Integral de $$$\sqrt{2 - 3 x}$$$

Halla $$$\int \sqrt{2 - 3 x}\, dx$$$.

Sea $$$u=2 - 3 x$$$.

Entonces $$$du=\left(2 - 3 x\right)^{\prime }dx = - 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - \frac{du}{3}$$$.

La integral se convierte en

$${\color{red}{\int{\sqrt{2 - 3 x} d x}}} = {\color{red}{\int{\left(- \frac{\sqrt{u}}{3}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{3}$$$ y $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\left(- \frac{\sqrt{u}}{3}\right)d u}}} = {\color{red}{\left(- \frac{\int{\sqrt{u} d u}}{3}\right)}}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=\frac{1}{2}$$$:

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

Recordemos que $$$u=2 - 3 x$$$:

$$- \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{9} = - \frac{2 {\color{red}{\left(2 - 3 x\right)}}^{\frac{3}{2}}}{9}$$

Por lo tanto,

$$\int{\sqrt{2 - 3 x} d x} = - \frac{2 \left(2 - 3 x\right)^{\frac{3}{2}}}{9}$$

Añade la constante de integración:

$$\int{\sqrt{2 - 3 x} d x} = - \frac{2 \left(2 - 3 x\right)^{\frac{3}{2}}}{9}+C$$

$$$\int \sqrt{2 - 3 x}\, dx = - \frac{2 \left(2 - 3 x\right)^{\frac{3}{2}}}{9} + C$$$A