Integral de $$$\sqrt{4 - 2 t}$$$

Halla $$$\int \sqrt{4 - 2 t}\, dt$$$.

Sea $$$u=4 - 2 t$$$.

Entonces $$$du=\left(4 - 2 t\right)^{\prime }dt = - 2 dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = - \frac{du}{2}$$$.

Por lo tanto,

$${\color{red}{\int{\sqrt{4 - 2 t} d t}}} = {\color{red}{\int{\left(- \frac{\sqrt{u}}{2}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(u \right)} = \sqrt{u}$$$:

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

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

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

Recordemos que $$$u=4 - 2 t$$$:

$$- \frac{{\color{red}{u}}^{\frac{3}{2}}}{3} = - \frac{{\color{red}{\left(4 - 2 t\right)}}^{\frac{3}{2}}}{3}$$

Por lo tanto,

$$\int{\sqrt{4 - 2 t} d t} = - \frac{\left(4 - 2 t\right)^{\frac{3}{2}}}{3}$$

Simplificar:

$$\int{\sqrt{4 - 2 t} d t} = - \frac{2 \sqrt{2} \left(2 - t\right)^{\frac{3}{2}}}{3}$$

Añade la constante de integración:

$$\int{\sqrt{4 - 2 t} d t} = - \frac{2 \sqrt{2} \left(2 - t\right)^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{4 - 2 t}\, dt = - \frac{2 \sqrt{2} \left(2 - t\right)^{\frac{3}{2}}}{3} + C$$$A