Ensino Superior ⇒ Integral Definida

[ailon]

Galera, qual o valor dessa integral definida: [tex3]\int\limits_{5}^{8}\frac{dy}{\sqrt{3y+1}} dx[/tex3] obrigado.

[Rafa2604]

[tex3]\;\; \int\limits_{5}^{8}\frac{1}{\sqrt{3y+1}} dy \; \rightarrow \; \; u = 3y+1 \;\; du = 3dy \;\; u = 3.5+1 = 16 \;\; u = 3.8+1 = 25 \; \; dy = \frac{1}{3}du \;\; \\ \;\; \rightarrow \; \int\limits_{16}^{25}\frac{1}{\sqrt{u}}\; \frac{1}{3} du = \frac{1}{3} \; \int\limits_{16}^{25} u ^{-\frac{1}{2}}\; du = \frac{1}{3} \left[ \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{16}^{25} = \frac{2}{3} [ \sqrt{u} ]_{16}^{25} = \frac{2}{3}[\sqrt{25}-\sqrt{16}] = \frac{2}{3}[5-4] = \frac{2}{3}[/tex3]