# Fundamental Theorem of Calculus/Second Part/Proof 1

## 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

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:

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

$\blacksquare$