evaluate $\displaystyle \int_0^1(\int_x^1 \frac{e^{-y}}{y} dy) dx $
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Change the order of integration. Makes things much easier! Draw the region carefully to get the new limits.
if we reverse the order of $\displaystyle dx $ and $\displaystyle dy $ , the integration becomes $\displaystyle \int_0^1(\int_0^y \frac{e^{-y}}{y} dx) dy $ so the final answer comes to be $\displaystyle 1 - \frac{1}{e} $. thanks
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