Fundamental Theorem of Calculus/Second Part/Proof 2
Theorem
Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.
Then:
 $f$ has a primitive on $\left[{a \,.\,.\, b}\right]$
 If $F$ is any primitive of $f$ on $\left[{a \,.\,.\, b}\right]$, then:
 $\displaystyle \int_a^b f \left({t}\right) \ \mathrm d t = F \left({b}\right)  F \left({a}\right) = \left[{ F \left({t}\right) }\right]_a^b$
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 $\left[{x_{k1} \,.\,.\, x_k}\right]$ where:
 $ a = x_0 < x_1 \cdots < x_{k1} < x_k = b$
Fix such a subdivision of the interval $\left[{a \,.\,.\, b}\right]$; call it $P$.
Next, we observe the following telescoping sum identity:
\((1):\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \sum_{i \mathop = 1}^{k}\left({ F \left({x_i}\right)  F \left({ x_{i1} }\right)}\right)\)

\(=\)

\(\displaystyle \)

\(\)

\(\displaystyle \)

\(\displaystyle F \left({b}\right)  F \left({a}\right)\)

\(\displaystyle \)

\(\displaystyle \)



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 $I_i = \left[{x_{i1} \,.\,.\, x_i}\right]$ there is some $c_i$ such that:
 $F' \left({c_i}\right) = \dfrac {F \left({x_i}\right)  F \left({x_{i1}}\right)} {x_{i}  x_{i1}}$
It follows that:

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle F \left({x_i}\right)  F \left({x_{i1} }\right)\)

\(=\)

\(\displaystyle \)

\(\)

\(\displaystyle \)

\(\displaystyle F' \left({c_i}\right) \left({ x_{i}  x_{i1} }\right)\)

\(\displaystyle \)

\(\displaystyle \)



\((2):\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle F \left({b}\right)  F \left({a}\right)\)

\(=\)

\(\displaystyle \)

\(\)

\(\displaystyle \)

\(\displaystyle \sum_{i \mathop = 1}^k F' \left({c_i}\right) \left({ x_{i}  x_{i1} }\right)\)

\(\displaystyle \)

\(\displaystyle \)

By equation $\left({1}\right)$



\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(=\)

\(\displaystyle \)

\(\)

\(\displaystyle \)

\(\displaystyle \sum_{i \mathop = 1}^{k} f \left({c_i}\right) \left({ x_{i}  x_{i1} }\right)\)

\(\displaystyle \)

\(\displaystyle \)

Because $F' = f$


From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \left[{x_{i1} \,.\,.\, x_i}\right]$):
 $\displaystyle \inf_{x \in I_i} \ f \left({x}\right) \le f \left({c_i}\right) \le \sup_{x \in I_i} \ f \left({x}\right)$
From the definitions of upper and lower sums, we conclude for any subdivision $P$:
 $\displaystyle L \left({P}\right) \le \sum_{i \mathop = 1}^{k} f \left({c_i}\right) \left({ x_{i}  x_{i1} }\right) \le U \left({P}\right)$
Lastly, from the definition of a definite integral and from $\left({2}\right)$, we conclude:
 $\displaystyle F \left({b}\right)  F \left({a}\right) = \int_a^b f \left({t}\right) \ \mathrm d t$
$\blacksquare$
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