anonymous
  • anonymous
compute the integral : ...
Calculus1
katieb
  • katieb
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anonymous
  • anonymous
|dw:1435505559058:dw|
anonymous
  • anonymous
u=x^2??? du= 2x
LynFran
  • LynFran
let u=-x^2 du=-2x

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anonymous
  • anonymous
then, -1/2du=(x)dx
LynFran
  • LynFran
im thinkin sumthing more like this...|dw:1435513166335:dw|
anonymous
  • anonymous
then, plug the u back in?
anonymous
  • anonymous
\[\int_0^1xe^{-x^2}\,dx=-\frac{1}{2}\int_0^1 e^{-x^2}\,d(-x^2)=-\frac{1}{2}\int_0^{-1}e^u\,du=\frac{1}{2}\int_{-1}^0 e^u\,du\]
anonymous
  • anonymous
do we plug the value for u back in
LynFran
  • LynFran
yes when u finish
anonymous
  • anonymous
|dw:1435507110146:dw|
LynFran
  • LynFran
but isnt the topic definite integral where we find the upper and lower limits
anonymous
  • anonymous
no, this question gave me 1 and 0
LynFran
  • LynFran
ok im not sure

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