# Fundamental Theorem of Calculus/First Part/Proof 2

## Theorem

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

Let $F$ be a real function which is defined on $\closedint a b$ by:

$\ds \map F x = \int_a^x \map f t \rd t$

Then $F$ is a primitive of $f$ on $\closedint a b$.

## Proof

 $\ds \dfrac \d {\d x} \map F x$ $=$ $\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_a^{x + \Delta x} \map f t \rd t - \int_a^x \map f t \rd t}$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^a \map f t \rd t + \int_a^{x + \Delta x} \map f t \rd t}$ because $\displaystyle \int_a^b \map f x \rd x = - \int_b^a \map f x \rd x$ $\ds$ $=$ $\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}$ Sum of Integrals on Adjacent Intervals for Continuous Functions

Suppose $\Delta x > 0$ and both $x$ and $x + \Delta x$ are in the closed interval $\closedint a b$.

By hypothesis, $f$ is continuous on the closed interval $\closedint a b$

Thus it is also continuous on the closed interval $\closedint x {x + \Delta x}$.

Thus the conditions of the Mean Value Theorem for Integrals are fulfilled.

So, by the Mean Value Theorem for Integrals, there exists some $k \in \closedint x {x + \Delta x}$ such that:

$\displaystyle \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$

Then:

 $\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \paren {\int_x^{x + \Delta x} \map f t \rd t}$ $=$ $\ds \lim_{\Delta x \mathop \to 0} \frac 1 {\Delta x} \map f k \Delta x$ $\ds$ $=$ $\ds \lim_{\Delta x \mathop \to 0} \map f k$

By the definition of $k$:

$x \le k \le x + \Delta x$

which means that $k \to x$ as $\Delta x \to 0$.

So:

 $\ds \lim_{\Delta x \mathop \to 0} \map f k$ $=$ $\ds \lim_{k \mathop \to x} \map f k$ $\ds$ $=$ $\ds \map f x$ because $f$ is continuous

For $\Delta x < 0$, consider $k \in \closedint {x + \Delta x} x$, and the argument is similar.

Hence the result, by the definition of primitive.

$\blacksquare$