Derivatives at the Endpoints of an Integration Interval

Why Endpoints Need Special Attention

An accumulation function records the signed integral of an integrand as an endpoint moves. At an interior input, that endpoint can move slightly to the left or right. At a boundary of the domain, only one of those directions is available. This changes how we define and calculate the derivative there.

For an accumulation function $F(x)=\int_a^x f(t)\,dt$ on $[a,b]$, the value at $a$ is determined by moving to the right, and the value at $b$ is determined by moving to the left. If $f$ is continuous at the relevant endpoint, the corresponding one-sided derivative is $f$ evaluated there. The distinction between one-sided and two-sided derivatives matters: the endpoint formula does not, by itself, provide inputs on both sides.

One-Sided Derivatives at a Boundary

Let $F$ be defined on a closed interval $[a,b]$. At the left endpoint $a$, the available nearby inputs have the form $a+h$ with $h>0$. At the right endpoint $b$, they have the form $b-h$ with $h>0$. These directions define the relevant difference quotients.

Definition: The right-hand derivative of $F$ at $a$ is the limit $$ F'_+(a)=\lim_{h\to 0^+}\frac{F(a+h)-F(a)}{h}, $$ if this limit exists. The left-hand derivative of $F$ at $b$ is $$ F'_-(b)=\lim_{h\to 0^+}\frac{F(b-h)-F(b)}{-h}, $$ if this limit exists. The denominator in the second quotient is $(b-h)-b=-h$, the actual change in input.

Both quotients measure change in output divided by change in input. The signs in the left-hand quotient are important: as the input moves from $b$ to $b-h$, the change in input is negative.

At an interior point $x$, the usual derivative is a two-sided limit: the input approaches $x$ from both the left and the right. At an endpoint of a function whose domain is only $[a,b]$, inputs from outside the interval are not available. Some conventions call the corresponding one-sided limit an endpoint derivative; it should not be confused with an ordinary two-sided derivative.

The Endpoint Formula for an Accumulation Function

Suppose $f$ is integrable on $[a,b]$, and define $F(x)=\int_a^x f(t)\,dt$ for $x\in[a,b]$. Continuity at an endpoint here means continuity relative to the interval: near $a$, values of $f(t)$ for $t\geq a$ approach $f(a)$; near $b$, values for $t\leq b$ approach $f(b)$.

Theorem: If $f$ is continuous at $a$ from within $[a,b]$, then $F'_+(a)$ exists and equals $f(a)$. If $f$ is continuous at $b$ from within $[a,b]$, then $F'_-(b)$ exists and equals $f(b)$.

The endpoint hypotheses are local: continuity at $a$ is what gives the formula at $a$, and continuity at $b$ is what gives it at $b$. The function must also be integrable on the interval so that the accumulation function is defined.

Why Continuity Gives the Endpoint Derivative

At the left endpoint, for $h>0$, the definition of $F$ gives $F(a)=0$ and $F(a+h)=\int_a^{a+h}f(t)\,dt$. Therefore the right-hand difference quotient is an average value of $f$ over a short interval:

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

To show that this average approaches $f(a)$, subtract $f(a)$ and use the integral estimate for the absolute value:

\left| \frac{1}{h}\int_a^{a+h}f(t)\,dt-f(a) \right| = \left| \frac{1}{h}\int_a^{a+h}\bigl(f(t)-f(a)\bigr)\,dt \right| \leq \frac{1}{h}\int_a^{a+h}|f(t)-f(a)|\,dt.

Given any $\varepsilon>0$, continuity from the right at $a$ gives a $\delta>0$ such that $0\leq t-a<\delta$ implies $|f(t)-f(a)|<\varepsilon$. For $0<h<\delta$, every $t\in[a,a+h]$ satisfies that bound. Consequently,

\frac{1}{h}\int_a^{a+h}|f(t)-f(a)|\,dt \leq \frac{1}{h}\int_a^{a+h}\varepsilon\,dt = \varepsilon.

Thus the difference between the quotient and $f(a)$ can be made arbitrarily small, which proves that $F'_+(a)=f(a)$.

At the right endpoint, use a positive $h$ and write $F(b-h)-F(b)=-\int_{b-h}^b f(t)\,dt$. Dividing by the input change $-h$ gives

\frac{F(b-h)-F(b)}{-h} = \frac{1}{h}\int_{b-h}^b f(t)\,dt.

This is the average of $f$ over $[b-h,b]$. Continuity from the left at $b$ makes that average approach $f(b)$, by the same estimate: the absolute difference from $f(b)$ is at most $\frac{1}{h}\int_{b-h}^b |f(t)-f(b)|\,dt$, which is less than $\varepsilon$ for sufficiently small $h>0$. Hence $F'_-(b)=f(b)$.

Worked Examples at Both Endpoints

Worked Example: A Polynomial Integrand

Let $F(x)=\int_0^x(3t^2-2)\,dt$ for $x\in[0,2]$. Find the endpoint derivatives from their difference quotients.

First evaluate the integral:

$$ F(x)=[t^3-2t]_0^x=x^3-2x. $$

At the left endpoint $0$, take $h>0$:

\frac{F(0+h)-F(0)}{h} = \frac{(h^3-2h)-0}{h} = h^2-2 \longrightarrow -2.

At the right endpoint $2$, take $h>0$. Since $F(2)=4$,

F(2-h)=(2-h)^3-2(2-h)=4-10h+6h^2-h^3, $$ $$ \frac{F(2-h)-F(2)}{-h} = \frac{-10h+6h^2-h^3}{-h} = 10-6h+h^2 \longrightarrow 10.

