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Accumulation Functions · Tutorial 1018 of 1495

Splitting Accumulations At Piecewise Breakpoints

See how to split a variable-limit integral at piecewise breakpoints and combine the signed contributions into a correct accumulation formula.

Intermediate 9 min read

What You'll Learn

  • Identify every formula breakpoint between the fixed basepoint and the variable endpoint.
  • Partition the integral into subintervals where one formula applies.
  • Add signed subintegrals while respecting the orientation of the interval.
  • Build piecewise formulas for an accumulation function.
  • Check formulas at the basepoint, breakpoints, and endpoints.
  • Avoid errors when subtracting antiderivative values or crossing a breakpoint.

Why Breakpoints Matter

A variable-limit integral records accumulated signed area as its endpoint moves. If the integrand has different formulas on different parts of its domain, one antiderivative formula cannot generally be used across the whole interval. The interval must first be divided wherever the integrand changes formula.

Each part is then integrated using the formula valid there. The resulting values are added with their signs: contributions below the horizontal axis count negatively, and reversing the orientation of an interval reverses the sign of its integral.

Key idea: To evaluate an accumulation with a piecewise-defined integrand, split the interval at every formula breakpoint it crosses, evaluate each subintegral using the appropriate formula, and add the signed results.

Definitions and the Splitting Rule

Suppose \(f\) is bounded and piecewise continuous on an interval containing \(a\) and \(x\). Its accumulation function with fixed basepoint \(a\) is defined by

$$ F(x)=\int_a^x f(t)\,dt. $$

The variable \(t\) is the integration variable: it runs across the interval of accumulation. The value \(x\) is the endpoint that determines how far the accumulation extends. If \(a\leq x\), the interval is traversed from \(a\) to \(x\); if \(x<a\), orientation reverses the sign.

A breakpoint is a value at which the formula used to define \(f\) changes. If \(a\leq c\leq x\) and \(c\) is a breakpoint, the splitting rule is

$$ \int_a^x f(t)\,dt = \int_a^c f(t)\,dt+\int_c^x f(t)\,dt. $$

The same rule can be applied repeatedly when there are several breakpoints between the endpoints. On each resulting subinterval, use the formula for \(f\) that applies there. A value assigned to \(f\) at a single breakpoint does not change the integral; for a bounded piecewise-continuous function, changing its value at finitely many individual points leaves its integral unchanged.

Why Splitting Works

The splitting rule follows from how the definite integral is formed. For \(a\leq c\leq x\), take a partition of \([a,x]\) that includes \(c\). A Riemann sum over the whole interval is the sum of the terms from \([a,c]\) and the terms from \([c,x]\). As the partition widths shrink, those two groups of terms approach the integrals on their respective subintervals. Their sum therefore approaches the integral over the whole interval.

$$ \int_a^x f(t)\,dt = \lim (\text{sum on }[a,c]+\text{sum on }[c,x]) = \int_a^c f(t)\,dt+\int_c^x f(t)\,dt. $$

For finitely many breakpoints \(c_1,\ldots,c_n\) in increasing order between \(a\) and \(x\), apply this argument at each breakpoint. The full integral is the sum of the subintegrals on the resulting pieces. If the endpoint lies to the left of the basepoint, first use the orientation convention \(\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt\), then partition the interval from \(x\) to \(a\).

Practical rule: List the basepoint, the variable endpoint, and every breakpoint between them in order. Integrate over each consecutive pair using the applicable formula, then add the results with the correct orientation.

