Calcular I= ∫∫ f(x,y)dxdy onde; f(x,y) = xe^xy ;R é o retangulo 1≤ x≤3, 0≤y≤1

Neste exercício, será calculada a seguinte integral:

\(\Longrightarrow \int \limits_0^1 \int \limits_1^3 f(x,y)\, dx \, dy = \int \limits_0^1 \int \limits_1^3 xe^{xy}\, dx \, dy\)

\(\Longrightarrow \int \limits_0^1 \int \limits_1^3 f(x,y)\, dx \, dy = \int \limits_1^3 x \Bigg ( \int \limits_0^1 e^{xy}\, dy \Bigg ) \, dx\)

                                  \( = \int \limits_1^3 x \Bigg ( {1 \over x} e^{xy}\bigg | _0^1 \Bigg ) \, dx\)

                                  \( = \int \limits_1^3 x {1 \over x}\Big ( e^{x\cdot 1} - e^{x\cdot 0} \Big ) \, dx\)

\(\Longrightarrow \int \limits_0^1 \int \limits_1^3 f(x,y)\, dx \, dy = \int \limits_1^3 \Big ( e^{x} - 1 \Big ) \, dx\)

Integrando em x, tem-se que:

\(\Longrightarrow \int \limits_0^1 \int \limits_1^3 f(x,y)\, dx \, dy = \Big ( e^{x} - x \Big ) \bigg| _1^3\)

                                  \( = \Big ( e^{3} - 3 \Big ) -\Big ( e^{1} - 1 \Big )\)

                                  \( = \Big (17,0855 \Big ) -\Big ( 1,7183 \Big )\)

\(\Longrightarrow \fbox {$ \int \limits_0^1 \int \limits_1^3 f(x,y)\, dx \, dy=15,3672 $}\)

0