Question

calculo diferencial e integral II

Answer 1

Resposta:

$\int\limits^1_4 {2\sqrt{x} } \, dx = 2\int\limits^1_4 {\sqrt{x} } \, dx = 2\int\limits^1_4 {x^{\frac{1}{2}} } \, dx = 2 \frac{x^{\frac{3}{2}} }{\frac{3}{2} }|_{1}^{4} = \frac{4}{3}x^{\frac{3}{2}} |_{1}^{4} = \frac{4}{3}(4^{\frac{3}{2} } - 1^{\frac{3}{2} } )= \frac{4}{3}(8 - 1) = \frac{28}{3}$