Integralen av $$$6 x \left(1 - x\right)$$$

Bestäm $$$\int 6 x \left(1 - x\right)\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=6$$$ och $$$f{\left(x \right)} = x \left(1 - x\right)$$$:

$${\color{red}{\int{6 x \left(1 - x\right) d x}}} = {\color{red}{\left(6 \int{x \left(1 - x\right) d x}\right)}}$$

Expand the expression:

$$6 {\color{red}{\int{x \left(1 - x\right) d x}}} = 6 {\color{red}{\int{\left(- x^{2} + x\right)d x}}}$$

Integrera termvis:

$$6 {\color{red}{\int{\left(- x^{2} + x\right)d x}}} = 6 {\color{red}{\left(\int{x d x} - \int{x^{2} d x}\right)}}$$

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

$$- 6 \int{x^{2} d x} + 6 {\color{red}{\int{x d x}}}=- 6 \int{x^{2} d x} + 6 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 6 \int{x^{2} d x} + 6 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

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

$$3 x^{2} - 6 {\color{red}{\int{x^{2} d x}}}=3 x^{2} - 6 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=3 x^{2} - 6 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Alltså,

$$\int{6 x \left(1 - x\right) d x} = - 2 x^{3} + 3 x^{2}$$

Förenkla:

$$\int{6 x \left(1 - x\right) d x} = x^{2} \left(3 - 2 x\right)$$

Lägg till integrationskonstanten:

$$\int{6 x \left(1 - x\right) d x} = x^{2} \left(3 - 2 x\right)+C$$

$$$\int 6 x \left(1 - x\right)\, dx = x^{2} \left(3 - 2 x\right) + C$$$A