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Integral Doble

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

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

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

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

$\displaystyle \phantom{I}={\int_{ -1 }^{ 1 } x-x^3 \;\textrm{d}x} $ (simplificación)

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

$\displaystyle \phantom{I}= {{1}\over{4}}-{{1}\over{4}} $ (substituting limits)

$\displaystyle \phantom{I}= 0 $ (simplificación)

     
Region for integration
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