Calcule a integral dupla f(x,y)= xy³ dydx nos intervalos (1,2) e (0, 2x)

Neste exercício, será calculada a seguinte integral:

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = \int \limits_1^2 \int \limits _0^{2x}xy^3\,dy\, dx\)

Integrando primeiro em \(dy\), tem-se o seguinte:

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = \int \limits_1^2x\Bigg ( \int \limits _0^{2x}y^3\,dy \Bigg ) \, dx\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = \int \limits_1^2x\Bigg ( {1 \over 4}y^4 \bigg | _0^{2x} \Bigg ) \, dx\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = {1 \over 4}\int \limits_1^2x\Big ( (2x)^4 -0^4 \Big ) \, dx\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = {2^4 \over 4}\int \limits_1^2x (x)^4 \, dx\)

Agora, integrando em \(dx\), o resultado é

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = 4\int \limits_1^2x^5 \, dx\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = 4 \Big ({1 \over 6}x^6 \bigg |_1^2 \Big )\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = {4 \over 6}\Big (2^6 - 1^6 \Big )\)

\(\Longrightarrow \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = {2 \over 3}\Big (64 - 1 \Big )\)

\(\Longrightarrow \fbox {$ \int \limits_1^2 \int \limits _0^{2x}f(x,y)\,dy\, dx = 42 $}\)