Fundamental Theorem of Calculus/First Part

Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a .. b}\right]$.

Let $F$ be a real function which is defined on $\left[{a .. b}\right]$ by:

$\displaystyle F \left({x}\right) = \int_a^x f \left({t}\right) \ \mathrm d t$

Then $F$ is a primitive of $f$ on $\left[{a .. b}\right]$.

Corollary

$\displaystyle \frac {\mathrm d}{\mathrm dx}\int_a^x f \left({t}\right) \ \mathrm d t = f\left({x}\right)$

Proof 1

To show that $F$ is a primitive of $f$ on $\left[{a .. b}\right]$, we need to establish the following:

• $F$ is continuous on $\left[{a .. b}\right]$
• $F$ is differentiable on the open interval $\left({a .. b}\right)$
• $\forall x \in \left[{a .. b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$.

Proof that F is Continuous

Since $f$ is continuous on $\left[{a .. b}\right]$, it follows from the Continuity Property that $f$ is bounded on $\left[{a .. b}\right]$.

Suppose that:

$\forall t \in \left[{a .. b}\right]: \left|{f \left({t}\right)}\right| < \kappa$

Let $x, \xi \in \left[{a .. b}\right]$.

From Sum of Integrals on Adjacent Intervals‎, we have that:

$\displaystyle \int_a^x f \left({t}\right) \ \mathrm d t + \int_x^\xi f \left({t}\right) \ \mathrm d t = \int_a^\xi f \left({t}\right) \ \mathrm d t$

That is:

$\displaystyle F \left({x}\right) + \int_x^\xi f \left({t}\right) \ \mathrm d t = F \left({\xi}\right)$

So:

$\displaystyle F \left({x}\right) - F \left({\xi}\right) = - \int_x^\xi f \left({t}\right) \ \mathrm d t = \int_\xi^x f \left({t}\right) \ \mathrm d t$

From the corollary to Upper and Lower Bounds of Integral:

$\left|{F \left({x}\right) - F \left({\xi}\right)}\right| < \kappa \left|{x - \xi}\right|$

Thus it follows that $F$ is continuous on $\left[{a .. b}\right]$.

$\Box$

Proof that F is Differentiable and f is its Derivative

• Second, that $F$ is differentiable on $\left({a .. b}\right)$ and that $\forall x \in \left[{a .. b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$:

Let $x, \xi \in \left[{a .. b}\right]$ such that $x \ne \xi$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} - f \left({\xi}\right)\)	\(=\)	\(\displaystyle \frac 1 {x - \xi} \left({F \left({x}\right) - F \left({\xi}\right) - \left({x - \xi}\right) f \left({\xi}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {x - \xi} \left({\int_\xi^x f \left({t}\right) \ \mathrm d t - \left({x - \xi}\right) f \left({\xi}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {x - \xi} \int_\xi^x \left({f \left({t}\right) - f \left({\xi}\right)}\right) \ \mathrm d t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Integral of Function plus Constant, putting $c = f \left({\xi}\right)$

Now, let $\epsilon > 0$.

If $\xi \in \left({a . . b}\right)$, then $f$ is continuous at $\xi$.

So for some $\delta > 0$, $\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$ provided $\left|{t - \xi}\right| < \delta$.

So provided $\left|{x - \xi}\right| < \delta$ it follows that $\left|{f \left({t}\right) - f \left({\xi}\right)}\right| < \epsilon$ for any $t$ in an interval whose endpoints are $x$ and $\xi$.

So from the corollary to Upper and Lower Bounds of Integral, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} - f \left({\xi}\right)}\right\vert\)	\(=\)	\(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert} \left\vert{\int_\xi^x \left({f \left({t}\right) - f \left({\xi}\right)}\right) \ \mathrm dt}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \frac 1 {\left\vert{x - \xi}\right\vert} \epsilon \left\vert{x - \xi}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \epsilon\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

provided $0 < \left|{x - \xi}\right| < \delta$.

But that's what this means:

$\displaystyle \frac {F \left({x}\right) - F \left({\xi}\right)} {x - \xi} \to f \left({\xi}\right)$ as $x \to \xi$

So $F$ is differentiable on $\left({a .. b}\right)$, and:

$\forall x \in \left[{a .. b}\right]: F^{\prime} \left({x}\right) = f \left({x}\right)$

$\blacksquare$

Proof 2

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D_x F\left({x}\right)\)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} \frac 1 {\Delta x} \left({\int_a^{x+\Delta x} f \left({t}\right) \ \mathrm d t - \int_a^x f \left({t}\right) \ \mathrm d t}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	definition of derivative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} \frac 1 {\Delta x} \left({\int_x^a f \left({t}\right) \ \mathrm d t + \int_a^{x+\Delta x} f \left({t}\right) \ \mathrm d t}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because $\int_a^b f \left({x}\right) \ \mathrm d x = - \int_b^a f \left({x}\right) \ \mathrm d x$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} \frac 1 {\Delta x} \left({\int_x^{x+\Delta x} f \left({t}\right) \ \mathrm d t}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Integrals on Adjacent Intervals

Suppose $\Delta x > 0$.

By the Mean Value Theorem for Integrals, there exists some $k \in \left[{x..x + \Delta x}\right]$ such that:

$\displaystyle \int_x^{x + \Delta x} f \left({x}\right) \ \mathrm d x = f \left({k}\right) \left({x + \Delta x - x}\right) = f\left({k}\right) \Delta x$

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \lim_{\Delta x \to 0} \frac 1 {\Delta x} \left({\int_x^{x+\Delta x} f \left({t}\right) \ \mathrm d t}\right)\)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} \frac 1 {\Delta x} f\left({k}\right) \Delta x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{\Delta x \to 0} f\left({k}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

By the definition of $k$:

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

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \lim_{\Delta x \to 0} f\left({k}\right)\)	\(=\)	\(\displaystyle \lim_{k \to x} \ f\left({k}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f\left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because $f$ is continuous

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

Hence the result, by the definition of primitive.

$\blacksquare$

Proof of Corollary

Follows from the main result and the definition of primitive.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 13.12$
• Roland E. Larson,  Robert P. Hostetler and Bruce H. Edwards: Calculus: 8th Edition (2005): $\S 4.4$