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## Integral Doble

We integrate the function $-10\,y+8\,x+2$ on the set given by $1 \leq x \leq 2$ and $0 \leq y \leq 2\,x$ . $$I=\iint_{M}f\,\mathrm{d}x\mathrm{d}y\ ,\qquad\qquad f= -10\,y+8\,x+2 \ , \qquad\qquad M=\begin{cases} 1 \leq x \leq 2 \\ 0 \leq y \leq 2\,x \end{cases}$$

$\displaystyle I=\int_{ 1 }^{ 2 }\int_{ 0 }^{ 2\,x } -10\,y+8\,x+2 \;\textrm{d}y\,\textrm{d}x$

$\displaystyle \phantom{I}=\int_{ 1 }^{ 2 }\left[ -y\,\left(5\,y-8\,x-2\right) \right]_{ 0 }^{ 2\,x } \;\textrm{d}x$

$\displaystyle \phantom{I}=\int_{ 1 }^{ 2 } \left(-1\right)\,\left(2\,x\right)\,\left(-2+\left(-8\,x\right)+5\, \left(2\,x\right)\right)-\left(-1\right)\,0\,\left(-2+\left(-8\,x \right)+5\,0\right) \;\textrm{d}x$ (substituting limits)

$\displaystyle \phantom{I}={\int_{ 1 }^{ 2 } 4\,x-4\,x^2 \;\textrm{d}x}$ (simplificación)

$\displaystyle \phantom{I}=\left[ 2\,x^2-{{4\,x^3}\over{3}} \right]_{ 1 }^{ 2 }$

$\displaystyle \phantom{I}= -{{2}\over{3}}-{{8}\over{3}}$ (substituting limits)

$\displaystyle \phantom{I}= -{{10}\over{3}}$ (simplificación)

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