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Integral Doble

We integrate the function $x^2-x\,y^2$ on the set given by $0 \leq y \leq 2$ and $y^2 \leq x \leq 2^{{{5}\over{2}}}\,\sqrt{y}$ . $$I=\iint_{M}f\,\mathrm{d}x\mathrm{d}y\ ,\qquad\qquad f= x^2-x\,y^2 \ , \qquad\qquad M=\begin{cases} 0 \leq y \leq 2 \\ y^2 \leq x \leq 2^{{{5}\over{2}}}\,\sqrt{y} \end{cases}$$

$\displaystyle I=\int_{ 0 }^{ 2 }\int_{ y^2 }^{ 2^{{{5}\over{2}}}\,\sqrt{y} } x^2-x\,y^2 \;\textrm{d}x\,\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 2 }\left[ {{x^3}\over{3}}-{{x^2\,y^2}\over{2}} \right]_{ y^2 }^{ 2^{{{5}\over{2}}}\,\sqrt{y} } \;\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 0 }^{ 2 } {{1}\over{3}}\,\left(2^{{{5}\over{2}}}\,\sqrt{y}\right)^3+\left(-{{ 1}\over{2}}\right)\,\left(2^{{{5}\over{2}}}\,\sqrt{y}\right)^2\,y^2- \left({{1}\over{3}}\,\left(y^2\right)^3+\left(-{{1}\over{2}}\right) \,\left(y^2\right)^2\,y^2\right) \;\textrm{d}y$ (substituting limits)

$\displaystyle \phantom{I}={\int_{ 0 }^{ 2 } {{y^6-96\,y^3+2^{{{17}\over{2}}}\,y^{{{3}\over{2}}}}\over{6}} \;\textrm{d}y}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{y^7}\over{42}}-4\,y^4+{{2^{{{17}\over{2}}}\,y^{{{5}\over{2}}} }\over{15}} \right]_{ 0 }^{ 2 }$

$\displaystyle \phantom{I}= {{7936}\over{105}}-0$ (substituting limits)

$\displaystyle \phantom{I}= {{7936}\over{105}}$ (simplificación)

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