Proof of FTC Part I

The Difference Quotient as a Local Average

In the previous tutorial, the Fundamental Theorem of Calculus, Part I, was stated for the accumulation function $F(x)=\int_a^x f(t)\,dt$. Its central claim is that the derivative of this function recovers the continuous integrand. The proof turns on one idea: a difference quotient of $F$ is an average value of $f$ over a short interval.

Assume throughout that $a<b$, that $f:[a,b]\to\mathbb{R}$ is continuous, and that

F(x)=\int_a^x f(t)\,dt,\qquad x\in[a,b].

The integral is well-defined because a continuous function on a closed bounded interval is Riemann integrable. For any $x$ and $x+h$ in $[a,b]$, additivity of the integral, together with the oriented-integral convention when $h<0$, gives

F(x+h)-F(x)=\int_x^{x+h}f(t)\,dt.

Therefore, whenever $h\ne0$,

\frac{F(x+h)-F(x)}{h} =\frac{1}{h}\int_x^{x+h}f(t)\,dt.

For $h>0$, the right side is the usual average of $f$ over $[x,x+h]$. For $h<0$, the integral changes sign when its limits are reversed, as does the denominator. The quotient still represents the average over the interval between $x$ and $x+h$. We will make precise why this average approaches $f(x)$ as the interval shrinks.

Continuity Controls Short-Interval Averages

The relevant estimate compares an average with the value of the function at a point in the interval. Its proof uses the absolute-value inequality and order preservation for the Riemann integral, established earlier in the course.

Lemma (Local Average Estimate): Let $f$ be continuous at $c$, and let $u<v$ lie in its domain. Suppose $f$ is Riemann integrable on $[u,v]$ and $c\in[u,v]$. Then $$ \left|\frac{1}{v-u}\int_u^v f(t)\,dt-f(c)\right| \leq \sup_{t\in[u,v]}|f(t)-f(c)|. $$

Proof. Linearity of the integral gives

\frac{1}{v-u}\int_u^v f(t)\,dt-f(c) =\frac{1}{v-u}\int_u^v\bigl(f(t)-f(c)\bigr)\,dt.

The Absolute-Value Inequality for the Riemann Integral now implies

\left|\frac{1}{v-u}\int_u^v f(t)\,dt-f(c)\right| \leq \frac{1}{v-u}\int_u^v|f(t)-f(c)|\,dt.

For every $t\in[u,v]$, the integrand in the last expression is at most $\sup_{s\in[u,v]}|f(s)-f(c)|$. By order preservation and the integral of a constant,

\frac{1}{v-u}\int_u^v|f(t)-f(c)|\,dt \leq \frac{1}{v-u}(v-u)\sup_{s\in[u,v]}|f(s)-f(c)|.

This is the asserted estimate. In particular, continuity at $c$ ensures that the right side is small whenever the whole interval $[u,v]$ is sufficiently close to $c$. $\square$

For the derivative proof, a direct version of this argument is especially useful. Subtract $f(x)$ from the difference quotient and use linearity:

\frac{F(x+h)-F(x)}{h}-f(x) =\frac{1}{h}\int_x^{x+h}\bigl(f(t)-f(x)\bigr)\,dt.

Although the integral is oriented when $h<0$, taking absolute values handles either sign. The Absolute-Value Inequality gives

\left|\frac{F(x+h)-F(x)}{h}-f(x)\right| \leq \frac{1}{|h|}\int_{\min\{x,x+h\}}^{\max\{x,x+h\}}|f(t)-f(x)|\,dt.

If continuity at $x$ ensures that $|f(t)-f(x)|<\varepsilon$ throughout the interval of integration, order preservation bounds the last expression by $\varepsilon$. The interval has length $|h|$, which cancels the denominator. This control works for positive and negative $h$.

Worked Example: A Difference Quotient for a Cubic Integrand

Let $f(t)=t^3-2t$ on $[-1,2]$, and define $F(x)=\int_{-1}^x f(t)\,dt$. Evaluating the integral gives

F(x)=\frac{x^4}{4}-x^2+\frac{3}{4}.

Indeed, an antiderivative is $t^4/4-t^2$; its value at $-1$ is $1/4-1=-3/4$, so subtracting that value adds $3/4$. For $x$ in the interior and $h\ne0$ small enough that $x+h\in[-1,2]$, expand the difference quotient:

\begin{aligned} \frac{F(x+h)-F(x)}{h} &=\frac{(x+h)^4-x^4}{4h} -\frac{(x+h)^2-x^2}{h}\\ &=x^3-2x+h\left(\frac{3}{2}x^2-1\right)+h^2x+\frac{h^3}{4}. \end{aligned}

The terms following $x^3-2x$ tend to zero as $h\to0$, so $F'(x)=x^3-2x=f(x)$. This calculation exhibits the same limiting behavior as the general proof: the average over a shrinking interval approaches the integrand at its starting point.

Proof of the Fundamental Theorem of Calculus, Part I

Theorem (Fundamental Theorem of Calculus, Part I): Let $a<b$, let $f:[a,b]\to\mathbb{R}$ be continuous, and define $$ F(x)=\int_a^x f(t)\,dt. $$ Then $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'(x)=f(x)$ for every $x\in(a,b)$. The right-hand derivative at $a$ exists and equals $f(a)$, and the left-hand derivative at $b$ exists and equals $f(b)$.

Proof. Since $f$ is continuous on the closed bounded interval $[a,b]$, it is bounded: there is $M\geq0$ such that $|f(t)|\leq M$ for every $t\in[a,b]$. The Increment Bound for an Accumulation Function from the previous tutorial gives

|F(y)-F(x)|\leq M|y-x| \qquad (x,y\in[a,b]).

