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a question - how to integrate f(x,y)=ye^xy , 0<x<1,0<y<1

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  • a question - how to integrate f(x,y)=ye^xy , 0<x<1,0<y<1

    how to integrate f(x,y)=ye^xy , 0<x<1,0<y<1
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  • #2
    answer to your question................. f(x,y)=ye^xy ; 0<x<1, 0<y<1 ------------------------------------------------------- Integral = I-------------------------- Upper-limit = UL------------------ Lower Limit = LL----------------- ------------------------------------------------------- f(x,y)=(I,LL:0,UL:1,for x)((I,LL:0,UL:1,for y)ye^xy dydx------------------------------------ Use integration by parts, enter the upper and lower limit------------and you would end-up with----(I,LL:0,UL:1,for x)[(1/x)e^(x)]dx---------------------now integrate [(1/x)e^(x)] with respect to x and enter the value of limits (LL=0,UL=1) and you would have (e^1)-e^0 which is same as (e^1)-1which is equal to 1.71828......that should be your answer.
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    • #3
      sorry the answer is e^1-2 i do have the solution book with me..
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      • #4
        I did that question again and I see my mistake but still I didn't get the answer you have...........can you look at it and let me know what mistake I made....------------------------------------------------------------ ye^xy w.r.t dydx both integrals limit (0 to 1)----------------------------------------integrate w.r.t dy we have------------------ [{(y^2)(e^(xy))} / 2] ------after inserting the limits we have----(e^x)/2 dx-----------now integrate w.r.t dx we get----------- .5e^x ----insert limits (0,1)----------.5e - .5------which is equal to .85 that should be the answer-----------------------if not can you check my work.
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        • #5
          yah the solution given below that is 27 feb 2005 is correct
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          • #6
            keep y fixed and integrated with x variable.you should have e^y - 1.now integrate (e^y -1)dy 0<y<1.if I'm not mistaking the answer should be e -2
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            • #7
              I got the same answer as you did
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