Question

qual o valor da area destacada na figura abaixo;

Answer 1

Boa noite João!

Solução!

$A=\displaystyle \int_{1}^{4} (\sqrt{x})dx\\\\\\\\\\ A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}} dx\\\\\\\\\\ A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\\ A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\ A=\frac{( x)^{ \frac{3}{2}} }{ \frac{3}{2}}~~\bigg|_{1}^{4} \\\\\\\\\\ A= \frac{ \sqrt{x^{3} } }{ \frac{3}{2} } ~~\bigg|_{1}^{4}$


$A= \frac{2}{3} \sqrt{ x^{3} } ~~\bigg|_{1}^{4} \\\\\\\ A= (\frac{2}{3} \sqrt{ x^{3} })-(\frac{2}{3} \sqrt{ x^{3} }) \\\\\\\ A= (\frac{2}{3} \sqrt{(4)^{3} })-(\frac{2}{3} \sqrt{(1)^{3} })\\\\\\\ A= (\frac{2}{3} \sqrt{64 })-(\frac{2}{3} \sqrt{1})\\\\\\\ A= (\frac{2}{3}.8)-(\frac{2}{3}.1)\\\\\\\ A= (\frac{16}{3})-(\frac{2}{3})\\\\\\\ A= (\frac{16-2}{3})\\\\\\\ A= \frac{14}{3}u.a$


$\boxed{Resposta:A= \frac{14}{3}u.a~~\boxed{Alternativa~~b}}$

Boa noite!
Bons estudos!