Integral de $$$\sqrt{t}$$$

Encontre $$$\int \sqrt{t}\, dt$$$.

Aplique a regra da potência $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=\frac{1}{2}$$$:

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

Portanto,

$$\int{\sqrt{t} d t} = \frac{2 t^{\frac{3}{2}}}{3}$$

Adicione a constante de integração:

$$\int{\sqrt{t} d t} = \frac{2 t^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{t}\, dt = \frac{2 t^{\frac{3}{2}}}{3} + C$$$A