The checks agree with the endpoint values of the integrand: $3(0)^2-2=-2$ and $3(2)^2-2=10$. Also, differentiating the antiderivative gives $F'(x)=3x^2-2$, as expected.

Worked Example: A Variable Lower Limit

Let $H(x)=\int_x^2(2t+1)\,dt$ for $x\in[0,2]$. Find the one-sided derivatives at $0$ and $2$.

Evaluating the integral gives

$$ H(x)=[t^2+t]_x^2=6-x^2-x. $$

At $0$, use $h>0$. Since $H(0)=6$,

\frac{H(h)-H(0)}{h} = \frac{(6-h^2-h)-6}{h} = -h-1 \longrightarrow -1.

At $2$, $H(2)=0$, and

H(2-h)=5h-h^2, \qquad \frac{H(2-h)-H(2)}{-h} = \frac{5h-h^2}{-h} = -5+h \longrightarrow -5.

The integrand has values $2(0)+1=1$ and $2(2)+1=5$ at the endpoints, so the calculated derivatives are their negatives. This sign is also checked by differentiating $H(x)=6-x^2-x$, which gives $H'(x)=-2x-1$.

Worked Example: A Square-Root Integrand

Let $A(x)=\int_{-1}^x\sqrt{t+1}\,dt$ for $x\in[-1,3]$. Calculate the endpoint derivatives directly.

An antiderivative gives

A(x)=\frac{2}{3}(x+1)^{3/2}.

At the left endpoint, $A(-1)=0$. For $h>0$,

\frac{A(-1+h)-A(-1)}{h} = \frac{\frac{2}{3}h^{3/2}}{h} = \frac{2}{3}\sqrt{h} \longrightarrow 0.

At the right endpoint, set $y=\sqrt{4-h}$, so $y\to2$ as $h\to0^+$. Since $A(3)=16/3$,

\frac{A(3-h)-A(3)}{-h} = \frac{2}{3}\frac{8-(4-h)^{3/2}}{h} = \frac{2}{3}\frac{8-y^3}{4-y^2} = \frac{2}{3}\frac{y^2+2y+4}{y+2} \longrightarrow 2.

The endpoint values of the integrand are $\sqrt{0}=0$ and $\sqrt{4}=2$, matching the quotients. Differentiating $A$ on the interior gives $A'(x)=\sqrt{x+1}$, consistent with those endpoint values.

Endpoint Derivatives and Interior Derivatives

At an interior point $c\in(a,b)$, the usual derivative uses inputs both below and above $c$. If $f$ is continuous there, the fundamental theorem gives $F'(c)=f(c)$. At $a$, only the right-hand quotient is available from the domain; at $b$, only the left-hand quotient is available. Continuity at the relevant endpoint supplies the corresponding one-sided result.

An endpoint derivative may match the derivative of a formula that extends beyond the interval, but the restricted function alone does not provide that extension. For instance, in the first example, the quotient at $0$ uses $h>0$. Substituting $h<0$ would require evaluating $F(h)$ outside its stated domain $[0,2]$. It is therefore not a valid two-sided derivative calculation for that domain.

Common Pitfall: Losing the Sign at the Right Endpoint

A common error is to divide $F(b-h)-F(b)$ by $h$ instead of by the actual input change $-h$. In the square-root example, the numerator $A(3-h)-A(3)$ is negative for small positive $h$. Dividing it by $h$ would give a limit of $-2$, which is the wrong sign. The correct quotient is

\frac{A(3-h)-A(3)}{(3-h)-3} = \frac{A(3-h)-A(3)}{-h} \longrightarrow 2.

Always calculate the denominator as “new input minus original input.” At the right endpoint, moving left makes that difference negative. For an accumulation function with a variable lower limit, the resulting endpoint derivative is also the negative of the integrand value.

Why the Endpoint Distinction Matters

Accumulation functions model quantities built up from a rate, such as total distance from a starting time or total change in a measured quantity. If the input is restricted to a time interval, its initial and final times are boundaries. The appropriate rate at those times is described by a one-sided derivative, because only one direction of change stays within the interval.

The endpoint formula is useful, but its conditions should be kept in view. Continuity from within the interval guarantees the endpoint derivative and identifies its value with the integrand at that endpoint. Without that continuity, this theorem does not guarantee the formula. In every calculation, identify the domain, choose the available direction, and use the signed change in input.

Check Your Understanding

Answer each question using the appropriate one-sided quotient and endpoint conditions.

For $F(x)=\int_1^x f(t)\,dt$ on $[1,4]$, which direction of approach defines the endpoint derivative at $1$?
Why is the denominator in the left-hand quotient at $b$ equal to $-h$ when the input is $b-h$?
If $f$ is continuous at $a$ from within $[a,b]$, what is the right-hand derivative of $F(x)=\int_a^x f(t)\,dt$ at $a$?
For $H(x)=\int_x^b f(t)\,dt$, what is the one-sided derivative at $b$ when $f$ is continuous there?
Why does a one-sided endpoint derivative not automatically establish an ordinary two-sided derivative for a function defined only on $[a,b]$?

Derivatives at the Endpoints of an Integration Interval

What You'll Learn

Why Endpoints Need Special Attention

One-Sided Derivatives at a Boundary

The Endpoint Formula for an Accumulation Function

Why Continuity Gives the Endpoint Derivative