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Beginning Antiderivative Question

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the antiderivative of [tex] F(x) \ = \ \sqrt{\frac{5}{x}} [/tex]

    2. Relevant equations

    [tex] F(x) \ = \ \sqrt{\frac{5}{x}} [/tex]

    3. The attempt at a solution

    1.) [tex] F(x) \ = \ \sqrt{\frac{5}{x}} [/tex]

    2.) [tex] F(x) \ = \ (\frac{5}{x})^\frac{1}{2} [/tex]

    3.) [tex] F(x) \ = \ \frac{5^\frac{1}{2}}{x^\frac{1}{2}} [/tex]

    4.) [tex] F(x) \ = \ (5^\frac{1}{2})(x^\frac{-1}{2}) [/tex]

    This is where I am stuck. I can find the antiderivative of the [tex] x^\frac{-1}{2} [/tex] term, which would be [tex] 2\sqrt{x} [/tex], but I'm not sure where to go with the [tex] 5^\frac{1}{2} [/tex] term.

    Any help would be greatly appreciated.
  2. jcsd
  3. Jan 19, 2009 #2


    Staff: Mentor

    You can use the power rule for antidifferentiation. If n != -1,
    \int x^{n} dx
    = \frac{x^{n + 1}}{n + 1} + C
  4. Jan 19, 2009 #3


    User Avatar
    Gold Member

    The sqrt(5) is merely a constant. Integrating functions is similar to differentiating functions. If I asked you to find d(kf(x))/dx where k is a constant, you would likely say kf'(x). Apply similar thinking when applying anti-differentiation.
  5. Jan 19, 2009 #4
    Well, I know that, for example, if asked to find the antiderivative of 5, the answer would be [tex] 5x + C [/tex].

    I was trying to use similar logic here by saying [tex] 5^\frac{1}{2} [/tex] would become something like [tex] \frac{10\sqrt{x^3}}{3} [/tex]. However, that didn't seem right.

    So, if I follow the logic that [tex] kf(x) [/tex] derives to [tex] k\frac{df}{dx} [/tex], and since [tex] 5^\frac{1}{2} [/tex] is simply a constant, then I would just leave it as is.

    Thus, the antiderivative would be [tex] (\sqrt{5})(2\sqrt{x}) [/tex].

    Am I understanding this correctly?
  6. Jan 19, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    You are understanding it correctly.
  7. Jan 19, 2009 #6
    Thank you very much everyone for all of your help!
  8. Jan 19, 2009 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    Which, of course, is the same as [tex]2\sqrt{5x}+ C[/tex] so that it is in the same form as the original function (and don't forget the "C").
  9. Jan 19, 2009 #8
    Yes, of course! Thank you, HallsofIvy.
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