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Fundamental Theorem of Calculus

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if f(x) is a differentiable real-valued function, and f'(x) is continuous and integrable. Then:

    [tex]\int_a^bf'(x)\,dx = f(b) - f(a)[/tex]

    2. Hint Provided

    Use the Cauchy Criterion for integrals, the mean value theorem applied to subintervals of a partition, and the relationship between Riemann integrals, Riemann sums, and Darboux sums.

    3. The attempt at a solution

    Let [tex]\epsilon > 0[/tex] be given. By the Cauchy Criterion for integrals, since f'(x) is integrable, there exists a partition P such that

    (1) [tex]U(f'(x), P) - L(f'(x), P) < \epsilon[/tex]

    where[tex]U(f'(x), P), L(f'(x), P)[/tex] represent the upper and lower darboux sums.
    Now consider each subinterval [tex][t_{k-1}, t_k][/tex] of [tex]P = \{t_1 \leq t_2 \leq \cdots t_n\}[/tex], each of width [tex]t_{k} - t_{k-1}[/tex]. Since [tex]f(x)[/tex] is differentiable, by the mean value theorem, there exists [tex]x_k \in (t_{k-1}, t_k)[/tex] such that

    [tex]f'(x_k) = \frac{f(t_k) - f(t_{k-1})}{t_k - t_{k-1}}[/tex]

    or equivalently,

    [tex]f'(x_k)(t_{k} - t_{k-1}) = f(t_k) - f(t_{k-1})[/tex]

    Taking the sum of both sides and noting that since f(x) is continuous and thus [tex]\bigcup [f(t_k) - f(t_{k-1})] = f(b) - f(a)[/tex] we obtain:

    [tex]\sum f'(x_k)(t_{k}-t_{k-1}) = f(b) - f(a)[/tex]

    Since the sum on the left is a Riemann sum, we now know:

    (2) [tex]L(f'(x), P) \leq f(b) - f(a) \leq U(f'(x), P)[/tex]

    We also know that

    [tex]L(f'(x), P) \leq \int_a^b f'(x)\,dx \leq U(f'(x), P)[/tex]


    (3) [tex]-U(f'(x), P) \leq -\int_a^b f'(x)\,dx \leq -L(f'(x), P)[/tex]

    Adding inequalities (2) and (3) gives:

    (4) [tex]L(f'(x), P) - U(f'(x), P) \leq (f(b) - f(a)) - \int_a^bf'(x)\,dx \leq U(f'(x), P) - L(f'(x), P)[/tex]

    Equation (4) along with equation (1) implies that:

    [tex]\left|(f(b) - f(a) - \int_a^bf'(x)\,dx\right| \leq U(f'(x), P) - L(f'(x), P) < \epsilon[/tex]

    Since [tex]\epsilon[/tex] is very small depending on the partition, it might as well be zero, and from this it follows that [tex]\int_a^bf'(x)\,dx = f(b) - f(a)[/tex]
    Last edited: Feb 23, 2012
  2. jcsd
  3. Feb 23, 2012 #2


    Staff: Mentor

    No, this is the right forum section.
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