MvCalc8

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I=\int _{1}^{2}[\int _{{x}}^{{x{\sqrt  {3}}}}xydy]dx\,

\int _{{x}}^{{x{\sqrt  {3}}}}xydy\,

=[{\frac  {xy^{2}}{2}}]_{{y=x}}^{{x{\sqrt  {3}}}}={\frac  {3x^{3}}{2}}-{\frac  {x^{3}}{2}}=x^{3}\,

Therefore,I=\int _{1}^{2}x^{3}dx\,

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

={\frac  {16}{4}}-{\frac  {1}{4}}={\frac  {15}{4}}\,

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