Reportar fallo

## Integral Doble

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

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

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

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

$\displaystyle \phantom{I}={\int_{ 1 }^{ 2 } -5\,x^4+8\,x^3+11\,x^2-2\,x \;\textrm{d}x}$ (simplificación)

$\displaystyle \phantom{I}=\left[ -x^5+2\,x^4+{{11\,x^3}\over{3}}-x^2 \right]_{ 1 }^{ 2 }$

$\displaystyle \phantom{I}= {{76}\over{3}}-{{11}\over{3}}$ (substituting limits)

$\displaystyle \phantom{I}= {{65}\over{3}}$ (simplificación)

Region for integration