## Hunus 3 years ago Question on integration by parts

1. Hunus

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

2. nitishanand

1-2/e

3. Hunus

I keep getting $-2/e-1$ but Wolfram says its the opposite

4. amistre64

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

5. amistre64

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

6. amistre64

oh, then 0 to 1 lol

7. DHASHNI

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

8. amistre64

-1 e^(-1) - e^(-1) + 0 e^(-0) + e^(-0) $-\frac{1}{e}-\frac{1}{e}+0+1$ $-\frac{2}{e}+1$

9. amistre64

the algebra is the killer in most cases

10. Hunus

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

11. Zarkon

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

12. Hunus

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