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Stuck with integration

  1. Apr 11, 2005 #1
    [tex]\int_a^b f(x)dx = a+2b[/tex]
    [tex]\int_a^b (f(x)+5)dx =?[/tex]
    [tex]\int_a^b (f(x)+5)dx =\int_a^b f(x)dx+\int_a^b 5dx[/tex]
    [tex]a+2b+\int_a^b 5dx[/tex]

    I'm stuck, what should be my next step?
     
  2. jcsd
  3. Apr 11, 2005 #2

    HallsofIvy

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    Are you serious? You can't integrate [tex]\int_a^b5dx[/tex]?

    Isn't that the same as [tex]5\int_a^b dx= 5(b-a)[/tex]?

    Isn't that about the first thing you learned in integration?
     
  4. Apr 11, 2005 #3
    hehe, thanks

    what's the rule for [tex]\int e^x[/tex]?
    is it [tex] e^x \int x[/tex]?
     
  5. Apr 11, 2005 #4

    dextercioby

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    Why?It's the exponential.Integration just adds a constant...Don't forget the differential of "x".

    Daniel.
     
  6. Apr 11, 2005 #5
    [tex]\int e^x[/tex] = [tex]\int e^x dx[/tex] ?
     
  7. Apr 11, 2005 #6

    dextercioby

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    No,no,the first notation is incorrect.It should always be

    [tex] \int \ \mbox{function} \cdot \mbox{element of integration} [/tex]

    Daniel.
     
  8. Apr 11, 2005 #7
    so if a question was [tex]\int e^{\frac{x}{2}} dx = e^{\frac{x}{2}} \int \frac{x}{2} dt[/tex]
     
  9. Apr 11, 2005 #8

    HallsofIvy

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    [tex]\int e^x dx= e^x+ C[/tex]

    because [tex]\frac{d e^x}{dx}= e^x[/tex], of course.
     
  10. Apr 11, 2005 #9

    dextercioby

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    No,if course not.U need to make a substitution

    [tex] \frac{x}{2}=u [/tex]

    Daniel.

    EDIT:BTW,knowledge of integration techniques assumed knowledge of differentiation methodes.
     
  11. Apr 11, 2005 #10
    [tex]\int e^{\frac{x}{2}} dx[/tex]
    [tex]u=x/2[/tex]
    [tex]du=1/2dx[/tex]
    [tex]\int e^{u} 2du[/tex]
    [tex]=2e^{\frac{x}{2}} [/tex]
     
  12. Apr 11, 2005 #11

    HallsofIvy

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    Let u= x/2. Then 2u= x so 2du= dx.
    [tex]\int e^{\frac{x}{2}}dx[/tex] becomes [tex]2\int e^u du= 2e^u+ C= 2e^{\frac{x}{2}}+ C[/tex]

    Surely you've learned simple substitutions.

    [tex]e^x\int \frac{x}{2}dt[/tex], on the other hand, is [tex]e^x(\frac{x^2}{4}+ C)[/tex].
     
    Last edited: Apr 11, 2005
  13. Apr 11, 2005 #12

    dextercioby

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    That's right up to a constant,which should never be forgotten when computing indefinite integrals.

    Daniel.
     
  14. Apr 11, 2005 #13
    is [tex]cos^2(x)=cos(x)cos(x)[/tex]
    so..
    [tex]\frac{d}{dx}cos^2(x)=-2sin(x)cos(x)[/tex]
     
  15. Apr 11, 2005 #14

    dextercioby

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    Yeah,why?U could apply the chain rule as well.

    Daniel.
     
  16. Apr 11, 2005 #15

    HallsofIvy

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    Yes, by golly!

    (I think we are posting a cross purposes now!)
     
  17. Apr 11, 2005 #16
    so if e is raised to any exponet that is not x, then I must use a subsitution when integrating?
     
  18. Apr 11, 2005 #17

    dextercioby

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    Yes.Always.Make that depends from case to case.Usually the antiderivatives of exponentials of "weird" arguments are not expressible in terms of "elementary functions".Simples example

    [tex] \int e^{-x^{2}} \ dx [/tex]

    Daniel.
     
    Last edited: Apr 11, 2005
  19. Apr 12, 2005 #18
    Just out of curiousity, when I did that particular integral on Mathematica's Online Integrator, I got:

    [tex] \int e^{-x^{2}} \ dx = \frac{1}{2}\sqrt{\pi} ERF[x] + C[/tex]

    Can someone tell me what Erf is?
     
  20. Apr 12, 2005 #19

    dextercioby

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    Sure

    [tex] \mbox{erf} \ (x)=:\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} \ dt [/tex]

    Daniel.
     
  21. Apr 12, 2005 #20
    Well, then that makes perfect sense. Is that something that is standardly used in Calculus? The constant seems to be random.
     
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