Mudança da ordem de integração

Sendo \(f: \mathbb{R^2} \to \mathbb{R} \) uma função positiva e integrável, o integral iterado \(\int_0^1\int_{\sqrt{1-x^2}}^1\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{y}\text{d}\text{x}\text{+}\int_1^2\int_{\sqrt{1-(x-2)^2}}^1\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{y}\text{d}\text{x}\) pode também ser dado, com ou sem mudança da ordem de integração por:

A)\(\int_0^1\int_{\sqrt{1-y^2}}^{2-\sqrt{1-y^2}}\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{x}\text{d}\text{y}\)

B)\(\int_0^1\int_{\sqrt{1-y^2}}^{2-\sqrt{1-y^2}}\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{y}\text{d}\text{x}\)

C)\(\int_1^0\int_{\sqrt{1-y^2}}^0\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{x}\text{d}\text{y}\text{+}\int_0^1\int_{2-\sqrt{1-y^2}}^2\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{x}\text{d}\text{y}\)

D)\(\int_1^0\int_0^{\sqrt{1-y^2}}\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{x}\text{d}\text{y}\text{+}\int_1^0\int_2^{2-\sqrt{1-y^2}}\text{f}\left(\begin{array}{c}x\\y\\\end{array}\right)\text{d}\text{x}\text{d}\text{y}\)

Para obter o zip que contém as instâncias deste exercício clique aqui(trocaOrdem)

Se deseja obter o código fonte que gera os exercícios contacte miguel.dziergwa@ist.utl.pt