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:

$\ds \int_a^b \map f t \rd t = \map F b - \map F a = \bigintlimits {\map F t} a b$

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:

$\ds \int_a^b \map f t \rd t$

$=$

$\ds \int_a^a \map f t \rd t + \int_a^b \map f t \rd t$

$\ds $

$=$

$\ds \map G a + \map G b$

From the definition of $G$

$\ds $

$=$

$\ds \map G b - \map G a$

$\map G a = 0$

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:

$\ds \int_a^b \map f t \rd t$

$=$

$\ds \paren {\map G b + c} - \paren {\map G a + c}$

$\ds $

$=$

$\ds \map F b - \map F a$

$\blacksquare$