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Homework Help: Integration Questions

  1. Feb 7, 2005 #1
    [tex] \int 2e^{2x} dx [/tex] Ok so I know that [tex] \int e^{u} du = e^{u} + C [/tex]. So we have [tex] 2\int e^{2x} dx [/tex]. Now [tex] u = 2x, du = 2dx [/tex]. So [tex] dx = \frac{1}{2} du [/tex]. So now we have [tex] \int e^{2x} du = e^{2x} + C [/tex] Was my method correct?

    [tex] \int e^{2x} dx [/tex] Ok for this [tex] u = 2x, du = 2dx [/tex]. So [tex] dx = \frac{1}{2} du [/tex]. So we have [tex] \frac{1}{2}\int e^{2x} du = \frac{1}{2} e^{2x} + C [/tex]. Is this correct?

    [tex] \int 5x^{3}e^{x^{2}} dx [/tex] So [tex] u = x^{2} du = 2xdx [/tex].
     
  2. jcsd
  3. Feb 7, 2005 #2
    first one yes,, can be verified by taking derivative of solution and seeing if you get back the integrand

    second one is good too, just a constant multiple of first.

    for third use integration by parts (hint: let [itex] u = x^2[/itex] and [itex] dv = x e^{x^2}[/itex])
     
    Last edited: Feb 7, 2005
  4. Feb 7, 2005 #3
    Ok so [tex] \int 5x^{3}e^{x^{2}} dx [/tex] so [tex] u = x^2 [/tex] which means [tex] du = 2xdx [/tex]. Does this mean that [tex] \frac{1}{2}\int e^{x^{2}} (2xdx) = \frac{1}{2} e^{x^{2}} + C [/tex] ?
     
  5. Feb 7, 2005 #4
    the formula for integration by parts is given by

    [tex] \int u dv = uv - \int v du[/tex]

    so for this case letting
    [tex] u = x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex]
    [tex]du = 2x dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex]

    and
    [tex]dv = x e^{x^2}dx \ \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex]
    [tex]\int dv = \int x e^{x^2}dx \ \ \ \ \ \ (4)[/tex]

    making the substitution [itex] w = x^2[/itex] in (4) and solving for v you get
    [tex] v = \frac{1}{2} e^{x^2} [/tex]

    plugging it all into the equation for seperation of variables then

    [tex] \int 5 x^3 e^{x^2} dx = 5[\frac{1}{2} x^2 e^{x^2} - \int 2x \frac{1}{2} e^{x^2} dx][/tex]

    try to do the rest
     
  6. Feb 7, 2005 #5
    however we are supposed to do substitution not parts

    also is:

    [tex] \int \frac{1}{x+1} dx = \Ln(x+1) + C [/tex]?
    [tex] \int \frac{1}{3-2x} dx = \frac{-1}{2} \Ln(3-2x) + C [/tex]
    [tex] \int \frac{e^{2x} + e^{x} + 1}{e^{x}} = e^{x} -e^{-x} + x + 2C [/tex]?
     
    Last edited: Feb 7, 2005
  7. Feb 7, 2005 #6

    dextercioby

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    You can't do it by substitution only.U have to apply part integration as well....

    Daniel.
     
  8. Feb 7, 2005 #7
    note: improper format for natural log in latex

    hint: try taking the derivative of your solutions, and see if they match the integrand
     
    Last edited: Feb 7, 2005
  9. Feb 7, 2005 #8

    dextercioby

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    What happened to the logarithms...? :confused:

    Is that
    [tex] \int \frac{dx}{xx} =\int \frac{dx}{x^{2}} [/itex]

    ???

    Daniel.
     
  10. Feb 7, 2005 #9

    dextercioby

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    HINT:[tex] d(\ln x)=\frac{dx}{x} [/tex]

    Daniel.
     
  11. Feb 7, 2005 #10
    actually it is [tex] \int \frac{1}{x\ln x} [/tex] Ok so [tex] \int \frac{\ln x}{x} dx [/tex]. So [tex] u = \ln x [/tex] and [tex] du = \frac{1}{x} dx [/tex]. That means [tex] dx = xdu [/tex]. Is this correct?

    Also for [tex] \int \frac{e^{2x} + e^{x} + 1}{e^{x}} = e^{x} -e^{-x} + x + 2C [/tex] I just divided through. Is this correct?

    Thanks
     
    Last edited: Feb 7, 2005
  12. Feb 7, 2005 #11

    dextercioby

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    If u put the "dx" in the integral,then yes... :smile:

    Daniel.
     
  13. Feb 7, 2005 #12

    dextercioby

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    Are u talking about the same integral??

    [tex] \int \frac{dx}{x\ln x} \neq \int -\ln x \frac{dx}{x} [/tex]

    Daniel.
     
  14. Feb 7, 2005 #13
    Ok so for [tex] \int \frac{1}{x\ln x} [/tex] should I rewrite or just make [tex] u = \ln x [/tex]?

    Thanks :smile:
     
  15. Feb 7, 2005 #14

    dextercioby

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    1.Put the "dx"... :tongue2:
    2.Yes,make that substitution.

    Daniel.
     
  16. Feb 7, 2005 #15
    ok so i got [tex] du = \frac{1}{x} dx [/tex] [tex] dx = xdu [/tex]. So now do I bring the x out in the front?I know the answer is [tex] \ln(\ln(x))) [/tex]

    thanks
     
    Last edited: Feb 7, 2005
  17. Feb 7, 2005 #16
    no need to do [tex]dx=xdu[/tex]. just substitute straight from [tex]\frac{1}{x}dx=du[/tex].

    edit: add a [tex]+C[/tex] to [tex]ln(lnx)[/tex].

    - kamataat
     
    Last edited: Feb 8, 2005
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