Integral iterada

por **JEOVA** » Sex Fev 01, 2013 01:45

Calcule a integral iterada
1) $\int_{2}^{4}\int_{-1}^{1}\left ( x^{2}+y^{2} \right )dydx 2)\int_{0}^{2}\int_{0}^{1}\left ( 2x+y)^{8} \right dxdy$

por **DanielFerreira** » Dom Fev 10, 2013 20:38

Jeova,
seja bem-vindo!
Procure postar apenas uma questão por tópico e explanar suas dúvidas/tentativas, ok?!

$\\ \int_{2}^{4}\int_{- 1}^{1}(x^2 + y^2) \; dydx = \\\\\\ \int_{2}^{4}\left [ x^2y + \frac{y^3}{3} \right ]_{- 1}^{1} \; dx = \\\\\\ \begin{cases} F(1) = x^2 \cdot 1 + \frac{1^3}{3} \Rightarrow \boxed{x^2 + \frac{1}{3}}\\\\ F(- 1) = x^2 \cdot (- 1) + \frac{(- 1)^3}{3} \Rightarrow \boxed{- x^2 - \frac{1}{3}}\end{cases} \\\\\\ \boxed{\boxed{F(1) - F(- 1) = 2x^2 + \frac{2}{3}}} \\\\\\ \int_{2}^{4}\left (2x^2 + \frac{2}{3} \right )dx = \\\\\\ \left [ \frac{2x^3}{3} + \frac{2x}{3} \right ]_{2}^{4} = \\\\\\ \begin{cases} G(4) = \frac{2 \cdot (4)^3}{3} + \frac{2 \cdot 4}{3}\Rightarrow \boxed{\frac{128 + 8}{3}} \\\\ G(2) = \frac{2 \cdot (2)^3}{3} + \frac{2 \cdot 2}{3} \Rightarrow \boxed{\frac{16 + 4}{3}}\end{cases} \\\\\\ G(4) - G(2) = \frac{128 + 8 - 16 - 4}{3} \\\\\\ \boxed{\boxed{\boxed{G(4) - G(2) = \frac{116}{3}}}}$

Integral iterada

Integral iterada

Integral iterada

Re: Integral iterada

Quem está online