Worked Examples

Worked Example: A Variable Endpoint Crossing One Breakpoint

Let

$$ f(t)= \begin{cases} t+2, & -2\leq t\leq 0,\\ 2-2t, & 0<t\leq 2, \end{cases} \qquad F(x)=\int_{-2}^x f(t)\,dt. $$

For \(-2\leq x\leq 0\), the first formula applies throughout:

$$ \begin{aligned} F(x) &=\int_{-2}^x(t+2)\,dt\\ &=\left[\frac{t^2}{2}+2t\right]_{-2}^x\\ &=\left(\frac{x^2}{2}+2x\right) -\left(\frac{(-2)^2}{2}+2(-2)\right)\\ &=\frac{x^2}{2}+2x-(2-4)\\ &=\frac{x^2}{2}+2x+2. \end{aligned} $$

For \(0<x\leq 2\), split at the breakpoint \(0\). The first portion contributes \(F(0)=2\):

$$ \begin{aligned} F(x) &=\int_{-2}^{0}(t+2)\,dt+\int_0^x(2-2t)\,dt\\ &=2+\left[2t-t^2\right]_0^x\\ &=2+(2x-x^2)\\ &=2+2x-x^2. \end{aligned} $$

At \(x=0\), the first formula gives \(0+0+2=2\), and the second gives \(2+0-0=2\). Both agree with \(\int_{-2}^{0}(t+2)\,dt=2\). At the basepoint, the first formula gives \(F(-2)=2-4+2=0\), as required. At \(x=2\), the second gives \(F(2)=2+4-4=2\).

As an additional check, differentiating the first expression gives \(x+2\), and differentiating the second gives \(2-2x\), exactly the corresponding formulas for \(f\) on the interiors of their intervals.

Worked Example: Accumulating Across Two Breakpoints

Define

$$ f(t)= \begin{cases} t+2, & -2\leq t<-1,\\ 1-t, & -1\leq t<1,\\ t-3, & 1\leq t\leq 3, \end{cases} \qquad F(x)=\int_{-2}^x f(t)\,dt. $$

For \(-2\leq x<-1\), only the first formula is needed:

$$ \begin{aligned} F(x) &=\int_{-2}^x(t+2)\,dt\\ &=\left[\frac{t^2}{2}+2t\right]_{-2}^x\\ &=\frac{x^2}{2}+2x-\left(2-4\right)\\ &=\frac{x^2}{2}+2x+2. \end{aligned} $$

The accumulated value at the first breakpoint is

$$ F(-1)=\frac{(-1)^2}{2}+2(-1)+2 =\frac12-2+2 =\frac12. $$

For \(-1\leq x<1\), begin with that accumulated value and integrate the middle formula:

$$ \begin{aligned} F(x) &=F(-1)+\int_{-1}^x(1-t)\,dt\\ &=\frac12+\left[t-\frac{t^2}{2}\right]_{-1}^x\\ &=\frac12+\left(x-\frac{x^2}{2}\right) -\left(-1-\frac12\right)\\ &=x-\frac{x^2}{2}+2. \end{aligned} $$

At \(x=1\), this gives \(F(1)=1-\frac12+2=\frac52\). For \(1\leq x\leq 3\), split at \(1\) and use the last formula:

$$ \begin{aligned} F(x) &=F(1)+\int_1^x(t-3)\,dt\\ &=\frac52+\left[\frac{t^2}{2}-3t\right]_1^x\\ &=\frac52+\left(\frac{x^2}{2}-3x\right) -\left(\frac12-3\right)\\ &=\frac{x^2}{2}-3x+5. \end{aligned} $$

Check the joins: at \(-1\), the first formula gives \(\frac12\), and the middle gives \(-1-\frac12+2=\frac12\). At \(1\), the middle gives \(\frac52\), and the last gives \(\frac12-3+5=\frac52\). At the basepoint, \(F(-2)=2-4+2=0\); at \(x=3\), \(F(3)=\frac92-9+5=\frac12\).

The total signed contribution across all three pieces also checks the final value: the first piece contributes \(\frac12\), the middle contributes \(2\), and the last contributes \(-2\). Their sum is \(\frac12+2-2=\frac12=F(3)\). Differentiating each displayed formula recovers \(t+2\), \(1-t\), and \(t-3\), respectively.

