Thanks, yeKcim. I can't draw to save my life
hehehehe
line $\displaystyle x=y $ is the same as line $\displaystyle y=x $ (you know that but i'm just pointing anyway )
as for integral ... put some work that you have done... and we'll see where is the problem
P.S. your result should be $\displaystyle \displaystyle \frac {e-1}{e} = 1 - \frac {1}{e} = 0.63212 $
yeKciM THANKYOU but are you 100% positive this is correct answer,,,i just want like one other peron to confirm it. i fully understand now that my original thinking is correct due to the diagram drawn above everything makes sense i think,,,,although still trying to do the other question i found
btw i lost my password y day,,,,,,,,,,,,,,,,,,,,created new account,,,,ive confused myslef again,,i cant believe it,,why did we choose to put dydx and then do the dx integral first i thought you cant do that? is the reason we put dy first because x>y?????
furthermore by doing the x integral first do we not have to change the limits in which case it doesnt work youd get x from y to infinity and y from 0 to infinty thanks?
]i.e you cannot do dx first when dy came fist$\displaystyle \int^{\infty}_{0} \int^{x}_{0}\frac {e^{-{\frac {x}{y}-y}}}{y} dy dx = \int _0 ^{\infty} dy \int _0 ^y \frac {e^{-{\frac {x}{y}-y}}}{y}\; dx$