Why Continuity Gives the Accumulation Derivative

From Accumulated Area to a Derivative

An accumulation function records how much of a function has been added up between a fixed starting point and a variable endpoint. Its defining formula involves an integral, while its derivative describes the function’s instantaneous rate of change. The key question is why those two descriptions are connected.

The answer is already present in the difference quotient. A change in the endpoint adds a thin interval to the accumulated integral. Dividing that change by the length of the interval gives an average value of the integrand there. When the integrand is continuous, its values on a sufficiently short interval near the endpoint are all close to its value at the endpoint. The average must then be close as well.

Definition: Let $f$ be integrable on the intervals under consideration, and fix a starting point $a$. The accumulation function based at $a$ is $$ A(x)=\int_a^x f(t)\,dt. $$ The integral is oriented: if $x<a$, then $\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt$.

We will examine the derivative at an interior point $c$ of the interval. The difference quotient uses a nonzero increment $h$ small enough that $c+h$ remains in the interval. The derivative exists at $c$ if this quotient approaches a finite limit as $h$ approaches zero from either side.

The Difference Quotient Is an Average

By the definition of $A$, and by additivity of definite integrals, the change in the accumulation function from $c$ to $c+h$ is the integral of $f$ over just that short interval:

A(c+h)-A(c) = \int_a^{c+h}f(t)\,dt-\int_a^c f(t)\,dt = \int_c^{c+h}f(t)\,dt.

Therefore, for $h\ne 0$, the difference quotient is

\frac{A(c+h)-A(c)}{h} = \frac{1}{h}\int_c^{c+h}f(t)\,dt.

When $h>0$, the integral covers an interval of length $h$, so the quotient is the average value of $f$ on that interval. When $h<0$, the integral is oriented in the negative direction, and division by the negative number $h$ again produces an average over an interval of length $|h|$. In either case, the quotient compares nearby values of $f$ with $f(c)$.

Why Continuity Forces the Average to Converge

Continuity of $f$ at $c$ means that for every $\varepsilon>0$, there is a $\delta>0$ such that $|f(t)-f(c)|<\varepsilon$ whenever $|t-c|<\delta$. Thus, once $h$ is sufficiently small, every point between $c$ and $c+h$ has an integrand value within $\varepsilon$ of $f(c)$.

Theorem: Suppose $f$ is integrable on the intervals used to define $A(x)=\int_a^x f(t)\,dt$, and suppose $f$ is continuous at an interior point $c$. Then $A$ is differentiable at $c$, and $$ A'(c)=f(c). $$

The continuity assumption supplies the local control needed for the proof. For any $\varepsilon>0$, choose $\delta>0$ so that $|f(t)-f(c)|<\varepsilon$ whenever $|t-c|<\delta$. If $0<|h|<\delta$, every point on the interval between $c$ and $c+h$ satisfies that inequality. Subtracting $f(c)$ from the difference quotient gives

\frac{A(c+h)-A(c)}{h}-f(c) = \frac{1}{h}\int_c^{c+h}\bigl(f(t)-f(c)\bigr)\,dt.

For either sign of $h$, the absolute value of an oriented integral is at most the integral of the absolute value over the interval with its endpoints in increasing order. That interval has length $|h|$. Consequently,

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

This holds for every $h$ satisfying $0<|h|<\delta$. It proves directly from the definition of a limit that the difference quotient approaches $f(c)$ as $h\to 0$. Hence $A'(c)=f(c)$. The bound also shows why both sides of the derivative limit work: it depends on $|h|$, not on $h$ being positive.

Worked Examples

Worked Example: A Polynomial Integrand

Let $f(t)=t^2+1$, let $A(x)=\int_0^x(t^2+1)\,dt$, and find the derivative of $A$ at $c=2$ from the difference quotient.

First evaluate the accumulation function:

A(x)=\int_0^x(t^2+1)\,dt = \left[\frac{t^3}{3}+t\right]_0^x = \frac{x^3}{3}+x.

For $h\ne0$, substitute $x=2+h$ and $x=2$ into the difference quotient:

\begin{aligned} \frac{A(2+h)-A(2)}{h} &= \frac{\frac{(2+h)^3}{3}+(2+h)-\left(\frac{8}{3}+2\right)}{h}\\ &= \frac{\frac{8+12h+6h^2+h^3}{3}+2+h-\frac{8}{3}-2}{h}\\ &= \frac{4h+2h^2+\frac{h^3}{3}+h}{h}\\ &= 5+2h+\frac{h^2}{3}. \end{aligned}

As $h\to0$, the quotient tends to $5$. The integrand’s value at the endpoint is $f(2)=2^2+1=5$, as the theorem predicts. As a check, differentiating the evaluated accumulation function gives $A'(x)=x^2+1$, so $A'(2)=5$.

Worked Example: A Trigonometric Integrand

Let $f(t)=\cos t$ and $A(x)=\int_0^x\cos t\,dt=\sin x$. We show from the difference quotient why $A'(0)=1$.

