Hi everyone,

I was hoping someone could please look over this proof for any errors. My assignment is to write a formal proof of the fundamental theorem of calculus. I am instructed to use in my proof the Cauchy Criterion for integrals and the mean value theorem. I have been provided a hint that I am suppose to apply the MVT to subintervals of a partition of the interval I am integrating over, and that to complete the proof I must evaluate several quantitative relationshipships between Riemann integrals, Riemann sums, and Darboux sums. Here is my proof.

If f(x) is a differentiable, real-valued function, and f'(x) is continuous and integrable, then:

\int_a^bf'(x)\,dx = f(b) - f(a)

Let \epsilon > 0 be given. Since f'(x) is integrable, there exists a partition P such that:

(1) U(f'(x), P) - L(f'(x), P) < \epsilon

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

f'(x_k) = \frac{f(t_k) - f(t_{k-1})}{t_k - t_{k-1}}

or equivalently,

f'(x_k)(t_{k} - t_{k-1}) = f(t_k) - f(t_{k-1})

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

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


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


(2) L(f'(x), P) \leq f(b) - f(a) \leq U(f'(x), P)


We also know that


L(f'(x), P) \leq \int_a^b f'(x)\,dx \leq U(f'(x), P)


Equivalently,

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

Adding inequalities (2) and (3) gives:

(4) 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)

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

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


Since \epsilon is arbitrary and thus given the right partition can be made so small that it might as well be zero, it follows that \int_a^bf'(x)\,dx = f(b) - f(a)