}}=12 y - \int{y^{2} d y} + {\color{red}{\left(\frac{y^{2}}{2}\right)}}$$

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

$$\frac{y^{2}}{2} + 12 y - {\color{red}{\int{y^{2} d y}}}=\frac{y^{2}}{2} + 12 y - {\color{red}{\frac{y^{1 + 2}}{1 + 2}}}=\frac{y^{2}}{2} + 12 y - {\color{red}{\left(\frac{y^{3}}{3}\right)}}$$

Alltså,

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

Förenkla:

$$\int{\left(- y^{2} + y + 12\right)d y} = \frac{y \left(- 2 y^{2} + 3 y + 72\right)}{6}$$

Lägg till integrationskonstanten:

$$\int{\left(- y^{2} + y + 12\right)d y} = \frac{y \left(- 2 y^{2} + 3 y + 72\right)}{6}+C$$

$$$\int \left(- y^{2} + y + 12\right)\, dy = \frac{y \left(- 2 y^{2} + 3 y + 72\right)}{6} + C$$$A