Question

Question on integration by parts

Answer 1

When integrating
\[\int\limits_{0}^{1} xe^{-x}dx\]
is the answer
\[-2/e -1\]
or\[1-2/e\]

Answer 2

1-2/e

Answer 3

I keep getting \[-2/e-1\] but Wolfram says its the opposite

Answer 4

if you did the int by parts correctly, you should only arrive at one solution ... how did you go about this?

Answer 5

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

Answer 6

oh, then 0 to 1 lol

Answer 7

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

Answer 8

-1 e^(-1) - e^(-1) + 0 e^(-0) + e^(-0)

\[-\frac{1}{e}-\frac{1}{e}+0+1\]

\[-\frac{2}{e}+1\]

Answer 9

the algebra is the killer in most cases

Answer 10

\[u=x\]
\[du=dx\]
\[v=-e^{-x}\]
\[dv=e^{-x}\]
\[\int\limits_{0}^{1} xe^{-x}dx=-xe^{-x}|_{0}^{1} + \int_{0}^{1}e^{-x}dx\]
\[\int\limits_{0}^{1} xe^{-x}dx=-1e^{-1} + -e^{-x}|_{0}^{1}\]
\[\int\limits_{0}^{1} xe^{-x}dx=-e^{-1} + -e^{-1}+1\]
\[\int\limits_{0}^{1} xe^{-x}dx=-2e^{-1}+1\]

Answer 11

@Hunus your integrand is positive (on (0,1)) ...your answer is negative...this tells you you are not correct.

Answer 12

Yea, I see where I went wrong on the \[-e^{-x}|_{0}^{1}\]
I put -e^-1 - 1 instead of -e^-1 -(-1)