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Multiplication in a Definite Integral Equation

  1. Jul 10, 2012 #1
    If

    [itex]f(x)=\int_0^\infty g(x) dx[/itex]

    and I wanted to multiply the integral by, say, a, what would I multiply the left side by? In other words,

    [itex]? \times f(x) = a \int_0^\infty g(x) dx[/itex]

    Thanks in advance!
     
  2. jcsd
  3. Jul 11, 2012 #2
    I'm a little confused, why/what do you want to multiply? If you want to keep the equation the same, you multiply both sides by the same thing, if you want to change the integral, you wouldn't do anything to the other side.
     
  4. Jul 11, 2012 #3
    Call f(x)=u and the integral v. Then, do the multiplication as you would in algebra, and then substitute the integral and the function back in.
     
  5. Jul 11, 2012 #4

    chiro

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    Hey drewfstr314 and welcome to the forums.

    If a is independent of the integral (which it should be), then yes that it what it does. It doesn't have to be a constant, it can be a function of any variable, but it can't be something that changes the integral: it has to be orthogonal or independent of the integral unless you specify otherwise in a special constraint of some sort.
     
  6. Jul 11, 2012 #5
    What if a did change the integral. I guess I mean something like

    [itex]? \times f(x) = h(x) \int_0^\infty \left( \frac{g(x)}{h(x)}\right) dx[/itex]

    since multiplying the integral by h(x) would make the integral of only g(x).
     
    Last edited: Jul 11, 2012
  7. Jul 11, 2012 #6
    Just so you know, if your definite integral integrates with respect to x, then your final answer won't be a function of x (other than being a constant function of x), so in your first equation, f(x) = C. However, if you multiply both sides by h(x), you can't put h(x) in the denominator inside the integral.
     
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