Integralen av $$$\frac{1}{16 y^{2}}$$$

Bestäm $$$\int \frac{1}{16 y^{2}}\, dy$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ med $$$c=\frac{1}{16}$$$ och $$$f{\left(y \right)} = \frac{1}{y^{2}}$$$:

$${\color{red}{\int{\frac{1}{16 y^{2}} d y}}} = {\color{red}{\left(\frac{\int{\frac{1}{y^{2}} d y}}{16}\right)}}$$

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

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

Alltså,

$$\int{\frac{1}{16 y^{2}} d y} = - \frac{1}{16 y}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{16 y^{2}} d y} = - \frac{1}{16 y}+C$$

$$$\int \frac{1}{16 y^{2}}\, dy = - \frac{1}{16 y} + C$$$A