From Accumulated Change to Instantaneous Rate
An accumulation function records the net change contributed by an integrand from a fixed starting point up to a variable endpoint. Its derivative describes how quickly that accumulated quantity changes as the endpoint moves.
The connection is intuitive: when the endpoint moves a short distance, the extra accumulated amount is the integral over that short interval. Dividing that amount by the interval’s length gives an average value of the integrand. If the integrand is continuous, its values over a sufficiently short interval are close to its value at the endpoint. The average therefore approaches that endpoint value.
This reasoning gives a precise way to derive the derivative, rather than merely assert it.
Accumulation Functions and Their Difference Quotients
Let \(f\) be continuous on an interval, and choose a fixed base point \(a\) in that interval. Define the accumulation function \(F\) by integrating \(f\) from \(a\) to the variable endpoint \(x\).
The variable \(t\) is a dummy variable of integration: it identifies the input of \(f\) inside the integral. The output \(F(x)\) depends on the endpoint \(x\).
To find \(F'(x)\), start with the definition of the derivative. For a nonzero increment \(h\), the difference quotient is the change in \(F\), divided by the change in its input:
By the definition of \(F\), the difference in accumulated values is the integral over the interval between the two endpoints. Additivity of definite integrals gives
Thus the difference quotient can be written as
This is the central expression. It is the average value of \(f\) over the short interval from \(x\) to \(x+h\). The signed integral and the signed interval length work together: if \(h\) is negative, both the orientation of the integral and the denominator reverse.
Why the Average Approaches the Endpoint Value
Assume \(f\) is continuous at \(x\). Continuity means that, given any \(\varepsilon>0\), there is a \(\delta>0\) such that \(f(t)\) differs from \(f(x)\) by less than \(\varepsilon\) whenever \(t\) is within \(\delta\) of \(x\).
When \(0<h<\delta\), every \(t\) in the interval from \(x\) to \(x+h\) satisfies that condition. Therefore, throughout this interval, \(f(t)\) lies between \(f(x)-\varepsilon\) and \(f(x)+\varepsilon\). Integrating these inequalities and dividing by the positive interval length \(h\) shows that the average value also lies between those bounds.
For a negative increment with \(-\delta<h<0\), reverse the orientation of the integral and the sign of the denominator together. The quotient becomes the average over the ordinary interval from \(x+h\) to \(x\), whose length is \(-h>0\). Every point in that interval is still within \(\delta\) of \(x\), so the same bounds apply.
In both cases, the difference between the average and \(f(x)\) is less than \(\varepsilon\) for sufficiently small nonzero \(h\). Since this is true for every \(\varepsilon>0\), the average approaches \(f(x)\) as \(h\) approaches zero.
But the expression on the left is exactly the difference quotient for \(F\). Taking its limit proves the derivative formula.
Worked Examples
Worked Example: A Polynomial Integrand
Let \(\displaystyle F(x)=\int_{-1}^{x}(t^2+1)\,dt\). Find \(F'(2)\) directly from the difference quotient.
At \(x=2\), the difference quotient is
Evaluate the integral using the antiderivative \(\frac{t^3}{3}+t\):
Taking the limit as \(h\to 0\) gives \(F'(2)=5\). The integrand’s value at the endpoint is \(2^2+1=5\), which checks the result.
Worked Example: A Trigonometric Integrand
Let \(\displaystyle G(x)=\int_0^x \cos t\,dt\). Find \(G'\!\left(\frac{\pi}{3}\right)\) using the average-value form of the difference quotient.
For a nonzero increment \(h\), the quotient at \(x=\frac{\pi}{3}\) is
The function \(\cos t\) is continuous at \(\frac{\pi}{3}\). Therefore, as \(h\to 0\), its average value over this shrinking interval approaches its value at that point. Compute that value:
Hence \(G'\!\left(\frac{\pi}{3}\right)=\frac{1}{2}\). The result checks against the integrand evaluated at the endpoint.
Worked Example: An Absolute-Value Integrand at Zero
Let \(\displaystyle H(x)=\int_0^x |t|\,dt\). Find \(H'(0)\) by checking increments from both sides.
When \(h>0\), \(|t|=t\) on the interval from \(0\) to \(h\). Thus
As positive \(h\) approaches zero, this quotient approaches \(0\).
When \(h<0\), \(|t|=-t\) on the interval from \(h\) to \(0\). Keeping the integral’s orientation from \(0\) to \(h\) gives
As negative \(h\) approaches zero, this quotient also approaches \(0\). The two-sided derivative is therefore \(H'(0)=0\), matching the continuous integrand value \(|0|=0\).
A Common Mistake: Confusing Change with Rate
A frequent error is to observe that the accumulated change over a short interval approaches zero and conclude that the derivative is zero. The derivative is not the change alone; it is the change divided by the input increment.
For example, let \(\displaystyle J(x)=\int_0^x (t+1)\,dt\), and examine the derivative at \(x=0\). The added accumulation over an increment \(h\) is
This change approaches zero as \(h\to 0\). But the difference quotient is
which approaches \(1\), not \(0\). The integrand’s value at the endpoint is \(0+1=1\), confirming the correct derivative. Always divide the accumulated change by the endpoint change before taking the limit.
When the Result Applies
Continuity of \(f\) at \(x\) is the condition that makes the average-value argument work: near \(x\), all integrand values are close to \(f(x)\). The theorem therefore guarantees the derivative at every interior point where the integrand is continuous.
The condition is sufficient, not a claim that every discontinuity prevents differentiability. The result established here is specifically that continuity at the point guarantees the derivative formula. At an endpoint of the domain, the available limit may be one-sided; the two-sided derivative statement above concerns interior points.
The key reasoning can be kept in three linked steps:
The change from \(F(x)\) to \(F(x+h)\) is \(\int_x^{x+h}f(t)\,dt\).
Divide that integral by \(h\), obtaining the average value of \(f\) on the short interval.
As the interval shrinks to \(x\), its average integrand value approaches \(f(x)\).
Check Your Understanding
Use the difference quotient and the average-value idea to answer these questions.
- For \(F(x)=\int_2^x (3t-1)\,dt\), write the difference quotient at a general point \(x\) in the form of an average value.
- Why does continuity of \(f\) at \(x\) imply that the average of \(f\) over the interval from \(x\) to \(x+h\) approaches \(f(x)\)?
- For \(G(x)=\int_0^x \sin t\,dt\), what does the theorem give for \(G'\!\left(\frac{\pi}{2}\right)\)?
- When \(h<0\), why does \(\frac{1}{h}\int_x^{x+h}f(t)\,dt\) still represent the average value over the interval?
- What is wrong with concluding that a derivative is zero merely because the accumulated change approaches zero?
