Why the Continuity Condition Matters
An accumulation function records the signed area contributed by an integrand as an endpoint moves. FTC Part I connects the rate of change of that accumulated quantity to the value of the integrand at the moving endpoint. The continuity hypothesis in that result is local: to guarantee the derivative at a particular point, the integrand must be continuous there. It need not be continuous everywhere between the fixed starting point and that point.
This distinction is useful when an integrand has a discontinuity somewhere in its domain. If it is still integrable over the interval and is continuous at a point \(x\), FTC Part I guarantees that the accumulation function is differentiable at \(x\), with derivative equal to the integrand’s value there.
Accumulation Functions and the Hypotheses
Let \(f\) be Riemann integrable on a closed interval \([a,b]\). For each \(x\in[a,b]\), define the accumulation function with fixed starting point \(a\) by integrating \(f\) from \(a\) to \(x\).
Riemann integrability on \([a,b]\) ensures that each integral in this definition exists. The theorem considered here concerns an interior point \(x\), so \(a<x<b\). At such a point, continuity of \(f\) means that the values of \(f(t)\) approach \(f(x)\) as \(t\) approaches \(x\).
The hypotheses do two different jobs. Integrability on the whole interval makes the accumulated values well-defined. Continuity at the particular point \(x\) ensures that averages of \(f\) over sufficiently short intervals near \(x\) are close to \(f(x)\). The first condition is global over the interval; the second is local at the point where the derivative is being considered.
Why Continuity at the Point Gives the Derivative
The derivative measures the change in \(F\) over a short interval, divided by the interval’s length. The change in \(F\) from \(x\) to \(x+h\) is the integral of \(f\) over that short interval. Thus, the difference quotient is an average value of \(f\) near \(x\). Continuity makes that average approach \(f(x)\).
For sufficiently small nonzero \(h\), with \(x+h\in[a,b]\), additivity of the definite integral gives \(F(x+h)-F(x)=\int_x^{x+h} f(t)\,dt\).
Divide the change by \(h\), then subtract the proposed derivative \(f(x)\). The result is \(\frac{1}{h}\int_x^{x+h}(f(t)-f(x))\,dt\).
For any \(\varepsilon>0\), continuity at \(x\) gives a \(\delta>0\) such that \(|f(t)-f(x)|<\varepsilon\) whenever \(|t-x|<\delta\). The integral error is then bounded by \(\varepsilon |h|\).
After dividing the error bound by \(|h|\), the difference between the difference quotient and \(f(x)\) is less than \(\varepsilon\). This proves that the quotient tends to \(f(x)\).
Here is the estimate in full. Since \(f\) is integrable, the integrals involved are defined. For \(0<|h|<\delta\), continuity gives \(|f(t)-f(x)|<\varepsilon\) throughout the interval between \(x\) and \(x+h\). The absolute value of an integral is at most the integral of the absolute value, so
The interval length in the estimate is \(|h|\), whether \(h\) is positive or negative. Because the bound holds for every \(\varepsilon>0\) when \(h\) is sufficiently small, the difference quotient tends to \(f(x)\). Therefore \(F'(x)=f(x)\). Notice that the argument uses values of \(f\) close to \(x\), not values near the fixed starting point \(a\).
Worked Example: A Continuous Integrand
Worked Example: A Continuous Integrand
Let \(f(t)=3t^2-2\) on \([-1,3]\), and define \(F(x)=\int_{-1}^{x}(3t^2-2)\,dt\). Find \(F'(2)\) using FTC Part I.
The polynomial \(f\) is continuous at \(x=2\), and \(f\) is integrable on \([-1,3]\). The theorem therefore applies. Evaluate the integrand at the point:
As a check, an antiderivative of \(3t^2-2\) is \(t^3-2t\), so
Differentiating gives \(F'(x)=3x^2-2\), and hence \(F'(2)=12-2=10\), as required.
Discontinuities Elsewhere Do Not Block the Theorem
Continuity on the whole integration interval is a convenient stronger hypothesis: if \(f\) is continuous everywhere on \([a,b]\), it is continuous at every interior point, so the theorem applies at every such point. But the local form of the theorem is more precise. A discontinuity away from \(x\) does not prevent the conclusion at \(x\), provided \(f\) remains integrable over the interval.
