anonymous
  • anonymous
I have question in substation rule
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
in substitution rule
myininaya
  • myininaya
all those are z's right?

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anonymous
  • anonymous
yep
myininaya
  • myininaya
did you let u=the bottom
myininaya
  • myininaya
\[u=e^z +z => du=(e^z+1 ) dz\]
anonymous
  • anonymous
yep i did in that way
myininaya
  • myininaya
\[\int\limits_{e^0+0}^{e^1+1}\frac{du}{u}\]
myininaya
  • myininaya
\[\ln|u||_1^{e^1+1}\]
myininaya
  • myininaya
\[\ln|e^1+1|-\ln|1|=\ln(e+1)-0=\ln(e+1)\]
anonymous
  • anonymous
yeah that is the right answer Thank you so much :)
myininaya
  • myininaya
Do you understand all the steps I took?
anonymous
  • anonymous
i didn't understand one part of it which is the one before last step of it
myininaya
  • myininaya
you mean when I plugged in the upper limit and then minus plugged in the lower limit?
anonymous
  • anonymous
yeah why did you put e to 1 + 1 and e to 0 +0?
myininaya
  • myininaya
so before the substitution the limits were 0
anonymous
  • anonymous
Oh I see Thank you so much that was very helpful :)
anonymous
  • anonymous
I understood each step that you had done
myininaya
  • myininaya
Awesome! :)

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