MvCalc13

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I=\int _{0}^{2}[\int _{{x^{2}}}^{{2x}}(2x+3y)dy]dx\,

Now \int _{{x^{2}}}^{{2x}}(2x+3y)dy\,

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

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

=10x^{2}-2x^{3}-{\frac  {3x^{4}}{2}}\,

Therefore,I=\int _{0}^{2}[10x^{2}-2x^{3}-{\frac  {3x^{4}}{2}}]dx\,

=[{\frac  {10x^{3}}{3}}-{\frac  {x^{4}}{2}}-{\frac  {3x^{5}}{10}}]_{0}^{2}\,

={\frac  {80}{3}}-{\frac  {16}{2}}-{\frac  {48}{5}}\,

={\frac  {800-240-288}{30}}\,

={\frac  {272}{30}}={\frac  {136}{15}}\,

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