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## Featured Proof

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 $\left[{x_{k-1} \,.\,.\, x_k}\right]$ where:

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

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

Next, we observe the following telescoping sum identity:

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

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 $I_i = \left[{x_{i-1} \,.\,.\, x_i}\right]$ 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:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle F \left({x_i}\right) - F \left({x_{i-1} }\right)$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle F' \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$(2):$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle F \left({b}\right) - F \left({a}\right)$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^k F' \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)$$ $$\displaystyle$$ $$\displaystyle$$ By equation $\left({1}\right)$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 1}^{k} f \left({c_i}\right) \left({ x_{i} - x_{i-1} }\right)$$ $$\displaystyle$$ $$\displaystyle$$ Because $F' = f$

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

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

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

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

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

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

$\blacksquare$