Variable Upper Bounds and the Chain Rule

A Moving Endpoint Produces a Derivative

An integral can describe how much of a quantity has accumulated between a fixed starting point and a changing endpoint. As the endpoint moves, the accumulated amount changes. The derivative measures that change.

For a continuous function $f$, define the accumulation function $F$ by integrating $f$ from a fixed number $a$ to the variable upper endpoint $x$. The Fundamental Theorem of Calculus, Part 1, says that the rate of change of this accumulation is the value of $f$ at the moving endpoint.

Definition: Let $f$ be continuous on an interval containing $a$ and $x$. The accumulation function with base point $a$ is $F(x)=\int_a^x f(t)\,dt$. The variable $t$ is a dummy variable of integration: it names the input of $f$ inside the integral and is not the endpoint $x$.

The fixed number $a$ determines where accumulation begins. The endpoint $x$ determines how far the integral extends. In particular, $F(a)=0$, since the integral over an interval of zero length is zero.

The Fundamental Theorem of Calculus, Part 1

The theorem applies when the integrand is continuous. At every interior point $x$ of the interval where $f$ is continuous, the accumulation function is differentiable, and its derivative is $f(x)$.

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

The key detail is that the value of the integrand is taken at the current upper endpoint. The integral combines values of $f$ across an interval, but its instantaneous rate of change depends on the value of $f$ at the endpoint.

Here the upper endpoint is $x$, so it changes at rate $1$ with respect to $x$. This is the simple chain-rule structure behind the formula: the rate of accumulation is the integrand at the endpoint, multiplied by the rate at which that endpoint moves, which is $1$. More general variable endpoints require additional chain-rule steps.

Why the Derivative Is the Endpoint Value

The theorem can be understood by examining a difference quotient. If $h$ is a small nonzero change in the endpoint, the change in accumulation is the integral from $x$ to $x+h$. Additivity across adjacent intervals gives that change exactly.

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

Because $f$ is continuous at $x$, its values near $x$ are close to $f(x)$. The average value of $f$ over the short interval from $x$ to $x+h$ must therefore be close to $f(x)$. As $h$ approaches zero, the difference quotient approaches $f(x)$. The following argument makes that reasoning precise and works for positive or negative $h$.

Write the difference quotient.
For $h\ne 0$, use additivity to express the change in $F$ as an integral over the short interval from $x$ to $x+h$.

Subtract the proposed derivative.
Since $h=\int_x^{x+h}1\,dt$, the difference between the quotient and $f(x)$ can be written as an average of $f(t)-f(x)$.

Use continuity.
Continuity at $x$ makes $|f(t)-f(x)|$ arbitrarily small whenever $t$ is sufficiently close to $x$. The whole average difference is then arbitrarily small.

\begin{aligned} \left| \frac{F(x+h)-F(x)}{h}-f(x) \right| &= \left| \frac{1}{h}\int_x^{x+h}(f(t)-f(x))\,dt \right|\\ &\leq \frac{1}{|h|}\int_{\min(x,x+h)}^{\max(x,x+h)} |f(t)-f(x)|\,dt\\ &< \frac{1}{|h|}\,\varepsilon |h|\\ &=\varepsilon. \end{aligned}

The bound uses the interval length $|h|$, so it remains valid whether $h$ is positive or negative. Since the difference between the quotient and $f(x)$ can be made smaller than every positive $\varepsilon$, the difference quotient tends to $f(x)$. Thus $F'(x)=f(x)$.

Worked Examples

Worked Example: A Polynomial Integrand

Let $F(x)=\int_1^x (t^2-3t+1)\,dt$. Find $F'(x)$, then calculate $F'(4)$.

The integrand $t^2-3t+1$ is a polynomial, so it is continuous for every real $t$. The Fundamental Theorem of Calculus, Part 1, therefore gives

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

At $x=4$, substitute into the derivative:

$$ F'(4)=4^2-3(4)+1=16-12+1=5. $$

Check by evaluating the integral with an antiderivative. An antiderivative of $t^2-3t+1$ is $\frac{t^3}{3}-\frac{3t^2}{2}+t$, so

\begin{aligned} F(x) &=\left[\frac{t^3}{3}-\frac{3t^2}{2}+t\right]_1^x\\ &=\frac{x^3}{3}-\frac{3x^2}{2}+x-\left(\frac{1}{3}-\frac{3}{2}+1\right)\\ &=\frac{x^3}{3}-\frac{3x^2}{2}+x+\frac{1}{6}. \end{aligned}

Differentiate this expression: $F'(x)=x^2-3x+1$, as predicted. Substituting $x=4$ again gives $5$.

Worked Example: An Exponential Integrand

Let $G(x)=\int_0^x e^t\,dt$. Find $G'(x)$ and $G'(0)$.

