Hunus Group Title Question on integration by parts 2 years ago 2 years ago

1. Hunus Group Title

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

2. nitishanand Group Title

1-2/e

3. Hunus Group Title

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

4. amistre64 Group Title

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

5. amistre64 Group Title

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 Group Title

oh, then 0 to 1 lol

7. DHASHNI Group Title

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 Group Title

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

9. amistre64 Group Title

the algebra is the killer in most cases

10. Hunus Group Title

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

11. Zarkon Group Title

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

12. Hunus Group Title

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