Question

Evaluate the following integral.

Answer 1

\[\int\limits_{0}^{1}xe ^{-x ^{2}}dx\]

Answer 2

substitute u=x^2

Answer 3

Should I substitute u=x^2 or u=-x^2?

Answer 4

better take u = -x^2 :)

Answer 5

both are correct

Answer 6

Well after that I got du=-2xdx

Answer 7

\[\int\limits_{0}^{1}xe^{u} \frac{ du }{ -2x }\]

Answer 8

\[-\frac{ 1 }{ 2 }\int\limits_{0}^{1}e^{u}du\]

Answer 9

yes, that's right @KadyLady

Answer 10

\[-\frac{ 1 }{ 2 }\left[ e^{u} \right]_{0}^{1}\]

Answer 11

then, subtitute back that u=-x^2

Answer 12

The answer is supposed to be \[\frac{ e-1 }{ 2e }\] but that's not what I'm getting.

Answer 13

What happens after \[-\frac{ 1 }{ 2 } e ^{-x^{2}}|_{0}^{1}\]

Answer 14

from ur answer :
-1/2(e^u) [0,1]
= -1/2(e^(-x^2)) [0,1]
= -1/2 * 1/e^(x^2) [0,1]
now, evaluate for the values of intervals

Answer 15

KadyLady your solution is correct. you just forgot to change the boundary condition of integral. without substitute x back, in the u form use the from 0 to -1 and you will get the answer.

Answer 16

Here i tried to show in the picture.

Answer 17

if u continue the steps before, u just put x=1 then x=0 then subtract of them :
-1/2 * 1/e - (-1/2 * 1/e^0)
= -1/2e + 1/2
= -1/2e + e/2e
= (-1+e)/2e
= (e-1)/2e

Answer 18

I understand both of your answers perfectly. I think @RadEn has plenty of medals earned already.

Answer 19

nopes... i just want help u :*