Worked Example: A Jump Away from the Point of Interest
On \([0,2]\), define \(f(t)=1\) when \(0\leq t<1\), and \(f(t)=2\) when \(1\leq t\leq2\). This step function is Riemann integrable, although it is discontinuous at \(t=1\). Define \(F(x)=\int_0^x f(t)\,dt\). Find \(F'(3/2)\).
The point \(3/2\) is not the discontinuity. In a neighborhood of \(3/2\), the integrand is constantly \(2\), so \(f\) is continuous there and \(f(3/2)=2\). FTC Part I gives
To check directly, for \(x\geq1\) the accumulated integral consists of an area of \(1\) from \(0\) to \(1\), followed by an area of \(2(x-1)\). Thus \(F(x)=1+2(x-1)=2x-1\) for \(x\geq1\). For \(|h|<1/2\), both \(3/2\) and \(3/2+h\) lie to the right of \(1\), so
The difference quotient is \(2\) for all such nonzero \(h\), confirming the derivative. The jump at \(1\) does not affect this calculation near \(3/2\).
Worked Example: A Different Integrand with a Remote Discontinuity
On \([-2,2]\), let \(f(t)=t\) for \(-2\leq t<0\), and \(f(t)=t+3\) for \(0\leq t\leq2\). The integrand has a jump at \(0\), but it is integrable. Define \(F(x)=\int_{-2}^{x}f(t)\,dt\). Find \(F'(1)\).
Since \(f(t)=t+3\) near \(1\), it is continuous at \(1\), with \(f(1)=1+3=4\). The theorem gives \(F'(1)=4\). Verify the accumulation formula at and near \(1\). For \(x\geq0\), split the integral at the discontinuity:
In particular, \(F(1)=-2+\frac12+3=\frac32\). For \(|h|<1\), \(1+h>0\), so the same formula holds at \(1+h\). Therefore
As \(h\) tends to \(0\), this quotient tends to \(4\), agreeing with \(f(1)\). The discontinuity at \(0\) is outside the small neighborhood used to find the derivative at \(1\).
A Common Mistake: Requiring Continuity Everywhere
A common error is to see any discontinuity between the starting point and the point of interest and conclude that FTC Part I cannot be used. For the step function in the earlier example, one might reason incorrectly: “\(f\) is discontinuous at \(1\), which lies between \(0\) and \(3/2\), so \(F'(3/2)\) is not given by \(f(3/2)\).” The conclusion does not follow. The relevant conditions are integrability on the full interval and continuity at \(3/2\); both hold, so \(F'(3/2)=f(3/2)=2\).
There is a related caution in the opposite direction. If \(f\) is discontinuous at a point \(x\), this theorem’s continuity hypothesis is not satisfied there. That means this result does not guarantee \(F'(x)=f(x)\); it does not, by itself, prove that \(F\) is not differentiable at \(x\). A failed sufficient condition is not automatically a proof that the conclusion fails.
How to Locate the Guaranteed Differentiability Points
When a problem asks where FTC Part I guarantees differentiability, use the following procedure. First identify the interval on which the integrand is integrable, so that the accumulation function is defined. Then inspect the integrand at each candidate interior point. Every point where the integrand is continuous is a point where the accumulation function is differentiable, and its derivative there equals the integrand’s value.
This result is useful because it separates two questions that can otherwise be confused: whether the integral defining \(F\) exists, and whether \(F\) has a derivative at a particular point. Integrability addresses the first question over the interval; continuity at the point addresses the second. The derivative reflects the integrand’s local value, even though the accumulated quantity includes contributions from a much larger interval.
Check Your Understanding
For each question, identify which hypothesis is global and which is local before applying the theorem.
- State the two hypotheses in the local continuity form of FTC Part I.
- If an integrand is integrable on \([a,b]\) and continuous at \(x\in(a,b)\), what does the theorem say about \(F'(x)\) for \(F(x)=\int_a^x f(t)\,dt\)?
- Can a discontinuity at a point strictly between \(a\) and \(x\) prevent the theorem from guaranteeing \(F'(x)=f(x)\)? Explain.
- For the step function equal to \(1\) before \(t=1\) and \(2\) from \(t=1\) onward, why can the theorem be applied at \(x=3/2\)?
- If \(f\) is not continuous at a point \(x\), what does the local continuity theorem establish—or fail to establish—about differentiability there?
