Fundamental Theorem of Calculus/First Part/Proof 2

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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.

$\blacksquare$


Sources