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

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3if you did the int by parts correctly, you should only arrive at one solution ... how did you go about this?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3e^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

DHASHNI
 3 years ago
Best ResponseYou've already chosen the best response.0let 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.31 e^(1)  e^(1) + 0 e^(0) + e^(0) \[\frac{1}{e}\frac{1}{e}+0+1\] \[\frac{2}{e}+1\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3the algebra is the killer in most cases

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

Zarkon
 3 years ago
Best 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.

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