Introduction
The Fundamental Theorem of Calculus Part I gives a direct formula for the derivative of an accumulation function at a point where its integrand is continuous. A removable discontinuity raises a useful question: what if the integrand approaches a definite value at a point, but its value there is different—or has not been assigned?
The accumulation function depends on integrals, and changing an integrand at just one point does not change its integral. As a result, the accumulation function can still have a derivative at a removable discontinuity. That derivative is determined by the value the integrand approaches, not necessarily by the value assigned at the point.
Removable Discontinuities and Accumulation Functions
Let \(f\) be defined on an interval around \(c\). The discontinuity at \(c\) is removable if the finite limit \(\lim_{x\to c}f(x)=L\) exists, but \(f\) is not continuous at \(c\). This can happen because \(f(c)\) is undefined, or because \(f(c)\ne L\). If \(f(c)\) is undefined, one can assign \(f(c)=L\) to obtain a continuous extension at \(c\).
For a fixed starting point \(a<c\), the corresponding accumulation function is
For \(A(c)\) and nearby values \(A(x)\) to be defined, the integrals from \(a\) through a neighborhood of \(c\) must exist. In what follows, assume \(f\) is Riemann integrable on \([a,c+\delta]\) for some \(\delta>0\), and that \(c>a\). A function with a removable discontinuity and otherwise continuous behavior on this interval satisfies this integrability condition.
A basic property of the Riemann integral is that changing the value of an integrable function at one point does not change its integral. Consequently, if two integrands differ only at \(c\), their accumulation functions agree wherever both are defined. This fact does not, by itself, assert that the integrand is continuous at \(c\).
The Derivative at a Removable Discontinuity
At nearby points where \(f\) is continuous, FTC Part I gives \(A'(x)=f(x)\). At the removable discontinuity itself, the conclusion is slightly different: the derivative, if established under the conditions below, equals the limit of \(f\) at that point.
Why the Theorem Works
The derivative of \(A\) at \(c\) is a limit of average values of \(f\) over short intervals next to \(c\). If every nearby value of \(f\) is close to \(L\), then those averages are close to \(L\) as well. Here is the full difference-quotient argument.
For nonzero \(h\) small enough that the relevant integrals are defined, additivity of the integral gives \(A(c+h)-A(c)=\int_c^{c+h}f(t)\,dt\). Thus the quotient is \(\frac{1}{h}\int_c^{c+h}f(t)\,dt\).
For \(h>0\), the difference between the quotient and \(L\) is \(\frac{1}{h}\int_c^{c+h}(f(t)-L)\,dt\). For \(h<0\), the same identity holds; the reversed orientation of the integral is accounted for by the negative \(h\).
Given any \(\varepsilon>0\), the limit assumption gives a \(\rho>0\) such that \(0<|t-c|<\rho\) implies \(|f(t)-L|<\varepsilon\). For \(0<|h|<\rho\), the integral estimate therefore gives \(\left|\frac{1}{h}\int_c^{c+h}(f(t)-L)\,dt\right|\le\varepsilon\). The value at the single point \(t=c\) has no effect on this integral.
The difference between the difference quotient and \(L\) is bounded in absolute value by every prescribed \(\varepsilon>0\) for sufficiently small nonzero \(h\). Therefore the quotient tends to \(L\), so \(A'(c)=L\).
The estimate in Step 3 uses the standard integral bound: if \(|g(t)|\le\varepsilon\) throughout an interval of length \(|h|\), then the absolute value of the integral of \(g\) over that interval is at most \(\varepsilon|h|\). A possible exception at the single endpoint \(c\) does not alter the integral.
Worked Example: A Jump in the Assigned Value
Worked Example: A Jump in the Assigned Value
On \([0,3]\), define \(f(t)=t+1\) for \(t\ne1\) and \(f(1)=5\). Let \(A(x)=\int_0^x f(t)\,dt\). Find \(A'(1)\), and compare it with \(f(1)\) and with derivatives at nearby points.
For every \(x\in[0,3]\), changing the integrand’s value at the one point \(t=1\) does not change the integral. Therefore
Differentiating this explicit expression gives \(A'(x)=x+1\), so \(A'(1)=1+1=2\). The limit of the integrand at \(1\) is \(\lim_{t\to1}f(t)=2\), while its assigned value is \(f(1)=5\). Thus \(A'(1)=2\ne5=f(1)\), in agreement with the theorem.
For any \(x\ne1\) in the interior of the interval, \(f\) is continuous at \(x\). FTC Part I applies there and gives \(A'(x)=f(x)=x+1\). The formula also equals \(2\) in the limit as \(x\to1\), even though it does not equal the assigned value \(f(1)=5\).
Check: differentiating \(A(x)=x^2/2+x\) gives \(A'(x)=x+1\) directly, and substituting \(x=1\) gives \(A'(1)=2\). The integral and the derivative agree with the theorem’s predicted limiting value.
