Integraal van $$$f x - g x$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \left(f x - g x\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(f x - g x\right)d x}}} = {\color{red}{\left(\int{f x d x} - \int{g x d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=f$$$ en $$$f{\left(x \right)} = x$$$:

$$- \int{g x d x} + {\color{red}{\int{f x d x}}} = - \int{g x d x} + {\color{red}{f \int{x d x}}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=g$$$ en $$$f{\left(x \right)} = x$$$:

$$\frac{f x^{2}}{2} - {\color{red}{\int{g x d x}}} = \frac{f x^{2}}{2} - {\color{red}{g \int{x d x}}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Dus,

$$\int{\left(f x - g x\right)d x} = \frac{f x^{2}}{2} - \frac{g x^{2}}{2}$$

Vereenvoudig:

$$\int{\left(f x - g x\right)d x} = \frac{x^{2} \left(f - g\right)}{2}$$

Voeg de integratieconstante toe:

$$\int{\left(f x - g x\right)d x} = \frac{x^{2} \left(f - g\right)}{2}+C$$

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