Funktion $$$120040 - \frac{6002 x}{5}$$$ integraali

Määritä $$$\int \left(120040 - \frac{6002 x}{5}\right)\, dx$$$.

Integroi termi kerrallaan:

$${\color{red}{\int{\left(120040 - \frac{6002 x}{5}\right)d x}}} = {\color{red}{\left(\int{120040 d x} - \int{\frac{6002 x}{5} d x}\right)}}$$

Sovella vakiosääntöä $$$\int c\, dx = c x$$$ käyttäen $$$c=120040$$$:

$$- \int{\frac{6002 x}{5} d x} + {\color{red}{\int{120040 d x}}} = - \int{\frac{6002 x}{5} d x} + {\color{red}{\left(120040 x\right)}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{6002}{5}$$$ ja $$$f{\left(x \right)} = x$$$:

$$120040 x - {\color{red}{\int{\frac{6002 x}{5} d x}}} = 120040 x - {\color{red}{\left(\frac{6002 \int{x d x}}{5}\right)}}$$

Sovella potenssisääntöä $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ käyttäen $$$n=1$$$:

$$120040 x - \frac{6002 {\color{red}{\int{x d x}}}}{5}=120040 x - \frac{6002 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{5}=120040 x - \frac{6002 {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{5}$$

Näin ollen,

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = - \frac{3001 x^{2}}{5} + 120040 x$$

Sievennä:

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = \frac{3001 x \left(200 - x\right)}{5}$$

Lisää integrointivakio:

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = \frac{3001 x \left(200 - x\right)}{5}+C$$

$$$\int \left(120040 - \frac{6002 x}{5}\right)\, dx = \frac{3001 x \left(200 - x\right)}{5} + C$$$A