Fundamental Theorem of Calculus/Second Part/Proof 1

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 $G$ be defined on $\left[{a \,.\,.\, b}\right]$ by:
 * $\displaystyle G \left({x}\right) = \int_a^x f \left({t}\right) \ \mathrm d t$

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

Therefore, we can compute:

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: