Integralen av $$$1 - \frac{x}{20}$$$

Bestäm $$$\int \left(1 - \frac{x}{20}\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(1 - \frac{x}{20}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\frac{x}{20} d x}\right)}}$$

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

$$- \int{\frac{x}{20} d x} + {\color{red}{\int{1 d x}}} = - \int{\frac{x}{20} d x} + {\color{red}{x}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{20}$$$ och $$$f{\left(x \right)} = x$$$:

$$x - {\color{red}{\int{\frac{x}{20} d x}}} = x - {\color{red}{\left(\frac{\int{x d x}}{20}\right)}}$$

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

$$x - \frac{{\color{red}{\int{x d x}}}}{20}=x - \frac{{\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{20}=x - \frac{{\color{red}{\left(\frac{x^{2}}{2}\right)}}}{20}$$

Alltså,

$$\int{\left(1 - \frac{x}{20}\right)d x} = - \frac{x^{2}}{40} + x$$

Förenkla:

$$\int{\left(1 - \frac{x}{20}\right)d x} = \frac{x \left(40 - x\right)}{40}$$

Lägg till integrationskonstanten:

$$\int{\left(1 - \frac{x}{20}\right)d x} = \frac{x \left(40 - x\right)}{40}+C$$

$$$\int \left(1 - \frac{x}{20}\right)\, dx = \frac{x \left(40 - x\right)}{40} + C$$$A