The function $e^t$ is continuous for every real $t$. Therefore, by the theorem,

G'(x)=e^x, \qquad G'(0)=e^0=1.

To check, evaluate the integral using the antiderivative $e^t$:

G(x)=\left[e^t\right]_0^x=e^x-e^0=e^x-1.

Differentiating gives $G'(x)=e^x$, so at $x=0$, $G'(0)=1$. This check also confirms that $G(0)=0$, as expected for an accumulation function based at $0$.

Worked Example: A Continuous Integrand with a Corner

Let $H(x)=\int_{-1}^x |t|\,dt$. Find $H'(x)$, including at $x=0$.

The integrand $|t|$ is continuous everywhere, including at $t=0$. Thus the theorem applies at every interior point and gives

$$ H'(x)=|x|. $$

We can check the result directly. For $x<0$, $|t|=-t$ between $-1$ and $x$, so

H(x)=\int_{-1}^x (-t)\,dt =\left[-\frac{t^2}{2}\right]_{-1}^x =\frac{1-x^2}{2}.

For $x\geq 0$, split at $0$, where the expression for $|t|$ changes:

\begin{aligned} H(x) &=\int_{-1}^0(-t)\,dt+\int_0^x t\,dt\\ &=\left[-\frac{t^2}{2}\right]_{-1}^0+\left[\frac{t^2}{2}\right]_0^x\\ &=\frac{1}{2}+\frac{x^2}{2}. \end{aligned}

For $x<0$, differentiating $\frac{1-x^2}{2}$ gives $-x=|x|$. For $x>0$, differentiating $\frac{1}{2}+\frac{x^2}{2}$ gives $x=|x|$. At $x=0$, both formulas give $H(0)=\frac12$. The left difference quotient is $\frac{H(h)-H(0)}{h}=\frac{-h^2/2}{h}=-h/2$ for $h<0$, which tends to $0$. The right difference quotient is $\frac{h^2/2}{h}=h/2$ for $h>0$, which also tends to $0$. Hence $H'(0)=0=|0|$.

A Common Mistake: Differentiating the Integrand

The derivative of an accumulation function is not generally the derivative of its integrand. Instead, the Fundamental Theorem of Calculus says to evaluate the integrand at the endpoint.

Common mistake: For $J(x)=\int_1^x(t^2+2)\,dt$, one might incorrectly claim $J'(x)=2x$ by differentiating $t^2+2$. That differentiates the integrand, not the accumulation function.

The correct application of the theorem is

$$ J'(x)=x^2+2. $$

For a direct check, an antiderivative of $t^2+2$ is $\frac{t^3}{3}+2t$. Therefore

J(x)=\left[\frac{t^3}{3}+2t\right]_1^x =\frac{x^3}{3}+2x-\frac{7}{3}, \qquad J'(x)=x^2+2.

The dummy variable $t$ is replaced by $x$ when the integrand is evaluated at the upper endpoint. It is not differentiated as though it were the variable of the accumulation function.

When the Rule Applies

The continuity condition is essential to the stated theorem. If $f$ is continuous at $x$, then the average value of $f$ over a sufficiently short interval next to $x$ approaches $f(x)$, which is why the accumulation function has derivative $f(x)$. When continuity is absent, this theorem does not guarantee the derivative formula at that point; other analysis may be needed.

When using the rule, keep these points in view:

The lower endpoint $a$ is fixed in this result.
The upper endpoint is $x$, so its rate of change with respect to $x$ is $1$.
The integrand must be continuous at the point where the derivative is being found.
The answer is the integrand evaluated at $x$, not the value of the integral and not the derivative of the integrand.

This rule connects area and rate of change. An integral records accumulated change over an interval; differentiating that accumulation recovers the current rate at its endpoint. That connection is useful whenever a quantity is defined by accumulating a continuous rate.

Check Your Understanding

Use the Fundamental Theorem of Calculus, Part 1, and show enough work to make your reasoning clear.

If $A(x)=\int_2^x (3t^2-1)\,dt$, what is $A'(x)$? State why the theorem applies.
For $B(x)=\int_0^x \cos(t)\,dt$, find $B'(\frac{\pi}{3})$.
For $C(x)=\int_{-2}^x |t|\,dt$, find $C'(x)$, including at $x=0$. What continuity fact allows the theorem to apply?
Identify the error in the claim that the derivative of $\int_1^x (t^3+4)\,dt$ is $3x^2$, and give the correct derivative.
In one sentence, explain why the derivative is determined by the integrand at the upper endpoint rather than by the whole accumulated integral.

Variable Upper Bounds and the Chain Rule

What You'll Learn

A Moving Endpoint Produces a Derivative

The Fundamental Theorem of Calculus, Part 1

Why the Derivative Is the Endpoint Value