When Does an Accumulation Function Have a Derivative?
An accumulation function records the signed integral of an integrand from a fixed starting point to a variable endpoint. The Fundamental Theorem of Calculus, Part 1, connects the derivative of this function to the value of the integrand. Its key condition is continuity: where the integrand is continuous, the theorem guarantees the derivative relationship.
At a discontinuity, that guarantee is unavailable. The accumulation function might fail to be differentiable there, or it might be differentiable with a derivative that does not equal the integrand’s assigned value. To determine what happens, examine the difference quotient rather than applying the theorem without its condition.
The condition is local: continuity at \(c\) is enough for the conclusion at \(c\). The theorem does not require that \(f\) be continuous everywhere in the whole interval in order to guarantee the derivative relationship at this particular point, provided the integrals defining \(A\) exist.
Why Continuity Gives the Derivative
For a nonzero change \(h\), the difference quotient for \(A\) at \(c\) can be rewritten as an average value of \(f\) over the short interval between \(c\) and \(c+h\). This follows from additivity of definite integrals:
If \(f\) is continuous at \(c\), its values throughout a sufficiently short interval around \(c\) are close to \(f(c)\). Therefore their average is also close to \(f(c)\). The following argument makes that reasoning precise.
Subtract \(f(c)\) from the quotient and use linearity of the integral to write the error as an average of \(f(t)-f(c)\).
For any \(\varepsilon>0\), continuity at \(c\) gives a \(\delta>0\) such that \(|f(t)-f(c)|<\varepsilon\) whenever \(|t-c|<\delta\).
When \(0<|h|<\delta\), all points between \(c\) and \(c+h\) satisfy the continuity bound, so the absolute error in the average is less than \(\varepsilon\).
In detail, for \(0<|h|<\delta\), the triangle inequality for integrals gives
This works for positive and negative \(h\); reversing the bounds changes the sign of the integral, while the absolute value accounts for that sign. Since the difference between the quotient and \(f(c)\) can be made smaller than any positive \(\varepsilon\), the quotient tends to \(f(c)\). Thus \(A'(c)=f(c)\).
Worked Example: A Continuous Integrand
Worked Example: A Continuous Integrand
Let \(f(t)=t^2+1\) on \([-1,2]\), and define \(A(x)=\int_{-1}^{x}(t^2+1)\,dt\). Find \(A(x)\), then verify the derivative relationship at \(x=1\).
An antiderivative of \(t^2+1\) is \(t^3/3+t\), so
Differentiate the expression for \(A\):
At \(x=1\), \(A'(1)=1^2+1=2\). Also, \(f(1)=1^2+1=2\), so \(A'(1)=f(1)\), as the theorem guarantees. Differentiating the computed accumulation function verifies the result.
What a Jump Discontinuity Can Do
At a jump, the averages of \(f\) immediately to the left and right of the point can approach different numbers. If the two one-sided difference quotients approach different values, the ordinary two-sided derivative does not exist.
Worked Example: A Jump Where the Accumulation Function Is Not Differentiable
On \([0,2]\), let \(f(t)=2\) for \(0\leq t<1\) and \(f(t)=5\) for \(1\leq t\leq2\). Define \(A(x)=\int_0^x f(t)\,dt\). Find \(A\) and test differentiability at the jump \(x=1\).
For \(0\leq x\leq1\), the integrand is \(2\) between \(0\) and \(x\), so \(A(x)=2x\). For \(1\leq x\leq2\), split at the jump and add the two integrals:
Both formulas give \(A(1)=2\). For \(h<0\) close to zero, \(1+h<1\), and the left-hand difference quotient is
For \(h>0\) close to zero, the right-hand quotient is
Because the left and right limits are \(2\) and \(5\), they are unequal. Therefore \(A'(1)\) does not exist. In particular, the value \(f(1)=5\) cannot be used as the derivative there.
More generally, if an integrand has finite one-sided limits \(L_-\) and \(L_+\) at an interior jump \(c\), the corresponding left- and right-hand difference quotients for its accumulation function approach \(L_-\) and \(L_+\), respectively. If those limits differ, the two-sided derivative cannot exist. This follows because each quotient is an average of the integrand on one side of \(c\), and values on that side approach the corresponding one-sided limit.
A Discontinuity Can Change the Derivative Relationship Without Preventing Differentiability
The absence of continuity does not imply that the accumulation function must be nondifferentiable. An integrand’s value at a single point does not change its definite integral. But the derivative, if it exists there, need not match the integrand’s value at that point.
Worked Example: An Isolated Change in the Integrand
On \([0,2]\), define \(f(t)=2\) for \(t\ne1\), and \(f(1)=7\). Let \(A(x)=\int_0^x f(t)\,dt\). Find \(A\), and compare \(A'(1)\) with \(f(1)\).
The integrand is bounded and has only one discontinuity, so it is Riemann integrable. Changing the value at the single point \(t=1\) does not affect any of these integrals. Thus, for \(0\leq x\leq2\),
It follows that \(A'(x)=2\), including at \(x=1\). But \(f(1)=7\), so
Here the accumulation function is differentiable at the discontinuity, but its derivative does not equal the assigned value of the integrand. The Fundamental Theorem’s continuity condition is not satisfied at \(1\), so the theorem does not guarantee the claimed equality there.
Common Mistake: Using the Integrand Value at a Discontinuity
A common error is to state \(A'(c)=f(c)\) at every point simply because \(A(x)\) is defined as an integral. For the jump example, this would give the incorrect claim \(A'(1)=f(1)=5\). But the left-hand quotient at \(1\) equals \(2\), while the right-hand quotient equals \(5\), so the derivative does not exist.
The correction is to check the condition before applying the theorem. If \(f\) is continuous at \(c\), the theorem applies there. If \(f\) is discontinuous at \(c\), calculate the relevant difference quotient or use the explicit formula for \(A\). The discontinuity alone does not determine the outcome: as the isolated-change example shows, the derivative may exist, yet differ from \(f(c)\).
Check Your Understanding
Use the continuity condition and, where needed, the difference quotient to answer these questions.
- If \(f\) is continuous at \(c\) and \(A(x)=\int_a^x f(t)\,dt\), what does the Fundamental Theorem of Calculus, Part 1, guarantee at \(c\)?
- For the jump integrand with values \(2\) to the left of \(1\) and \(5\) at and to the right of \(1\), what are the left- and right-hand difference quotient limits for \(A\) at \(1\)?
- In the isolated-change example, why does setting \(f(1)=7\) not change \(A(x)\)?
- Can \(A\) be differentiable at a discontinuity of \(f\)? Explain using an example from this tutorial.
- Why is it not valid to apply the derivative relationship automatically at a point where \(f\) is discontinuous?
