Question

Alguem pode responder integral definida ?

Answer 1

$\int\limits^4_0 {x} \, dx \frac{1}{\sqrt{6x+1}}dx =$
$\frac{1}{6}\int \frac{1}{\sqrt{u}}du= \frac{1}{6}\int \:u^{-\frac{1}{2}}du= \frac{1}{6}\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}= \frac{1}{6}\frac{\left(6x+1\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}= \frac{1}{3}\sqrt{6x+1}+C$
$\lim _{x\to \:4-}\left(\frac{1}{3}\sqrt{6x+1}\right) = 5/3 \lim _{x\to \:0+}\left(\frac{1}{3}\sqrt{6x+1}\right) = 1/3$
então
$\int _0^4\frac{1}{\sqrt{6x+1}}dx=\frac{4}{3}$