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
Let $f$ have a primitive $F$ which is continuous on the closed interval $\left[{a..b}\right]$.

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


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

By repeatedly adding and subtracting like quantities:


 * $F \left({x_k}\right) \underbrace{- F \left({x_{k-1}}\right) + F \left({x_{k-1}}\right)}_0 \ldots \underbrace{- F \left({x_1}\right)+ F\left({x_1}\right)}_0 - F \left({x_0}\right)$

$\implies$

$(A) \quad F \left({b}\right) - F \left({a}\right) = \sum_{i=1}^{k} F \left({x_i}\right) - F \left({x_{i-1}}\right)$

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

Because $F$ is differentiable, $F$ is continuous.

By the Mean Value Theorem, in every subinterval $\left[{x_{k-1}..x_k}\right]$ there is some $c_i$ where:
 * $F' \left({c_i}\right) = \dfrac {F \left({x_i}\right) - F \left({x_{i-1}}\right)} {\Delta x_i}$

where $\Delta x_i \equiv x_{i} - x_{i-1}$.

Multiplying both sides by $\Delta x_i$:


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

Substituting $F'(c_i)\Delta x_i$ into $(A)$:


 * $\displaystyle F \left({b}\right) - F \left({a}\right) = \sum_{i=1}^{k} F' \left({c_i}\right) \Delta x_i$

Because $F' = f$:


 * $\displaystyle F \left({b}\right) - F \left({a}\right) = \sum_{i=1}^{k} f \left({c_i}\right) \Delta x_i$

Because $f$ and $F$ are both continuous, we can take the limit $||\Delta|| \to 0$ of both sides, where $||\Delta||$ is the magnitude of the largest subinterval $ \left[{x_{k-1} .. x_k}\right]$:


 * $\displaystyle \lim_{||\Delta|| \to 0} F \left({b}\right) - F \left({a}\right) = \lim_{||\Delta|| \to 0} \sum_{i=1}^{k} f(c_i)\Delta x_i$

The $LHS$ is a constant. The $RHS$ is the definition of the integral:


 * $\displaystyle F \left({b}\right) - F \left({a}\right) = \int_a^b f \left({x}\right) \mathrm d x$