Question

Considere a seguinte integral:

Answer 1

Resposta:

26/3

Explicação passo a passo:

$\displaystyle\int_{0}^{4}\sqrt{2x+1} ~dx\\\\2x+1=u\\\\p/x=0\implies u=1\\\\p/x=4 \implies u=9\\\\2dx=du \implies dx=\frac{1}{2}du \\\\\displaystyle\int_{1}^{9}\sqrt{u} }*\frac{1}{2}du=\frac{1}{2} \displaystyle\int_{1}^{9}u^{\frac{1}{2} }=\frac{1}{2} *\frac{u^{\frac{3}{2} } }{\frac{3}{2} } ]{ {{9} \atop {1}} \right. =\frac{1}{2} *\frac{2}{3} \sqrt{u^3} \left \left ]{ {{9} \atop {1}} \right.=\frac{1}{3}(\sqrt{9^3-1^3} )=\\\\\frac{1}{3} (\sqrt{3^6-1})=\frac{1}{3} (3^3-1)$

$=\frac{1}{3}*26=\frac{26}{3}$