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Hunus
Question on integration by parts
When integrating \[\int\limits_{0}^{1} xe^{-x}dx\] is the answer \[-2/e -1\] or\[1-2/e\]
I keep getting \[-2/e-1\] but Wolfram says its the opposite
if you did the int by parts correctly, you should only arrive at one solution ... how did you go about this?
e^-x x -e^-x: -x e^-x -1 e^-x : -e^-x 0 -e^-x : 0 I get: -x e^(-x) - e^(-x) if i did it right
let u =x dv=e^(-x)dx du=dx v=-e^(-x) =>-xe^(-x)-e^(-x) on limits the ans is -1/e-1/e-1=-2/e-1
-1 e^(-1) - e^(-1) + 0 e^(-0) + e^(-0) \[-\frac{1}{e}-\frac{1}{e}+0+1\] \[-\frac{2}{e}+1\]
the algebra is the killer in most cases
\[u=x\] \[du=dx\] \[v=-e^{-x}\] \[dv=e^{-x}\] \[\int\limits_{0}^{1} xe^{-x}dx=-xe^{-x}|_{0}^{1} + \int_{0}^{1}e^{-x}dx\] \[\int\limits_{0}^{1} xe^{-x}dx=-1e^{-1} + -e^{-x}|_{0}^{1}\] \[\int\limits_{0}^{1} xe^{-x}dx=-e^{-1} + -e^{-1}+1\] \[\int\limits_{0}^{1} xe^{-x}dx=-2e^{-1}+1\]
@Hunus your integrand is positive (on (0,1)) ...your answer is negative...this tells you you are not correct.
Yea, I see where I went wrong on the \[-e^{-x}|_{0}^{1}\] I put -e^-1 - 1 instead of -e^-1 -(-1)