Question #176658

Find the first quandrant area under the curve y=xe^-x.


1
Expert's answer
2021-04-13T13:30:43-0400

Let us find the first quandrant area under the curve y=xexy=xe^{-x}. For this firstly let us sketch the graph of y=xex:y=xe^{-x}:




It follows that the area AA is


A=0+xexdx=lima+0axexdx=A=\int_0^{+\infty}xe^{-x}dx=\lim\limits_{a\to +\infty}\int_0^{a}xe^{-x}dx=


u=x,dv=exdx,du=dx,v=ex|u=x, dv=e^{-x}dx,du=dx, v=-e^{-x}|


=lima+(xex0a+0aexdx)=lima+(aeaex0a)=lima+(aeaea+1)=lima+a1ea+1==\lim\limits_{a\to +\infty}(-xe^{-x}|_0^a+\int_0^{a}e^{-x}dx) =\lim\limits_{a\to +\infty}(-ae^{-a}-e^{-x}|_0^a)= \lim\limits_{a\to +\infty}(-ae^{-a}-e^{-a}+1)= \lim\limits_{a\to +\infty}\frac{-a-1}{e^a}+1=


| let us use the L'Hôpital's Rule |


=lima+1ea+1=1=\lim\limits_{a\to +\infty}\frac{-1}{e^a}+1=1


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