Quick integration question

  1. I need to figure out,

    [tex] \int_0^h \frac{1}{2\sqrt{hx}}dx [/tex]

    If h is a constant,

    how do i do this?

    my book shows that I can pull out,

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

    How does the 2 from [tex]\frac{1}{2\sqrt{hx}} [/tex] come out with the [tex]\sqrt{h}[/tex]?

    I thought I would've only been able to pull out 1/root h,

    like this,

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


    why does 2 root h get assigned constant? instead of only h
  2. jcsd
  3. rock.freak667

    rock.freak667 6,231
    Homework Helper

    1/2 is a constant, 1/√h is a constant

    it must follow that

    1/2√h is constant as well.
  4. Mark44

    Staff: Mentor

    The basic idea is that [itex]\int k*f(x) dx = k*\int f(x) dx[/itex].

    The rest in your problem is just algebra.
    [tex]\frac{1}{2\sqrt{hx}} = \frac{1}{2*\sqrt{h}\sqrt{x}} = \frac{1}{2\sqrt{h}} \frac{1}{\sqrt{x}}[/tex]

    Integration is being done with respect to x (i.e., with x as the variable), so h is just another constant in this process.
  5. cheers
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