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Fundamental theorem of calculus

  1. Sep 24, 2008 #1
    There are two theorems:

    The fundamental theorem of calculus: [tex]\int_{a}^{b}F'(x) = F(b) - F(a)[/tex]

    And the theorem that states if f is continious on [a,b]and g:[a,b]->R is defined by
    g(x) = [tex]\int_{a}^{x}f(t) dt[/tex], then g is differentiable on (a,b) and;

    g'(x) = [tex]\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)[/tex]

    "There’s one question that asks if it is possible to readily show that each one implies the other, by a few lines of manipulation."

    Does anyone know how to do that?

    Let f be cts on [a,b], and let F(x) = [tex]\int f(x) dx[/tex] be called F, the antiderivative of f. Then

    [tex]\int_{a}^{b}f(x) dx = F(b) - F(a)[/tex]

    By theorem [tex]\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)[/tex], we know [tex]\int_{a}^{x}f(t) dt = F(b) - F(a)[/tex] & F(x) has the same derivative, f(x). Hence there will be a constant c such that [tex]\int_{a}^{x}f(t)dt = F(x)+c[/tex] When x = a, we get

    F(a) + c = [tex]\int_{a}^{a}f(t)dt = 0[/tex] so c = -F(a)

    Hence, [tex]\int_{a}^{x}f(t)dt [/tex] = F(x)-F(a)
    When x = b this yields

    [tex]\int_{a}^{b}f(t)dt = F(b)-F(a)[/tex]

    I'm not sure if this really answers the question though.
  2. jcsd
  3. Sep 24, 2008 #2


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    do you know the mean value theorem?
  4. Sep 24, 2008 #3
    A little, how can I use it to show that in those two theorems each one implies the other one?
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