Accumulation When the Rate Changes Formula
An accumulation function records the signed change contributed by an integrand as the upper endpoint moves. When the integrand is piecewise-defined, its formula changes at specified breakpoints. The accumulation must account for each piece over the portion of the interval where that piece applies.
This requires two related checks. First, split the integral wherever the integrand changes formula, so each part is evaluated using the correct rule. Then, when finding the derivative of the accumulation function, use the appropriate piece on each interval and check what happens at every breakpoint. A breakpoint is not automatically a point where the derivative fails to exist; the neighboring formulas determine the result.
The variable \(t\) is the integration variable, while \(x\) is the variable upper endpoint. The value \(A(x)\) is a signed accumulation: portions where \(f(t)\) is negative contribute negatively. Reversing the order of the bounds changes the sign, so an \(x\) below \(a\) gives \(A(x)=-\int_x^a f(t)\,dt\).
Splitting at a Breakpoint
Suppose \(c\) is a breakpoint between \(a\) and \(x\). Additivity of definite integrals gives \(\int_a^x f(t)\,dt=\int_a^c f(t)\,dt+\int_c^x f(t)\,dt\). This identity lets us calculate the accumulation using the formula for \(f\) on each relevant interval. If \(x\) lies before \(c\), only the first piece is needed; if it lies after \(c\), both pieces contribute.
For several breakpoints, split once at each breakpoint crossed between \(a\) and \(x\). The result is a piecewise formula for \(A\). To find \(A'(x)\) at an interior point of one piece, the Fundamental Theorem of Calculus applies when the integrand is continuous near that point: the derivative is the integrand's formula there. At a breakpoint, check the neighboring formulas rather than extending either one without justification.
Worked Example: A Continuous Integrand with One Breakpoint
Let \(f(t)=2t\) for \(0\le t\le1\), and \(f(t)=2\) for \(1<t\le3\). Define \(A(x)=\int_0^x f(t)\,dt\). Find \(A(x)\) and \(A'(x)\), including at \(x=1\).
For \(0\le x\le1\), use the first formula throughout the integral: \[ A(x)=\int_0^x2t\,dt=[t^2]_0^x=x^2-0=x^2. \]
For \(1<x\le3\), split at \(t=1\): \[ A(x)=\int_0^1 2t\,dt+\int_1^x2\,dt. \] The first part is \([t^2]_0^1=1-0=1\). The second is \([2t]_1^x=2x-2\). Therefore \[ A(x)=1+(2x-2)=2x-1. \]
Thus \(A(x)=x^2\) for \(0\le x\le1\), and \(A(x)=2x-1\) for \(1<x\le3\). On the interiors of the pieces, differentiation gives \(A'(x)=2x\) for \(0<x<1\), and \(A'(x)=2\) for \(1<x<3\).
At the breakpoint, the function formulas agree: the first gives \(A(1)=1^2=1\), and the second formula gives \(2(1)-1=1\). Their derivative formulas also agree there: the left-hand derivative is \(2(1)=2\), and the right-hand derivative is \(2\). Hence \(A'(1)=2\). This matches the integrand value \(f(1)=2\). As a value check, \(A(2)=2(2)-1=3\), and splitting the integral gives \(\int_0^1 2t\,dt+\int_1^2 2\,dt=1+2=3\).
Why the Breakpoint Derivative Can Be Checked
The derivative at a breakpoint can be justified directly from the definition of derivative. Let \(c\) be an interior breakpoint, and suppose the formulas on the two sides fit together continuously at \(c\), with common value \(L\). For \(h\ne0\) sufficiently small, the difference quotient for the accumulation function is \[ \frac{A(c+h)-A(c)}{h} =\frac{1}{h}\int_c^{c+h} f(t)\,dt. \] This equality follows by additivity, including when \(h<0\), because reversing the bounds supplies the corresponding negative sign.
To see why the quotient tends to \(L\), subtract \(L\) and use continuity of the piecewise integrand at \(c\): \[ \left|\frac{1}{h}\int_c^{c+h}f(t)\,dt-L\right| =\left|\frac{1}{h}\int_c^{c+h}(f(t)-L)\,dt\right| \le \sup_{t\text{ between }c\text{ and }c+h}|f(t)-L|. \] The inequality holds for either sign of \(h\): the absolute value of an integral is at most the interval length times the largest absolute value of its integrand, and that interval has length \(|h|\). As \(h\) approaches zero, continuity makes the supremum on the shrinking interval approach zero. Therefore the difference quotient tends to \(L\), so \(A'(c)=L\).
This argument explains why checking the breakpoint matters. The Fundamental Theorem gives the derivative on each interval where the relevant formula is continuous. At a breakpoint, the derivative follows when the two sides approach the same integrand value; it is not enough merely to differentiate the formulas away from the breakpoint.
Worked Example: Two Linear Pieces Meeting at a Breakpoint
Let \(f(t)=t+1\) for \(-1\le t\le1\), and \(f(t)=3-t\) for \(1<t\le3\). Define \(A(x)=\int_{-1}^x f(t)\,dt\). Find the accumulation function and its derivative, checking the breakpoint \(x=1\).
