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## Integral Doble

We integrate the function $y-1$ on the set given by $0 \leq y \leq 4$ and $-\sqrt{y} \leq x \leq \sqrt{y}$ . $$I=\iint_{M}f\,\mathrm{d}x\mathrm{d}y\ ,\qquad\qquad f= y-1 \ , \qquad\qquad M=\begin{cases} 0 \leq y \leq 4 \\ -\sqrt{y} \leq x \leq \sqrt{y} \end{cases}$$

$\displaystyle I=\int_{ 0 }^{ 4 }\int_{ -\sqrt{y} }^{ \sqrt{y} } y-1 \;\textrm{d}x\,\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 4 }\left[ x\,y-x \right]_{ -\sqrt{y} }^{ \sqrt{y} } \;\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 4 } \left(-1\right)\,\sqrt{y}+\sqrt{y}\,y-\left(\left(-1\right)\,\left( -\sqrt{y}\right)+\left(-\sqrt{y}\right)\,y\right) \;\textrm{d}y$ (substituting limits)

$\displaystyle \phantom{I}={\int_{ 0 }^{ 4 } \sqrt{y}\,\left(2\,y-2\right) \;\textrm{d}y}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{4\,y^{{{3}\over{2}}}\,\left(3\,y-5\right)}\over{15}} \right]_{ 0 }^{ 4 }$

$\displaystyle \phantom{I}= {{224}\over{15}}-0$ (substituting limits)

$\displaystyle \phantom{I}= {{224}\over{15}}$ (simplificación)

Region for integration