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anonymous

  • 5 years ago

I have question in substation rule

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    in substitution rule

  3. myininaya
    • 5 years ago
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    all those are z's right?

  4. anonymous
    • 5 years ago
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    yep

  5. myininaya
    • 5 years ago
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    did you let u=the bottom

  6. myininaya
    • 5 years ago
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    \[u=e^z +z => du=(e^z+1 ) dz\]

  7. anonymous
    • 5 years ago
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    yep i did in that way

  8. myininaya
    • 5 years ago
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    \[\int\limits_{e^0+0}^{e^1+1}\frac{du}{u}\]

  9. myininaya
    • 5 years ago
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    \[\ln|u||_1^{e^1+1}\]

  10. myininaya
    • 5 years ago
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    \[\ln|e^1+1|-\ln|1|=\ln(e+1)-0=\ln(e+1)\]

  11. anonymous
    • 5 years ago
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    yeah that is the right answer Thank you so much :)

  12. myininaya
    • 5 years ago
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    Do you understand all the steps I took?

  13. anonymous
    • 5 years ago
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    i didn't understand one part of it which is the one before last step of it

  14. myininaya
    • 5 years ago
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    you mean when I plugged in the upper limit and then minus plugged in the lower limit?

  15. anonymous
    • 5 years ago
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    yeah why did you put e to 1 + 1 and e to 0 +0?

  16. myininaya
    • 5 years ago
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    so before the substitution the limits were 0<z<1 we wanted everything in terms of u after we made substitution if u=e^z+z, then e^0+0<z<e^1+1 ---- if z=0, then u=e^0+0 if z=1, then u=e^1+1

  17. anonymous
    • 5 years ago
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    Oh I see Thank you so much that was very helpful :)

  18. anonymous
    • 5 years ago
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    I understood each step that you had done

  19. myininaya
    • 5 years ago
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    Awesome! :)

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spraguer (Moderator)
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