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Why do we say the upper limit is the variable?

  1. Nov 26, 2013 #1
    While doing work, I always see notation like:

    F(x) = ∫0xf(t)dt - first equation

    F(x) = ∫1xf(t)dt

    What is the exact reasoning we always use the upper limit? What is the lower limit was a variable? From FTC 2, the answer for the first equation is F(x) - F(0) right yet is written as F(x) when F(0) has a value? An explanation would be very helpful.

    How would this be assessed: F(x) = ∫0x2f(t)dt [1] ?

    From examples, I see that taking the derivative of [1] requires that you apply chain rule, so that you end up multiplying the derivative of this by 2x, in this example, since that is the derivative of the variable introduced with respect to the original variable. I'm just slightly confused on how we would assess this function when not taking the derivative and if someone could provide the general answer for a function in this form:

    F(x) = ∫g(x)f(x)f(t)dt

    with an explanation that would be great! Thank you so much! :)
  2. jcsd
  3. Nov 26, 2013 #2


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    You are confused by the notation. Those equations give definitions for ##F(x)##, but here ##F(x)## is not the antiderivative of ##f(x)##. As you note, ##F(x)## is related to the antiderivative by a term involving the evaluation of the antiderivative at the other limit.

    It's helpful to use a different notation for the antiderivative, so let ##\mathcal{F}(x)## be the antiderivative of ##f(x)##. Then, if we define

    $$F(x) = \int^{a(x)}_{b(x)} f(t) dt,$$

    by the fundamental theorem, we can say that ##F(x)## is related to the antiderivative ##\mathcal{F}(x)## by

    $$ F(x) = \mathcal{F}(a(x)) - \mathcal{F}(b(x)).$$
  4. Nov 26, 2013 #3


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    If the lower limit was the variable, for example
    [tex] F(x) = \int_{x}^{0} f(t) dt = - \int_{0}^{x} f(t) dt [/tex]
    we would get that
    [tex] F'(x) = -f(x) [/tex]
    which is just a little less convenient is all.

    For the last part, if
    [tex] H(x) = \int_{0}^{x} f(t) dt [/tex]
    [tex] F(x) = \int_{g(x)}^{h(x)} f(t) dt [/tex]
    [tex] F(x) = H(g(x)) - H(h(x)) [/tex]
    and you know what H'(x) is, so you can do the chain rule on the right hand side.
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