Differentiating an Accumulation Function

Accumulating a Quantity

An integral with a fixed starting point and a changing endpoint describes an amount accumulated as the endpoint moves. For example, if $f(t)$ is a rate, then integrating $f$ from a fixed time $a$ to a time $x$ gives the net amount accumulated between those times.

The resulting function depends on $x$. A central result of calculus says that, when $f$ is continuous, the rate at which this accumulated amount changes at $x$ is simply the value of $f$ at $x$. The endpoint value matters because a small increase in the endpoint adds a thin interval of accumulation near that endpoint.

Definition: Let $f$ be integrable on the interval between a fixed number $a$ and each $x$ under consideration. The function $$ A(x)=\int_a^x f(t)\,dt $$ is called an accumulation function. The integration variable $t$ is a dummy variable; $x$ is the input of the function and the variable upper endpoint.

The fixed lower limit $a$ sets the starting point for the accumulation. Changing $x$ changes how far the integral extends. The value $A(x)$ records the signed accumulated amount, not necessarily a positive area.

The Endpoint Derivative Rule

The first part of the Fundamental Theorem of Calculus gives the derivative of this function. The continuity condition is important: assume $f$ is continuous on an interval containing $a$ and the $x$-values being considered. At every interior point $x$ of that interval, the accumulation function is differentiable.

Theorem: If $f$ is continuous on an interval containing $a$, and $$ A(x)=\int_a^x f(t)\,dt, $$ then, for each interior point $x$ of that interval, $$ A'(x)=f(x). $$

The rule is direct: evaluate the integrand at the variable upper endpoint. The result is not the integral itself, nor is it the derivative of the integrand. The integral defines $A$; the theorem says how to differentiate $A$.

Worked Example: A Polynomial Integrand

Let $$ A(x)=\int_2^x (3t^2-4t+1)\,dt. $$ The integrand is continuous for every real $t$, so the endpoint derivative rule applies for every real $x$. Evaluate the integrand at $t=x$:

$$ A'(x)=3x^2-4x+1. $$

For a check, evaluate the integral using the antiderivative $t^3-2t^2+t$:

\begin{aligned} A(x) &=[t^3-2t^2+t]_2^x\\ &=(x^3-2x^2+x)-(2^3-2(2^2)+2)\\ &=x^3-2x^2+x-(8-8+2)\\ &=x^3-2x^2+x-2. \end{aligned}

Differentiating this expression gives $A'(x)=3x^2-4x+1$, as predicted. At $x=2$, the derivative is $3(2^2)-4(2)+1=12-8+1=5$, which equals the integrand evaluated at the endpoint $2$.

Worked Example: A Trigonometric Integrand

Consider $$ B(x)=\int_{\pi/2}^x \sin t\,dt. $$ Since $\sin t$ is continuous, the theorem gives the derivative by substituting the endpoint into the integrand:

B'(x)=\sin x.

To check, an antiderivative of $\sin t$ is $-\cos t$, so

\begin{aligned} B(x) &=[-\cos t]_{\pi/2}^x\\ &=-\cos x-(-\cos(\pi/2))\\ &=-\cos x. \end{aligned}

Differentiating $-\cos x$ gives $\sin x$. For instance, at $x=\pi/6$, the theorem gives $B'(\pi/6)=\sin(\pi/6)=1/2$, matching the derivative of the explicit expression at that point.

Worked Example: An Exponential Integrand

Let $$ C(x)=\int_0^x e^{2t}\,dt. $$ The integrand is continuous, so

C'(x)=e^{2x}.

As a check, an antiderivative of $e^{2t}$ is $\frac12 e^{2t}$, and therefore

\begin{aligned} C(x) &=\left[\frac12 e^{2t}\right]_0^x\\ &=\frac12 e^{2x}-\frac12 e^0\\ &=\frac12(e^{2x}-1). \end{aligned}

Differentiate the expression: $C'(x)=\frac12(2e^{2x})=e^{2x}$, which agrees with the theorem. At $x=0$, $C'(0)=e^0=1$.