At $c=0$, the quotient is

\frac{A(0+h)-A(0)}{h} = \frac{1}{h}\int_0^h\cos t\,dt.

Since $\cos t$ is continuous at zero and $\cos 0=1$, the theorem’s estimate applies. More explicitly, the difference from $1$ is bounded by

\left| \frac{1}{h}\int_0^h\cos t\,dt-1 \right| = \left| \frac{1}{h}\int_0^h(\cos t-1)\,dt \right| \leq \sup_{t\text{ between }0\text{ and }h}|\cos t-1|.

As $h\to0$, continuity makes the supremum on the right approach zero: all points between $0$ and $h$ approach $0$, and $\cos t$ approaches $\cos 0=1$. Thus the difference quotient tends to $1$, so $A'(0)=1$. Checking the explicit formula $A(x)=\sin x$, we have $A'(x)=\cos x$, and $A'(0)=\cos 0=1$.

Worked Example: A Continuous Integrand with a Corner

Let $f(t)=|t|$ and $A(x)=\int_0^x|t|\,dt$. The function $f$ has a corner at zero, but it is continuous there. We find $A'(0)$ directly.

For $x\geq0$, the integral is $\int_0^x t\,dt=x^2/2$. For $x<0$, the orientation and the identity $|t|=-t$ on $[x,0]$ give

A(x) = -\int_x^0|t|\,dt = -\int_x^0(-t)\,dt = -\left[-\frac{t^2}{2}\right]_x^0 = -\frac{x^2}{2}.

Thus $A(x)=x|x|/2$ for both signs of $x$, and $A(0)=0$. For every $h\ne0$,

\frac{A(0+h)-A(0)}{h} = \frac{h|h|/2}{h} = \frac{|h|}{2}.

Since $|h|/2\to0$, the derivative is $A'(0)=0=f(0)$. This is also confirmed by differentiating the piecewise formula: for $x>0$, $A'(x)=x$, while for $x<0$, $A'(x)=-x$; both expressions approach zero at zero. The example shows that the integrand need not itself be differentiable at $c$; continuity at $c$ is enough for this result.

A Common Mistake: Ignoring the Division by the Interval Length

A shrinking interval makes its integral shrink in many familiar situations. But the difference quotient divides that integral by the interval length, which is also shrinking. It is therefore not valid to conclude that the quotient tends to zero merely because the numerator does.

For a concrete example, take $f(t)=1$ and $A(x)=\int_0^x1\,dt=x$. A mistaken argument would say that $A(h)-A(0)=h\to0$, so the derivative is zero. The actual calculation is

\frac{A(h)-A(0)}{h} = \frac{h-0}{h} = 1 \qquad (h\ne0).

Thus $A'(0)=1$, which agrees with $f(0)=1$. The proof above avoids this mistake by estimating the difference between the average value and $f(c)$, rather than estimating the integral change alone.

Why the Argument Matters

The result connects local information about the integrand to the rate of change of accumulated quantity. On a short interval, the average value of a continuous function is close to its value at the endpoint. As the interval shrinks, that average settles on the endpoint value, making it the derivative of the accumulation function.

Notice what the proof needs: the integral change can be written as an integral over the short interval, and continuity keeps the integrand close to $f(c)$ throughout that interval. The proof does not require $f$ to be differentiable. Nor does it rely on guessing a formula for $A$ and differentiating it. The difference quotient itself explains the derivative.

Key takeaway: The difference quotient of an accumulation function is an average of the integrand over a shrinking interval. Continuity makes every value in that average close to the endpoint value, so the average converges to $f(c)$.

Check Your Understanding

Use the difference-quotient argument and the continuity condition to answer these questions.

For $A(x)=\int_1^x f(t)\,dt$, rewrite $\bigl(A(c+h)-A(c)\bigr)/h$ as an integral over an interval with endpoints $c$ and $c+h$.
Why must the proof handle negative as well as positive values of $h$?
In the continuity estimate, where does the factor $|h|$ come from, and why does it cancel?
Can the integrand have a corner at $c$ while the accumulation function is differentiable there? Explain using the theorem’s condition.
Why does knowing that $A(c+h)-A(c)\to0$ by itself fail to determine the limit of the difference quotient?

Why Continuity Gives the Accumulation Derivative

What You'll Learn

From Accumulated Area to a Derivative

The Difference Quotient Is an Average

Why Continuity Forces the Average to Converge

Worked Examples

Worked Example: A Polynomial Integrand

Worked Example: A Trigonometric Integrand

Worked Example: A Continuous Integrand with a Corner

A Common Mistake: Ignoring the Division by the Interval Length

Why the Argument Matters

Check Your Understanding

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Vol. 2: Solutions Manual

Why Continuity Gives the Accumulation Derivative

What You'll Learn

From Accumulated Area to a Derivative

The Difference Quotient Is an Average

Why Continuity Forces the Average to Converge

Worked Examples

Worked Example: A Polynomial Integrand

Worked Example: A Trigonometric Integrand

Worked Example: A Continuous Integrand with a Corner

A Common Mistake: Ignoring the Division by the Interval Length

Why the Argument Matters

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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