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## Integral Doble

We integrate the function $-{{y^2}\over{16}}-{{x^2}\over{9}}+4$ on the set given by $0 \leq x \leq 3$ and $0 \leq y \leq 2$ . $$I=\iint_{M}f\,\mathrm{d}x\mathrm{d}y\ ,\qquad\qquad f= -{{y^2}\over{16}}-{{x^2}\over{9}}+4 \ , \qquad\qquad M=\begin{cases} 0 \leq x \leq 3 \\ 0 \leq y \leq 2 \end{cases}$$

$\displaystyle I=\int_{ 0 }^{ 3 }\int_{ 0 }^{ 2 } -{{y^2}\over{16}}-{{x^2}\over{9}}+4 \;\textrm{d}y\,\textrm{d}x$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 3 }\left[ -{{y^3}\over{48}}-{{x^2\,y}\over{9}}+4\,y \right]_{ 0 }^{ 2 } \;\textrm{d}x$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 3 } 4\,2+\left(-{{1}\over{9}}\right)\,x^2\,2+\left(-{{1}\over{48}} \right)\,2^3-\left(4\,0+\left(-{{1}\over{9}}\right)\,x^2\,0+\left(- {{1}\over{48}}\right)\,0^3\right) \;\textrm{d}x$ (substituting limits)

$\displaystyle \phantom{I}={\int_{ 0 }^{ 3 } {{47}\over{6}}-{{2\,x^2}\over{9}} \;\textrm{d}x}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{47\,x}\over{6}}-{{2\,x^3}\over{27}} \right]_{ 0 }^{ 3 }$

$\displaystyle \phantom{I}= {{43}\over{2}}-0$ (substituting limits)

$\displaystyle \phantom{I}= {{43}\over{2}}$ (simplificación)

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