Worked Example: A Variable Endpoint to the Left of the Basepoint

Let

$$ g(t)= \begin{cases} 1, & -2\leq t<0,\\ t-1, & 0\leq t\leq 2, \end{cases} \qquad G(x)=\int_1^x g(t)\,dt. $$

First suppose \(-2\leq x\leq 0\). Since the variable endpoint is left of the basepoint, the interval crosses \(0\) in the reverse direction. Split there, using the formula valid on each piece:

$$ \begin{aligned} G(x) &=\int_1^0(t-1)\,dt+\int_0^x 1\,dt\\ &=\left[\frac{t^2}{2}-t\right]_1^0+[t]_0^x\\ &=\left(0-\left(\frac12-1\right)\right)+(x-0)\\ &=\frac12+x. \end{aligned} $$

For \(0\leq x\leq 2\), the interval from \(1\) to \(x\) stays within the second piece:

$$ \begin{aligned} G(x) &=\int_1^x(t-1)\,dt\\ &=\left[\frac{t^2}{2}-t\right]_1^x\\ &=\frac{x^2}{2}-x-\left(\frac12-1\right)\\ &=\frac{x^2}{2}-x+\frac12. \end{aligned} $$

At the breakpoint, both expressions give \(G(0)=\frac12\). At the basepoint, the second gives \(G(1)=\frac12-1+\frac12=0\). For a direct orientation check at \(x=-1\), \(\int_{-1}^1g(t)\,dt=1-\frac12=\frac12\), so \(\int_1^{-1}g(t)\,dt=-\frac12\); the first formula gives \(-1+\frac12=-\frac12\). Differentiating the expressions on the interiors gives \(1\) and \(x-1\), as required.

Common Mistakes and How to Avoid Them

A frequent error is to find an antiderivative and substitute only the variable endpoint, forgetting to subtract its value at the lower endpoint. For instance,

$$ \int_{-2}^x(t+2)\,dt \ne \frac{x^2}{2}+2x. $$

The correct calculation subtracts the lower-endpoint value:

$$ \left[\frac{t^2}{2}+2t\right]_{-2}^x = \frac{x^2}{2}+2x-(2-4) = \frac{x^2}{2}+2x+2. $$

The incorrect expression gives \(-2\) at \(x=-2\), even though an integral with identical endpoints is \(0\). Checking the basepoint exposes the error immediately.

Another mistake is to use one formula for \(f\) on an interval that crosses a breakpoint. In that situation, split first. Also keep track of orientation: when \(x<a\), the accumulation from \(a\) to \(x\) is the negative of the accumulation from \(x\) to \(a\). Finally, signed contributions can cancel. Do not replace them with unsigned geometric areas unless the question asks for total area.

Why the Method Is Useful

Piecewise-defined rates arise naturally when a model changes at specified times or thresholds. An accumulation may include contributions from several regimes, such as a rate that is constant during one period and varies during another. Splitting at the transition points makes the calculation reflect the model and prevents applying a formula outside the interval where it is valid.

It also provides a reliable way to assemble an accumulation function. Compute the value already accumulated at each breakpoint, then use that value as the starting contribution for the next piece. Check that adjacent expressions agree at each join and that the result is zero at the fixed basepoint. These checks help catch sign errors, missing contributions, and incorrect endpoint substitutions.

Check Your Understanding

For each question, identify the relevant pieces before calculating.

  1. If \(f(t)=2\) for \(0\leq t<1\) and \(f(t)=t\) for \(1\leq t\leq 3\), how would you split \(\int_0^x f(t)\,dt\) when \(1<x\leq 3\)?
  2. Why must an antiderivative evaluation \([H(t)]_a^x\) subtract \(H(a)\)?
  3. For a variable endpoint to the left of the basepoint, what orientation rule should you use?
  4. What should you check when comparing two formulas for an accumulation function at a breakpoint?
  5. Why can two subintegrals have opposite signs even though both are part of one accumulation?

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