Fundamental Theorem of Calculus/Second Part/Proof 1

Proof
Let $G$ be defined on $\closedint a b$ by:
 * $\ds \map G x = \int_a^x \map f t \rd t$

We have:
 * $\ds \map G a = \int_a^a \map f t \rd t = 0$ from Integral on Zero Interval
 * $\ds \map G b = \int_a^b \map f t \rd t$ from the definition of $G$ above.

Therefore, we can compute:

By the first part of the theorem, $G$ is a primitive of $f$ on $\closedint a b$.

By Primitives which Differ by Constant‎, we have that any primitive $F$ of $f$ on $\closedint a b$ satisfies $\map F x = \map G x + c$, where $c$ is a constant.

Thus we conclude: