Introduction
An accumulation function records the signed integral of an integrand from a fixed starting point to a variable endpoint. When the endpoint moves across a jump in the integrand, the accumulated quantity can remain continuous even though its rate of change changes abruptly.
The key question is not simply what value the integrand takes at the jump. The derivative of the accumulation function is determined by what the integrand approaches from each side. We will find those two one-sided derivatives directly from difference quotients.
Jump Discontinuities and Accumulation
Let \(f\) be integrable on intervals near an interior point \(c\), and let \(a\) be a fixed starting point. The accumulation function is
Suppose \(f\) has finite one-sided limits at \(c\). If those limits are different, \(f\) has a jump discontinuity at \(c\).
The value \(f(c)\) need not equal either one-sided limit. In fact, changing the value of an integrable function at just one point does not change its integral. The accumulation function's behavior at \(c\) is therefore governed by the values of \(f\) arbitrarily close to \(c\), not by the value assigned exactly at \(c\).
Worked Example: A Constant Integrand With a Jump
Worked Example: A Constant Integrand With a Jump
Define \(f(t)=2\) for \(t<1\) and \(f(t)=5\) for \(t>1\), with any value at \(t=1\). Let \(A(x)=\int_0^x f(t)\,dt\). Find the formula for \(A\), then compare its one-sided difference quotients at \(x=1\).
For \(0\le x\le1\), the integrand is \(2\), so
For \(x>1\), split the accumulation at the jump:
At the jump, \(A(1)=2\). For \(h<0\) sufficiently close to \(0\),
For \(h>0\),
The left-hand difference quotient is \(2\), while the right-hand one is \(5\). Thus \(A'(1)\) does not exist. The two formulas for \(A\) agree at \(1\), so \(A\) is continuous there; its slopes on the two sides are different.
Check: The left formula gives \(A(1)=2\), and the right formula gives \(5(1)-3=2\). Differentiating each branch gives \(2\) on the left and \(5\) on the right, matching the calculated quotients.
Why the One-Sided Quotients Give the One-Sided Limits
For \(h>0\), the change in accumulation from \(c\) to \(c+h\) is the integral over that short interval. Dividing by \(h\) gives the average value of \(f\) there. For \(h<0\), the orientation of the integral reverses along with the sign of \(h\), again producing an average over the interval between \(c+h\) and \(c\).
Here is the full justification. For \(h>0\), additivity of the integral gives
Take any \(\varepsilon>0\). Because \(f(t)\to L_+\) as \(t\to c^+\), there is a \(\delta>0\) such that \(0<t-c<\delta\) implies \(|f(t)-L_+|<\varepsilon\). Thus for \(0<h<\delta\), the average differs from \(L_+\) by at most \(\varepsilon\):
Therefore the right-hand difference quotients approach \(L_+\).
For \(h<0\), the quotient can be written as
The final expression is the average of \(f\) on \([c+h,c]\). Since \(f(t)\to L_-\) as \(t\to c^-\), the same estimate, using \(-h>0\), shows that this average approaches \(L_-\).
A two-sided derivative exists precisely when the left-hand and right-hand difference quotients approach the same number. This happens exactly when \(L_-=L_+\). In that case, both sides approach their common value, which is \(A'(c)\). At a jump, the two limits differ, so the one-sided derivatives differ and the derivative does not exist.
Worked Example: Different Linear Branches
Worked Example: Different Linear Branches
Let \(f(t)=t+1\) for \(t<0\), and \(f(t)=3-t\) for \(t>0\). The value at \(0\) may be chosen arbitrarily. Define \(A(x)=\int_{-1}^x f(t)\,dt\). Find \(A(0)\) and the one-sided difference quotients there.
First,
For \(-1\le x\le0\), the same antiderivative gives
For \(x>0\), add the integral from \(0\) to \(x\):
For \(h<0\), substituting into the left formula gives
For \(h>0\), the right formula gives
The one-sided limits of the integrand are \(1\) from the left and \(3\) from the right, exactly as the quotients show. Since these differ, \(A'(0)\) does not exist.
Check: Both formulas for \(A\) give \(1/2\) at \(x=0\). Differentiating the left formula gives \(x+1\), which approaches \(1\); differentiating the right formula gives \(3-x\), which approaches \(3\).
Continuity Does Not Guarantee Differentiability
The accumulation function remains continuous at a jump when the integrand is integrable and bounded near that point. To see why, finite one-sided limits imply that \(f\) is bounded near \(c\): there are numbers \(M\) and \(\delta>0\) such that \(|f(t)|\le M\) for \(0<|t-c|<\delta\). For \(|h|<\delta\),
As \(h\to0\), the bound \(M|h|\to0\), so \(A(c+h)\to A(c)\). Continuity follows, even though unequal one-sided limits of \(f\) prevent differentiability of \(A\).
Worked Example: The Value at the Jump Is Not the Slope
Worked Example: The Value at the Jump Is Not the Slope
Let \(f(t)=-2\) for \(t<1\), \(f(t)=3\) for \(t>1\), and specifically set \(f(1)=100\). Let \(A(x)=\int_0^x f(t)\,dt\). Determine whether \(A'(1)\) exists.
For \(0\le x\le1\),
For \(x>1\),
For \(h<0\),
For \(h>0\),
Because the one-sided quotients approach different values, \(A'(1)\) does not exist. In particular, the value \(f(1)=100\) is not the derivative. The integrand's one-sided limits, \(-2\) and \(3\), determine the two one-sided slopes.
Check: The formulas for \(A\) both give \(-2\) at \(1\). Their derivatives are \(-2\) and \(3\), respectively, confirming the quotient calculations.
A Common Mistake
A tempting but incorrect step is to apply the formula \(A'(c)=f(c)\) at a jump. That formula requires continuity of \(f\) at the point in question; it cannot be applied when the left and right limits differ. In the preceding example, substituting \(f(1)=100\) would incorrectly claim that the derivative is \(100\). The correct calculation takes the limits of the two difference quotients, which are \(-2\) and \(3\), so no derivative exists.
When the integrand is continuous at the endpoint, the two one-sided limits agree, and the usual accumulation derivative result applies. At a genuine jump, calculate the left and right quotients separately instead of substituting the point value.
Check Your Understanding
Use one-sided difference quotients or the theorem above to answer each question.
- If the left and right limits of an integrand at \(c\) are \(4\) and \(-1\), respectively, what are the one-sided derivatives of its accumulation function at \(c\)? Does the derivative exist?
- Why can an accumulation function be continuous at a jump even when its derivative does not exist there?
- If \(f(c)\) is changed while all other values of \(f\) stay fixed, does the accumulation function change? Explain.
- Suppose the one-sided limits of the integrand at \(c\) are both \(2\). What does the difference-quotient result say about the derivative of the accumulation function at \(c\)?
- For the integrand in the last worked example, what incorrect derivative would result from substituting \(f(1)\), and why is that method invalid?
