# Fundamental Theorem of Calculus/Second Part

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Then:

$(1): \quad f$ has a primitive on $\closedint a b$
$(2): \quad$ If $F$ is any primitive of $f$ on $\closedint a b$, then:
$\displaystyle \int_a^b \map f t \rd t = \map F b - \map F a = \bigintlimits {\map F t} a b$

## Proof 1

Let $G$ be defined on $\left[{a \,.\,.\, b}\right]$ by:

$\displaystyle G \left({x}\right) = \int_a^x f \left({t}\right) \rd t$

We have:

$\displaystyle G \left({a}\right) = \int_a^a f \left({t}\right) \rd t = 0$ from Integral on Zero Interval
$\displaystyle G \left({b}\right) = \int_a^b f \left({t}\right) \rd t$ from the definition of $G$ above.

Therefore, we can compute:

 $\displaystyle \int_a^b f \left({t}\right) \rd t$ $=$ $\displaystyle \int_a^a f \left({t}\right) \rd t + \int_a^b f \left({t}\right) \rd t$ Sum of Integrals on Adjacent Intervals for Continuous Functions $\displaystyle$ $=$ $\displaystyle G \left({a}\right) + G \left({b}\right)$ From the definition of $G$ $\displaystyle$ $=$ $\displaystyle G \left({b}\right) - G \left({a}\right)$ $G \left({a}\right) = 0$

By the first part of the theorem, $G$ is a primitive of $f$ on $\left[{a \,.\,.\, b}\right]$.

By Primitives which Differ by Constant‎, we have that any primitive $F$ of $f$ on $\left[{a \,.\,.\, b}\right]$ satisfies $F \left({x}\right) = G \left({x}\right) + c$, where $c$ is a constant.

Thus we conclude:

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

$\blacksquare$

## Proof 2

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:

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

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:

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

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

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

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:

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

$\blacksquare$