## Fundamental Theorem of Calculus

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