Worked Example: A Removable Discontinuity with a Zero Limit
Worked Example: A Removable Discontinuity with a Zero Limit
On \([0,4]\), define \(g(t)=|t-2|\) for \(t\ne2\) and \(g(2)=3\). Set \(G(x)=\int_0^x g(t)\,dt\). Find \(G'(2)\) and compare it with \(g(2)\).
The limit of \(g(t)\) at \(2\) is \(0\). To calculate \(G\), use \(g(t)=2-t\) for \(t<2\) and \(g(t)=t-2\) for \(t>2\). For \(0\le x\le2\),
In particular, \(G(2)=4-2=2\). For \(2\le x\le4\), split the integral at \(2\):
For \(h>0\), the right difference quotient at \(2\) is
For \(h<0\), use the formula for \(x\le2\):
Both expressions tend to \(0\) as \(h\to0\), so \(G'(2)=0\). This equals \(\lim_{t\to2}g(t)=0\), not \(g(2)=3\).
Check: differentiating the expression for \(x<2\) gives \(G'(x)=2-x=|x-2|\); differentiating the expression for \(x>2\) gives \(G'(x)=x-2=|x-2|\). The one-sided derivative values both tend to \(0\) at \(2\), matching the direct difference-quotient calculation. At \(2\), however, \(g\) is not continuous and its assigned value is \(3\).
Nearby Continuity Points and the Pointwise Formula
The two examples show how the conclusion changes depending on the point under consideration. If \(f\) is continuous at \(x\), FTC Part I applies to the accumulation function and states that \(A'(x)=f(x)\). At a removable discontinuity \(c\), that continuity hypothesis fails. The theorem proved above instead gives \(A'(c)=\lim_{t\to c}f(t)\), provided the integral and limit conditions hold.
| Point being considered | Condition | Derivative conclusion |
|---|---|---|
| A continuity point \(x\) of \(f\) | \(f\) is continuous at \(x\), and \(A\) is defined nearby | FTC Part I gives \(A'(x)=f(x)\) |
| A removable discontinuity \(c\) | The finite limit \(L=\lim_{t\to c}f(t)\) exists, and \(A\) is defined on an interval through \(c\) | The difference-quotient argument gives \(A'(c)=L\) |
If \(f(c)\) has been assigned a value different from \(L\), the pointwise identity \(A'(c)=f(c)\) is false. If \(f(c)\) is undefined, that identity is not even a meaningful equality. One may instead define a continuous extension by setting \(f(c)=L\); for the extended integrand, the pointwise formula applies at \(c\), and its value there agrees with the derivative.
A Hypothesis That Cannot Be Omitted
The limit and local integrability near \(c\) are not enough if the fixed-start integral from \(a\) to \(c\) does not exist. For example, take \(a=0\), \(c=1\), and let \(f(t)=1/t\) for \(t>0\), assigning any finite value at \(t=0\). The function is continuous near \(1\), with limit \(1\) there, and is integrable on a small closed interval around \(1\). But the improper integral from \(0\) to \(1\) diverges:
Thus \(A(1)=\int_0^1 f(t)\,dt\) is not defined as a finite integral, so the accumulation function \(A(x)=\int_0^x f(t)\,dt\) cannot be used to assert a derivative at \(1\). This is why the theorem requires the fixed-start integral to exist as well as the local limit condition.
Common Mistake: Substituting the Assigned Value
A common error is to assume that the derivative of an accumulation function always equals the integrand’s value at the point, without checking continuity. In the first example, that mistake would give:
But the computed accumulation function is \(A(x)=x^2/2+x\), so its derivative is \(A'(x)=x+1\), and therefore \(A'(1)=2\). The correction is to check the hypothesis before applying the pointwise FTC formula: \(f\) is not continuous at \(1\). The limiting-value result applies instead, and \(\lim_{t\to1}f(t)=2\).
Check Your Understanding
For each question, distinguish the integrand’s limit at the point from its assigned value, and check the hypotheses before using an FTC formula.
- If \(f\) has a removable discontinuity at \(c\), with \(\lim_{t\to c}f(t)=L\), what is \(A'(c)\) when \(A(x)=\int_a^x f(t)\,dt\) is defined through a neighborhood of \(c\)?
- For the first worked example, what are \(f(1)\), \(\lim_{t\to1}f(t)\), and \(A'(1)\)?
- Why does FTC Part I give \(A'(x)=f(x)\) at continuity points near \(c\), but not automatically at the removable discontinuity itself?
- Why does changing an integrand’s value at one point leave its Riemann integral unchanged?
- What fails in the example \(f(t)=1/t\) for \(t>0\) when the fixed starting point is \(a=0\) and the point of interest is \(c=1\)?
