Integralen av $$$\frac{\left(y - 4\right)^{2}}{3}$$$

Bestäm $$$\int \frac{\left(y - 4\right)^{2}}{3}\, dy$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(y \right)} = \left(y - 4\right)^{2}$$$:

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

Låt $$$u=y - 4$$$ vara.

Då $$$du=\left(y - 4\right)^{\prime }dy = 1 dy$$$ (stegen kan ses »), och vi har att $$$dy = du$$$.

Integralen blir

$$\frac{{\color{red}{\int{\left(y - 4\right)^{2} d y}}}}{3} = \frac{{\color{red}{\int{u^{2} d u}}}}{3}$$

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

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

Kom ihåg att $$$u=y - 4$$$:

$$\frac{{\color{red}{u}}^{3}}{9} = \frac{{\color{red}{\left(y - 4\right)}}^{3}}{9}$$

Alltså,

$$\int{\frac{\left(y - 4\right)^{2}}{3} d y} = \frac{\left(y - 4\right)^{3}}{9}$$

Lägg till integrationskonstanten:

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

$$$\int \frac{\left(y - 4\right)^{2}}{3}\, dy = \frac{\left(y - 4\right)^{3}}{9} + C$$$A