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Question Number 78700 by abdomathmax last updated on 20/Jan/20

calculate ∫∫_D  ((∣x−2∣)/y)dxdy with  D={(x,y)∈R^2 /0≤x≤3 and  1≤y≤e}

$${calculate}\:\int\int_{{D}} \:\frac{\mid{x}−\mathrm{2}\mid}{{y}}{dxdy}\:{with} \\ $$$${D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\mathrm{0}\leqslant{x}\leqslant\mathrm{3}\:{and}\:\:\mathrm{1}\leqslant{y}\leqslant{e}\right\} \\ $$

Commented by john santu last updated on 20/Jan/20

= ∫_1 ^( e)  ∫_0 ^( 3) (((∣x−2∣)/y) )dx dy  = ∫_(1 ) ^( e) (((2x−(1/2)x^2 )∣_0 ^2  +((1/2)x^2 −2x)∣_2 ^3 )/y) dy   = ∫_1 ^( e) ((  2+(1/2))/y) dy = (5/2)(ln y) ∣_1 ^e  = (5/2)

$$=\:\int_{\mathrm{1}} ^{\:{e}} \:\int_{\mathrm{0}} ^{\:\mathrm{3}} \left(\frac{\mid{x}−\mathrm{2}\mid}{{y}}\:\right){dx}\:{dy} \\ $$$$=\:\int_{\mathrm{1}\:} ^{\:{e}} \frac{\left(\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)\mid_{\mathrm{2}} ^{\mathrm{3}} }{{y}}\:{dy}\: \\ $$$$=\:\int_{\mathrm{1}} ^{\:{e}} \frac{\:\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}}{{y}}\:{dy}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\left({ln}\:{y}\right)\:\mid_{\mathrm{1}} ^{{e}} \:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$

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