Integralen av $$$\frac{1}{625} - x^{2}$$$

Bestäm $$$\int \left(\frac{1}{625} - x^{2}\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(\frac{1}{625} - x^{2}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{625} d x} - \int{x^{2} d x}\right)}}$$

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=\frac{1}{625}$$$:

$$- \int{x^{2} d x} + {\color{red}{\int{\frac{1}{625} d x}}} = - \int{x^{2} d x} + {\color{red}{\left(\frac{x}{625}\right)}}$$

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

$$\frac{x}{625} - {\color{red}{\int{x^{2} d x}}}=\frac{x}{625} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x}{625} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Alltså,

$$\int{\left(\frac{1}{625} - x^{2}\right)d x} = - \frac{x^{3}}{3} + \frac{x}{625}$$

Lägg till integrationskonstanten:

$$\int{\left(\frac{1}{625} - x^{2}\right)d x} = - \frac{x^{3}}{3} + \frac{x}{625}+C$$

$$$\int \left(\frac{1}{625} - x^{2}\right)\, dx = \left(- \frac{x^{3}}{3} + \frac{x}{625}\right) + C$$$A