Why the Derivative Equals the Endpoint Value

The theorem can be understood directly from the definition of a derivative. Fix an interior point $x$, and take a nonzero change $h$ small enough that $x+h$ remains in the interval. The difference quotient for $A$ is

\begin{aligned} \frac{A(x+h)-A(x)}{h} &=\frac{1}{h}\left(\int_a^{x+h}f(t)\,dt-\int_a^x f(t)\,dt\right)\\ &=\frac{1}{h}\int_x^{x+h}f(t)\,dt. \end{aligned}

The second equality follows by additivity of definite integrals. Subtract $f(x)$ from the quotient. Since the integral of the constant $f(x)$ from $x$ to $x+h$ is $h f(x)$, this gives

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

Take the absolute value. The absolute value of an integral is bounded by the interval length times an upper bound for the absolute value of its integrand. In particular, continuity of $f$ ensures that the following supremum is finite for sufficiently small $h$, and yields

\left| \frac{A(x+h)-A(x)}{h}-f(x) \right| \leq \sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|.

To see that the right-hand side tends to zero, let $\varepsilon>0$. Continuity of $f$ at $x$ gives a $\delta>0$ such that $|f(t)-f(x)|<\varepsilon$ whenever $|t-x|<\delta$. If $0<|h|<\delta$, every point $t$ between $x$ and $x+h$ satisfies $|t-x|<\delta$. Consequently, the supremum in the bound is at most $\varepsilon$. Since this holds for every positive $\varepsilon$, the bound tends to zero as $h\to0$. Thus the difference quotient tends to $f(x)$, which proves $A'(x)=f(x)$.

This argument uses continuity to control the integrand near $x$, not the derivative conclusion it is establishing. The small interval contributes an average value close to $f(x)$, so its contribution per unit change in the endpoint approaches $f(x)$.

Interpreting the Result and Checking Your Work

If $f$ represents a rate of change, $A(x)$ represents the net quantity accumulated from the fixed starting point to $x$. The equation $A'(x)=f(x)$ says that the instantaneous rate of change of the accumulated quantity equals the current rate. For example, if $f$ has units of litres per minute and $x$ is measured in minutes, then the integral has units of litres, while its derivative with respect to time has units of litres per minute, matching $f(x)$.

When applying the theorem, make these checks:

Identify the integrand $f(t)$ and the variable upper endpoint $x$.
Confirm that $f$ is continuous at the point where the derivative is being found.
Evaluate $f$ at the endpoint; do not integrate again to find the derivative.
If you also compute an explicit formula for the integral, differentiate it to verify the result.

The continuity requirement is not decorative. It is what allows the values of $f(t)$ near $x$ to be made uniformly close to $f(x)$, which is the key estimate in the proof. The theorem as stated guarantees the derivative at interior points where the continuity condition holds.

A Common Mistake

A frequent error is to differentiate the integrand instead of evaluating it at the endpoint. For example, for $$ D(x)=\int_0^x (t^2+1)\,dt, $$ a mistaken calculation might differentiate $t^2+1$ and then substitute $x$, giving $D'(x)=2x$. That is not the endpoint derivative rule: it would give $D'(2)=4$.

The correct calculation evaluates the original integrand at $t=x$:

D'(x)=x^2+1, \qquad D'(2)=2^2+1=5.

Check by evaluating the integral: $$ D(x)=\left[\frac{t^3}{3}+t\right]_0^x=\frac{x^3}{3}+x. $$ Differentiating gives $D'(x)=x^2+1$, and at $x=2$ this is $4+1=5$. The check confirms that the derivative is the integrand at the endpoint, not the derivative of the integrand.

Summary

For a fixed lower limit $a$, the accumulation function $A(x)=\int_a^x f(t)\,dt$ records the amount accumulated up to $x$. If $f$ is continuous, the first part of the Fundamental Theorem of Calculus gives $A'(x)=f(x)$ at interior points. The difference-quotient proof works because continuity makes $f(t)$ close to $f(x)$ throughout the short interval between $x$ and $x+h$.

Check Your Understanding

Answer each question using the fixed-start accumulation rule and its continuity condition.

If $A(x)=\int_1^x (t^3-2t)\,dt$, what is $A'(x)$, and what is $A'(2)$?
For $B(x)=\int_0^x \cos t\,dt$, what does the endpoint derivative rule give for $B'(x)$?
In the difference quotient proof, why is $A(x+h)-A(x)$ equal to $\int_x^{x+h}f(t)\,dt$?
Which property of $f$ ensures that the supremum of $|f(t)-f(x)|$ over the shrinking interval tends to zero?
For $C(x)=\int_0^x (t^2+4)\,dt$, what is the correct derivative, and what incorrect result would come from differentiating the integrand first?

Differentiating an Accumulation Function

What You'll Learn

Accumulating a Quantity