Question

integral de ∫_0^1▒x^3/√(1+x^2 ) a resposta e 2/3 - √2/3

Answer 1

Olá Odin 

Basta utilizar a substituição para a resolução da integral:

$\int\limits^1_0 { \frac{x^3}{ \sqrt{1+x^2} } } \, dx \ \ \ \boxed{u=1+x^2} \ \ \ \boxed{du=2x.dx}\ \ \ \ \boxed{x^2=u-1} \\ \\ \int\limits^1_0 { \frac{(u-1).x}{ \sqrt{u} } } \, \frac{du}{2x} \\ \\ \frac{1}{2} \int\limits^1_0 { \frac{u-1}{ \sqrt{u} } } \, du \\ \\ \frac{1}{2} \int\limits^1_0 {( \frac{u}{ \sqrt{u} }- \frac{1}{ \sqrt{u} } }) \, du \\ \\ \frac{1}{2} \int\limits^1_0 {( u^{ \frac{1}{2} }- u^{- \frac{1}{2} } }) \, du$

$\frac{1}{2} \frac{2}{3}.u^{ \frac{3}{2} } - \frac{1}{2}. \frac{2}{1}u^{ \frac{1}{2} } \|_0^1 \\ \\ \frac{1}{3} \sqrt{u^{3}} - \sqrt{u} \|_0^1 \\ \\ \frac{1}{3} \sqrt{(x^2+1)^{3}} - \sqrt{x^2+1} \|_0^1 \\ \\( \frac{1}{3} \sqrt{(1^2+1)^{3}} - \sqrt{1^2+1} ) -(\frac{1}{3} \sqrt{(0^2+1)^{3}} - \sqrt{0^2+1} ) \\ \\ ( \frac{1}{3} 2\sqrt{2} - \sqrt{2} ) -(\frac{1}{3} - 1 ) \\ \\ \boxed{ -\frac{ \sqrt{2} }{3} + \frac{2}{3} }$


Comenta aí depois o que achou !