MvCalc24

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I=\int _{{1}}^{{\log 8}}[\int _{{0}}^{{\log y}}e^{{x+y}}dx]dy\,

Now \int _{{0}}^{{\log y}}e^{{x+y}}dx=[e^{{x+y}}]_{{0}}^{{\log y}}\,

=[e^{{y+\log y}}-e^{y}]\,

Therefore,I=\int _{{1}}^{{\log 8}}[e^{{y+\log y}}-e^{y}]dy=\int _{{1}}^{{\log 8}}e^{y}(y-1)dy\,

=[ye^{y}-e^{y}-e^{y}]_{{1}}^{{\log 8}}=8\log 8-16+e\,

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