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The integral of e^x

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I thought sometimes the integral of e^x is xe^x. Under what circumstances is the integral of e^x = xe^x? I think it has something to do with u substitution.
     
  2. jcsd
  3. May 18, 2012 #2

    Ray Vickson

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    Under NO circumstances is the integral of e^x equal to x e^x. I cannot imagine why you think that would hold.

    RGV
     
  4. May 18, 2012 #3
    what about the derivative?
     
  5. May 18, 2012 #4

    sharks

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    The derivative of [itex]e^x[/itex] is: [tex]e^x .\frac{d(x)}{dx}[/tex]
     
  6. May 18, 2012 #5

    HallsofIvy

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    The derivative of [itex]xe^x[/itex] is, by the product rule [itex](x)'e^x+ (x)(e^x)'= 1(e^x)+ x(e^x)= xe^x+ e^x= (x+ 1)e^x[/itex]. As Ray Vickson said, the integral of [itex]xe^x[/itex] is NOT equal to itself and neither is the derivative.

    The only functions having the property that their derivative is equal to the function itself is a constant times [itex]e^x[/itex].
     
  7. May 18, 2012 #6

    sharks

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    I don't think anyone could have explained it better. This should resolve your confusion, robertjford80.
     
  8. May 18, 2012 #7
    Here's an example

    Screenshot2012-05-18at60502PM.png

    What's going on here? It clear says that the derivative of

    c1e(3/2)x[ = (3/2)c1e(3/2)x
     
  9. May 18, 2012 #8

    SammyS

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    What's going on with the derivative of c1e(3/2)x[ is mainly the chain rule.
     
  10. May 18, 2012 #9
    so the integral of e^2x is e^2x and the derivative of e^x is e^x but the derivative of e^2x is 2e^2x, is that right?
     
  11. May 18, 2012 #10

    sharks

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    Correct, except for the integral of ##e^{2x}## which is ##\frac{e^{2x}}{2}##
     
  12. May 18, 2012 #11
    <deleted>
     
  13. May 18, 2012 #12
    well, why don't you use the chain rule with e^x which would make it xe^x?


    {this referred to number nine's deleted post} i saw it before he deleted it.
     
  14. May 18, 2012 #13
    The chain rule is to multiply by the derivative, and the derivative of x is 1.

    If it helps, d/dx (ex) = 1 * ex
     
  15. May 18, 2012 #14
    thanks villyer, I hadn't thought about that.
     
  16. May 18, 2012 #15

    Ray Vickson

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    No, the first statement is not right, and is not what you asked originally. The indefinite integral of exp(a*x) for constant a is (1/a)*exp(a*x) + C; the derivative of exp(a*x) is a*exp(a*x). When a = 1 these both give just exp(x). For a = 2 they give (1/2) exp(2x) and 2 exp(2x), respectively.

    RGV
     
  17. May 18, 2012 #16

    sharks

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    That's exactly what i said before.
     
  18. May 18, 2012 #17
    if you're talking about post 4, then i don't think you provided enough info to convey that
     
  19. May 18, 2012 #18

    sharks

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    It's obvious that [itex]\frac{dx}{dx}=1[/itex] which gives [itex]e^x .1=e^x[/itex]. Unless, you didn't know that, but it's really a basic notion of the principles of differentiation.
    You should go over the basic principles, as it should help you to understand ##e^x## and the others more complicated that will follow.
     
  20. May 18, 2012 #19
    if it was obvious i would not have posted the question
     
  21. May 19, 2012 #20
    d(ex)/dx = ex

    And so

    ∫ex dx = ___________ ....(i)

    Note that integration is just reverse of differentiation.

    If you want to verify this , then its simple ! Differentiate the left hand side of equation (i) with respect to x and see if its equal to ex. It will work.

    If you want to prove it then analyze it by means of graph of f(x)=ex.

    And note if you do this :

    d(ex)/d(e) = xex-1

    But it can never be xex !
     
    Last edited: May 19, 2012
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