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Integral Doble

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

$\displaystyle I=\int_{ 0 }^{ 1 }\int_{ y }^{ 2\,y } \sin \left(y+x\right) \;\textrm{d}x\,\textrm{d}y$

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

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

$\displaystyle \phantom{I}={\int_{ 0 }^{ 1 } \cos \left(2\,y\right)-\cos \left(3\,y\right) \;\textrm{d}y}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{\sin \left(2\,y\right)}\over{2}}-{{\sin \left(3\,y\right)}\over{3 }} \right]_{ 0 }^{ 1 }$

$\displaystyle \phantom{I}= -0-{{2\,\sin \left(3\right)-3\,\sin \left(2\right)}\over{6}}$ (substituting limits)

$\displaystyle \phantom{I}= -{{2\,\sin \left(3\right)-3\,\sin \left(2\right)}\over{6}}$ (simplificación)

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