MvCalc7

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I=\int _{1}^{4}[\int _{{0}}^{{{\sqrt  {4-x}}}}xydy]dy\,

\int _{{0}}^{{{\sqrt  {4-x}}}}xydy\,

=[{\frac  {xy^{2}}{2}}]_{{y=0}}^{{{\sqrt  {4-x}}}}\,

=[{\frac  {x(4-x)}{2}}]={\frac  {4x-x^{2}}{2}}\,

Threrefore, I=\int _{1}^{4}(2x-{\frac  {x^{2}}{2}})dx\,

=[x^{2}-{\frac  {x^{3}}{6}}]_{1}^{4}\,

=[16-{\frac  {64}{6}}]-[1-{\frac  {1}{6}}]=15-({\frac  {32}{3}}-{\frac  {1}{6}})\,

=15-{\frac  {63}{6}}=15-{\frac  {21}{2}}={\frac  {9}{2}}\,

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