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    I'm asked to show that the root mean squared speed equals ... using a standard integral. I've been given an equation for f(v). (I can attach a photo later if it'll help)

    However, I do not understand why:
    <v^2> = inegrate[(v^2)*f(v)]dv
    ... the limits being infinity and 0. ('integrate' is where the integral symbol is meant to be)

    Why does the mean square speed, <v^2>, equal that integral? Is there a derivation to prove this? It's just not making sense.

    I've got this identity:
    <x^2> = integrate[(x^2)*P(x)]dx


    I substituted v for x, dv for dx, and thus P(x) would be replaced by P(v)..which in turn is f(v). BUT P(v) does NOT equal f(v)...does it..which is why I don't get how to use this indentity to get the equation at the top!

    Thank you!
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    For any discrete variable \bar{x} = \sum xP(x).
    Extending this to a continuous variable,  \bar{x} = \int xf(x). The summation or integral is over all possible values of x, denoted by \pm \infty, but the function likely will equal 0 for some of these values.
    Replace the x by v2
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    (Original post by morgan8002)
    For any discrete variable \bar{x} = \sum xP(x).
    Extending this to a continuous variable,  \bar{x} = \int xf(x). The summation or integral is over all possible values of x, denoted by \pm \infty, but the function likely will equal 0 for some of these values.
    Replace the x by v2
    Somehow it all makes sense with that now!! Cheers
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    (Original post by PhysicsGal)
    Somehow it all makes sense with that now!! Cheers
    You're welcome.
 
 
 
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