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HunusBest ResponseYou've already chosen the best response.0
When integrating \[\int\limits_{0}^{1} xe^{x}dx\] is the answer \[2/e 1\] or\[12/e\]
 2 years ago

HunusBest ResponseYou've already chosen the best response.0
I keep getting \[2/e1\] but Wolfram says its the opposite
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
if you did the int by parts correctly, you should only arrive at one solution ... how did you go about this?
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
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
 2 years ago

DHASHNIBest ResponseYou've already chosen the best response.0
let u =x dv=e^(x)dx du=dx v=e^(x) =>xe^(x)e^(x) on limits the ans is 1/e1/e1=2/e1
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
1 e^(1)  e^(1) + 0 e^(0) + e^(0) \[\frac{1}{e}\frac{1}{e}+0+1\] \[\frac{2}{e}+1\]
 2 years ago

amistre64Best ResponseYou've already chosen the best response.3
the algebra is the killer in most cases
 2 years ago

HunusBest ResponseYou've already chosen the best response.0
\[u=x\] \[du=dx\] \[v=e^{x}\] \[dv=e^{x}\] \[\int\limits_{0}^{1} xe^{x}dx=xe^{x}_{0}^{1} + \int_{0}^{1}e^{x}dx\] \[\int\limits_{0}^{1} xe^{x}dx=1e^{1} + e^{x}_{0}^{1}\] \[\int\limits_{0}^{1} xe^{x}dx=e^{1} + e^{1}+1\] \[\int\limits_{0}^{1} xe^{x}dx=2e^{1}+1\]
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.2
@Hunus your integrand is positive (on (0,1)) ...your answer is negative...this tells you you are not correct.
 2 years ago

HunusBest ResponseYou've already chosen the best response.0
Yea, I see where I went wrong on the \[e^{x}_{0}^{1}\] I put e^1  1 instead of e^1 (1)
 2 years ago
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