# Fundamental Theorem of Calculus

• Apr 2nd 2012, 02:25 PM
james121515
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 $\displaystyle f(x)$ is a differentiable, real-valued function, and $\displaystyle f'(x)$ is continuous and integrable, then:

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

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

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

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

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

or equivalently,

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

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

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

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

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

We also know that

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

Equivalently,

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

Adding inequalities (2) and (3) gives:

(4) $\displaystyle L(f'(x), P) - U(f'(x), P) \leq (f(b) - f(a)) - \int_a^bf'(x)\,dx \leq U(f'(x), P) - L(f'(x), P)$

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

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

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