MvCalc6

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I=\int _{1}^{2}[\int _{3}^{4}{\frac  {1}{(x+y)^{2}}}dx]dy\,

\int _{3}^{4}{\frac  {1}{(x+y)^{2}}}dx\,

=[-{\frac  {1}{x+y}}]_{{x=3}}^{{4}}=-{\frac  {1}{4+y}}+{\frac  {1}{3+y}}\,

Therefore,I=\int _{1}^{2}[{\frac  {1}{3+y}}-{\frac  {1}{4+y}}]dy\,

=[\log(3+y)-\log(4+y)]_{1}^{2}\,

=[\log {\frac  {3+y}{4+y}}]_{1}^{2}=\log {\frac  {5}{6}}-\log {\frac  {4}{5}}=\log {\frac  {25}{24}}\,

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