If $M>0$, then for any $\varepsilon>0$, taking $\delta=\varepsilon/M$ shows that $|x-y|<\delta$ implies $|F(y)-F(x)|<\varepsilon$. If $M=0$, the bound says $F(y)=F(x)$ for every $x,y$, so $F$ is constant. Thus $F$ is continuous on $[a,b]$.

Now fix $x\in(a,b)$. For every nonzero $h$ small enough that $x+h\in[a,b]$, the increment identity gives

\frac{F(x+h)-F(x)}{h}-f(x) =\frac{1}{h}\int_x^{x+h}\bigl(f(t)-f(x)\bigr)\,dt.

Let $\varepsilon>0$. By continuity of $f$ at $x$, there is $\delta>0$ such that $t\in[a,b]$ and $|t-x|<\delta$ imply $|f(t)-f(x)|<\varepsilon$. Choose $h\ne0$ with $|h|<\delta$ and $x+h\in[a,b]$. Every $t$ between $x$ and $x+h$ satisfies $|t-x|\leq|h|<\delta$. Applying the absolute-value inequality and order preservation to the displayed integral therefore gives

\left|\frac{F(x+h)-F(x)}{h}-f(x)\right| \leq \frac{1}{|h|}\int_{\min\{x,x+h\}}^{\max\{x,x+h\}}\varepsilon\,dt =\varepsilon.

As $h\to0$, this proves that the difference quotient tends to $f(x)$. Hence $F'(x)=f(x)$ for every $x\in(a,b)$.

At $a$, the domain permits positive increments $h>0$ with $a+h\leq b$. The same identity and estimate apply with $x=a$, using continuity of $f$ at $a$. Thus

\lim_{h\to0^+}\frac{F(a+h)-F(a)}{h}=f(a).

At $b$, the domain permits negative increments $h<0$ with $b+h\geq a$. Using continuity at $b$, the same estimate gives

\lim_{h\to0^-}\frac{F(b+h)-F(b)}{h}=f(b).

These are the right-hand derivative at $a$ and the left-hand derivative at $b$. The assumption $a<b$ ensures that each endpoint has an interval of allowed increments on its inward side. This completes the proof. $\square$

Worked Example: The Derivative at a Corner of the Integrand

On $[-1,1]$, let $f(t)=|t|$ and define $F(x)=\int_0^x |t|\,dt$, using an oriented integral when $x<0$. For $x\geq0$,

F(x)=\int_0^x t\,dt=\frac{x^2}{2}.

For $x<0$, $|t|=-t$ between $x$ and $0$, so

F(x)=-\int_x^0(-t)\,dt=-\frac{x^2}{2}.

At the origin, $f$ is continuous even though it is not differentiable there. The difference quotient of $F$ is

\frac{F(h)-F(0)}{h} = \begin{cases} h/2, & h>0,\\ -h/2, & h<0. \end{cases}

For $h>0$, this is $h/2$; for $h<0$, it is also $|h|/2$. In either case it tends to $0=f(0)$. The theorem needs continuity of the integrand, not differentiability of the integrand.

Worked Example: Checking Both Endpoint Derivatives

Let $f(t)=t+2$ on $[1,3]$, and set $G(x)=\int_1^x(t+2)\,dt$. Direct evaluation gives

G(x)=\frac{x^2}{2}+2x-\frac{5}{2}.

At the left endpoint, $G(1)=0$. For $h>0$ small enough that $1+h\leq3$,

\frac{G(1+h)-G(1)}{h} =\frac{3h+h^2/2}{h} =3+\frac{h}{2}\longrightarrow 3=f(1).

At the right endpoint, $G(3)=8$. For $h<0$ small enough that $3+h\geq1$,

\frac{G(3+h)-G(3)}{h} =\frac{5h+h^2/2}{h} =5+\frac{h}{2}\longrightarrow 5=f(3).

The increments have opposite signs at the two endpoints, but each quotient approaches the integrand’s value at that endpoint.

What the Proof Uses—and What It Does Not

The proof separates into two tasks. Boundedness of $f$ and the Increment Bound establish continuity of the accumulation function. The derivative formula requires a more local fact: continuity at the point forces the values of $f$ across a sufficiently short interval to stay close to $f(x)$. Consequently, their average stays close to $f(x)$ as well.

The orientation convention is essential when an increment is negative. Writing $\int_x^{x+h} f(t)\,dt$ as though it were an ordinary integral with increasing limits would lose a minus sign. The oriented integral and the signed denominator work together, so the difference-quotient argument applies on both sides of an interior point.

Finally, the interval hypothesis $a<b$ is needed for the endpoint statements. On an interval of positive length, there are allowed increments into the interval from each endpoint. If $a=b$, the domain is a single point and neither of those one-sided difference quotients is available. The next part of the Fundamental Theorem of Calculus will use the derivative relationship proved here to evaluate definite integrals from antiderivatives.

Check Your Understanding

Use the proof and examples to answer the following questions.

How does additivity of the integral express $F(x+h)-F(x)$ when $h<0$?
Why does continuity of $f$ make the average of $f$ over a shrinking interval approach its value at the fixed point?
Which earlier result gives continuity of the accumulation function once $f$ is known to be bounded?
What is the allowed sign of the increment in the one-sided derivative at $a$, and at $b$?
Why must the theorem assume $a<b$ when it asserts endpoint derivatives?

What You'll Learn