Fundamental Theorem of Calculus/Second Part/Proof 2

Proof
As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$.

$\left[{a \,.\,.\, b}\right]$ can be divided into any number of closed subintervals of the form $\left[{x_{k-1} \,.\,.\, x_k}\right]$ where:


 * $ a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$

Fix such a finite subdivision of the interval $\left[{a \,.\,.\, b}\right]$; call it $P$.

Next, we observe the following telescoping sum identity:

Because $F' = f$, $F$ is differentiable.

By Differentiable Function is Continuous, $F$ is also continuous.

Therefore we can apply the Mean Value Theorem on $F$.

It follows that in every closed subinterval $I_i = \left[{x_{i - 1} \,.\,.\, x_i}\right]$ there is some $c_i$ such that:


 * $F' \left({c_i}\right) = \dfrac {F \left({x_i}\right) - F \left({x_{i - 1} }\right)} {x_i - x_{i - 1} }$

It follows that:

From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \left[{x_{i - 1} \,.\,.\, x_i}\right]$):


 * $\displaystyle \inf_{x \mathop \in I_i} \ f \left({x}\right) \le f \left({c_i}\right) \le \sup_{x \mathop \in I_i} \ f \left({x}\right)$

From the definitions of upper and lower sums, we conclude for any finite subdivision $P$:


 * $\displaystyle L \left({P}\right) \le \sum_{i \mathop = 1}^{k} f \left({c_i}\right) \left({x_i - x_{i - 1} }\right) \le U \left({P}\right)$

Lastly, from the definition of a definite integral and from $\left({2}\right)$, we conclude:


 * $\displaystyle F \left({b}\right) - F \left({a}\right) = \int_a^b f \left({t}\right) \rd t$