For \(-1\le x\le1\), integrate the first piece: \[ A(x)=\int_{-1}^x(t+1)\,dt =\left[\frac{t^2}{2}+t\right]_{-1}^{x} =\left(\frac{x^2}{2}+x\right)-\left(\frac12-1\right) =\frac{x^2}{2}+x+\frac12 =\frac{(x+1)^2}{2}. \]
In particular, \(A(1)=\frac{(1+1)^2}{2}=\frac42=2\). For \(1<x\le3\), split at \(1\): \[ A(x)=\int_{-1}^1(t+1)\,dt+\int_1^x(3-t)\,dt. \] The first integral is \(A(1)=2\). For the second, \[ \int_1^x(3-t)\,dt =\left[3t-\frac{t^2}{2}\right]_1^x =3x-\frac{x^2}{2}-\left(3-\frac12\right) =3x-\frac{x^2}{2}-\frac52. \] Thus \(A(x)=2+3x-\frac{x^2}{2}-\frac52=3x-\frac{x^2}{2}-\frac12\) when \(1<x\le3\).
On the interval interiors, differentiating gives \(A'(x)=x+1\) for \(-1<x<1\), and \(A'(x)=3-x\) for \(1<x<3\). At \(x=1\), the left accumulation formula gives \(2\), and the right formula gives \(3-\frac12-\frac12=2\). The derivative formulas both give \(2\) at the breakpoint: \(1+1=2\) and \(3-1=2\). Therefore \(A'(1)=2=f(1)\).
Check a value beyond the breakpoint. At \(x=2\), the formula gives \(A(2)=6-2-\frac12=\frac72\). Directly, the first piece contributes \(2\), and the second contributes \(\int_1^2(3-t)\,dt=4-\frac52=\frac32\). Their sum is \(2+\frac32=\frac72\), as required.
When the Integrand Jumps at a Breakpoint
If the formulas on the two sides do not approach the same value, the accumulation function can still be continuous, while its derivative at the breakpoint may fail to exist. The one-sided difference quotients reveal the issue. This is why checking a breakpoint requires more than substituting the breakpoint into one selected branch.
Worked Example: A Jump and Its Effect on the Derivative
Let \(f(t)=1\) for \(0\le t<2\), and \(f(t)=3\) for \(2\le t\le4\). Define \(A(x)=\int_0^x f(t)\,dt\). For \(0\le x\le2\), \(A(x)=\int_0^x1\,dt=x\). For \(2<x\le4\), split at \(2\): \[ A(x)=\int_0^2 1\,dt+\int_2^x3\,dt=2+3(x-2)=3x-4. \] At the breakpoint both accumulation formulas give \(A(2)=2\), so \(A\) is continuous there.
For \(h<0\) close to zero, \(A(2+h)=2+h\), so \[ \frac{A(2+h)-A(2)}{h}=\frac{(2+h)-2}{h}=1. \] For \(h>0\) close to zero, \(A(2+h)=3(2+h)-4=2+3h\), so \[ \frac{A(2+h)-A(2)}{h}=\frac{(2+3h)-2}{h}=3. \] The one-sided difference quotients have different limits, \(1\) and \(3\). Consequently, \(A'(2)\) does not exist. Differentiating the two accumulation formulas separately gives the same one-sided results: \(1\) on the left and \(3\) on the right. This example shows why a breakpoint must be checked from both sides.
A Reliable Procedure
For a piecewise-defined integrand, keep the integral construction and the derivative check separate. The integral construction determines the accumulation values; the derivative check determines whether and how the resulting function changes at each point.
Identify which integrand formula applies on each interval from the base point to the variable endpoint.
Use additivity to break the accumulation integral at every breakpoint between the base point and the endpoint.
Evaluate each definite integral with the formula valid on that piece, then combine the results to obtain a formula for the accumulation function.
Apply the Fundamental Theorem where the integrand is continuous near the point, using that interval's formula.
Compare accumulation values and the left- and right-hand derivative behavior. If the integrand formulas meet continuously, the accumulation derivative equals their common value.
Common Pitfall: Using the Wrong Piece
A common error is to apply one branch of the integrand beyond the interval where it is defined. In the first worked example, for \(x=2\) the applicable integrand is \(f(x)=2\), because \(2\) lies in the second piece. Using the first formula \(2x\) at \(x=2\) would incorrectly give \(f(2)=4\). The correct accumulation derivative is \(A'(2)=2\), as is also confirmed by differentiating \(A(x)=2x-1\) on that interval.
At a breakpoint, another error is to report a derivative after differentiating just one neighboring formula. For example, in the jump example, the left formula gives slope \(1\) and the right formula gives slope \(3\); neither alone is the two-sided derivative. A derivative at an interior breakpoint exists only if the two-sided difference quotient has one common limit. For a continuous piecewise integrand, checking that its neighboring formulas agree at the breakpoint verifies that the accumulation derivative equals that common value.
Why Piecewise Accumulation Matters
Many rates change according to conditions: a rate may follow one rule before a particular time and another afterward. An accumulation function combines the contributions from all the intervals crossed, while its derivative describes the current rate on the interval containing the input. Correctly handling breakpoints keeps the accumulated total consistent even when the rate formula changes.
The key habits are straightforward: split the definite integral at each breakpoint, retain all earlier contributions when writing a later branch of the accumulation function, and check both sides before assigning a derivative at a breakpoint. These checks prevent a correct interval formula from being mistakenly treated as valid everywhere.
Check Your Understanding
Use the piecewise formulas on their stated intervals, and check both sides whenever a question concerns a breakpoint.
- If an accumulation integral crosses two breakpoints, at which points should it be split?
- Suppose the integrand is continuous on both sides of a breakpoint and the two formulas have the same value there. What is the derivative of the accumulation function at that breakpoint?
- For \(f(t)=2\) on \(0\le t<1\) and \(f(t)=5\) on \(1\le t\le3\), write an accumulation formula with base point \(0\) on each interval.
- In the preceding question, compare the left- and right-hand derivative values at \(x=1\). Does the accumulation function have a derivative there?
- Why must an accumulation formula for an endpoint beyond a breakpoint include the contribution from the interval before that breakpoint?
