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Homework Help: Anti derivative of sqrt{2x+1}

  1. May 24, 2005 #1
    I know the anti derivative of [tex]\sqrt{2x+1}[/tex] is [tex]1/3(2x+1)^(3/2)[/tex] but I can't develop a concrete method for finding the anti derivative of functions like this. How would you go about finding the anti derivative of this function?
     
  2. jcsd
  3. May 24, 2005 #2

    dextercioby

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    Make the substitution

    [tex] 2x+1=u [/tex]

    Daniel.
     
  4. May 24, 2005 #3
    Right. To go further with that substitution method...

    [tex]\int \sqrt{2x+1}dx[/tex]

    [tex]u = 2x+1[/tex]

    [tex]du = 2dx[/tex]

    [tex]\frac{1}{2}du = dx[/tex]


    [tex]\int \sqrt{2x+1}dx = \frac{1}{2}\int \sqrt{u}du[/tex]

    Take it from there.

    Jameson
     
    Last edited: May 25, 2005
  5. May 24, 2005 #4
    with [tex] 2x+1=u [/tex]
    you get [tex] \sqrt{u} = u^(^1^/^2^) [/tex]
    and using the [tex]x^n -> \frac{x^n^+^1}{n+1}[/tex] rule I get [tex]2/3 (u)^(^3^/^2^)[/tex] which is [tex]2/3 (2x+1)^(^3^/^2^)[/tex] which isn't correct, so what am i missing here?

    [edit- oh. I was trying to work with the take of anti derivatives to come up with the answer so i could use it to integrate, but it looks like you have to use a little integration sometimes to get antiderivatives (indefinte integrals) sometimes. thanks.]
     
    Last edited: May 24, 2005
  6. May 24, 2005 #5
    Look at my post... you need to factor out that one-half.
     
  7. May 24, 2005 #6
    An even better/more appropriate excercise: Expand it to the family

    [tex] \int \sqrt{ax+b} \ dx[/tex]

    [tex] u = ax+b, du = a dx[/tex]

    [tex] \frac{1}{a} \int \sqrt{u} \ du [/tex]

    [tex] \frac{1}{a} \ \frac{2}{3}\ u^{\frac{3}{2}} [/tex]

    [tex] \frac{2}{3a} \ u^{\frac{3}{2}} [/tex]

    With a = 2, b = 3 you get

    [tex] \frac{1}{3} \ u^{\frac{3}{2}} [/tex]
     
    Last edited: May 24, 2005
  8. May 24, 2005 #7

    Integral

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    Jameson,
    Your line:
    [tex]\int \sqrt{2x+1}dx = \frac{1}{2}\int udu[/tex]

    Should read
    [tex]\int \sqrt{2x+1}dx = \frac{1}{2}\int \sqrt{u}du[/tex]
     
  9. May 25, 2005 #8
    Ah yes, an important typo. I'll fix it. Thanks.
     
  10. May 25, 2005 #9
    The anti-derivative and the integral are the same thing... I'm confused about what you mean.
     
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