Why Split a Definite Integral?
A definite integral over an interval can often be understood more clearly by dividing that interval into smaller pieces. A useful dividing point might be where the function crosses the horizontal axis, where its formula changes, or where it has a discontinuity. The value of the integral is preserved, but the contributions from the pieces can be easier to interpret or calculate separately.
This is especially helpful when a graph lies above the horizontal axis on one part of the interval and below it on another. The definite integral records signed contributions: regions above the axis contribute positively, while regions below it contribute negatively. Splitting at a root makes those contributions visible instead of hiding them in one net value.
The key is to choose split points for a reason. Splitting an interval at an arbitrary point is valid when the relevant integrals exist, but a strategic point helps reveal structure or simplify the work.
Interval Additivity and Split Points
For split points \(a=c_0<c_1<\cdots<c_n=b\), the corresponding identity is
This formula requires the integral to exist on the original interval; in the usual proper Riemann-integral setting, the restrictions to the subintervals are then integrable as well. A discontinuity does not automatically prevent splitting. For example, a bounded function with only finitely many discontinuities on a closed interval is Riemann integrable. By contrast, splitting does not make an unbounded function properly Riemann integrable.
A split point at an endpoint adds no new piece, and a zero-length subinterval contributes zero. In practice, we choose interior points where the description or interpretation of the function changes.
Where Strategic Split Points Come From
Three common sources of useful split points are:
- Roots: points \(c\) where \(f(c)=0\). A root can separate intervals on which the graph has different signs.
- Discontinuities: points where the function has a jump or another discontinuity. Splitting can keep the behavior on either side distinct.
- Changes in formula: points where a piecewise-defined function switches from one expression or graph segment to another.
These points are candidates, not automatic solutions. A root need not be a sign change: for instance, \(x^2=0\) at \(x=0\), but \(x^2\) is nonnegative on both sides of zero. A discontinuity may also require more care than merely naming it: the integral must exist, and the portions must be evaluated with their correct values and signs.
Worked Example: Splitting at a Sign Change
Worked Example: Splitting at a Sign Change
Evaluate \(\int_0^2 (x-1)\,dx\) by splitting at the root of \(f(x)=x-1\), and interpret the two contributions.
First find the root:
On \([0,1]\), \(x-1\le 0\), so the graph is below the horizontal axis. It forms a right triangle with base \(1-0=1\) and height \(1\). Its signed contribution is negative:
On \([1,2]\), \(x-1\ge 0\). This region is a triangle with base \(2-1=1\) and height \(1\), so its signed contribution is positive:
By interval additivity, the contributions combine as follows:
Check: The two triangles have equal area and opposite signs, so they cancel. The net signed integral is \(0\), even though the total geometric area is \(\frac12+\frac12=1\).
Why Splitting at a Root Clarifies Signed Contributions
The preceding example illustrates a general use of roots. Suppose \(f\) is continuous on \([a,b]\), and \(f(c)=0\) for some \(c\) strictly between \(a\) and \(b\). Splitting at \(c\) separates the contributions on either side:
Continuity matters in this statement about signs. If a continuous function has no zero on an interval, it cannot change from positive to negative there: by the intermediate value theorem, such a change would require a zero between the two values. Thus, when the roots have been identified, checking the sign on each interval between consecutive roots tells us whether its signed contribution is positive or negative. The root itself marks the boundary between intervals; it contributes no separate area.
This argument explains the method, not a requirement to split at every root. If the integral is already simple, an extra split may add work. If several roots divide the interval into regions with different signs, splitting can make cancellation and net accumulation easier to see.
Worked Example: Splitting Where the Formula Changes
Worked Example: Splitting Where the Formula Changes
Let \(f\) be defined on \([0,2]\) by \(f(x)=x\) for \(0\le x\le 1\) and \(f(x)=2-x\) for \(1<x\le 2\). Find \(\int_0^2 f(x)\,dx\) by splitting where the formula changes.
The formula switches at \(x=1\), so use the subintervals \([0,1]\) and \([1,2]\). On the first interval the graph rises from height \(0\) to height \(1\). The area is a triangle with base \(1\) and height \(1\):
On the second interval the graph falls from height \(1\) to height \(0\). This is another triangle with base \(2-1=1\) and height \(1\):
Therefore, additivity gives
Check: The graph is a triangle with base \(2\) and height \(1\). Its area is \(\frac12(2)(1)=1\), matching the sum of the two pieces. The endpoint value at \(x=1\) agrees with both formulas, so there is no ambiguity at the split.
Worked Example: Splitting at a Discontinuity
Worked Example: Splitting at a Discontinuity
Define \(g\) on \([0,3]\) by \(g(x)=2\) for \(0\le x<1\), and \(g(x)=-1\) for \(1\le x\le 3\). Evaluate \(\int_0^3 g(x)\,dx\) by splitting at the jump.
The function has a jump at \(x=1\). It is bounded and has only this one discontinuity, so it is Riemann integrable on \([0,3]\). Split the interval into \([0,1]\) and \([1,3]\). On the first interval, the graph has height \(2\) and width \(1\), giving a positive rectangular contribution:
On the second interval, the graph has height \(-1\) and width \(3-1=2\). The rectangle lies below the axis, so its signed contribution is negative:
Thus,
Check: The positive rectangular area is \(2\), and the negative rectangular area has magnitude \(2\). They cancel in the signed integral. The value assigned at the single point \(x=1\) does not affect the integral.
A Common Mistake: Adding Geometric Areas Without Signs
A frequent error is to split an integral at a root, calculate the ordinary areas of the regions, and add both as positive numbers. That computes total geometric area, not the signed integral.
For the function \(f(x)=x-1\) on \([0,2]\), the incorrect calculation is:
The correction is to attach a sign to each contribution. The left triangle is below the axis, while the right triangle is above it:
The value \(1\) is the total geometric area. The value \(0\) is the net signed accumulation. Which quantity is wanted depends on the question, but they must not be confused.
Choosing Split Points Efficiently
A useful procedure is to locate structural changes first and then decide which ones matter for the calculation. A root is especially useful when signs need to be separated; a formula change is useful when each piece has a simpler description; and a discontinuity is useful when the behavior on the two sides should be treated separately.
Find roots, discontinuities, and points where the graph or formula changes within the interval.
Confirm that the definite integral exists on the interval and that splitting applies to the pieces.
On every subinterval, identify the applicable formula and whether the graph is above or below the axis.
Use interval additivity, retaining negative signs for contributions below the axis.
Compare the result with the graph or geometry and distinguish net signed accumulation from total geometric area.
Not every possible split point is helpful. A split should make the pieces easier to describe, calculate, or interpret. If it does none of these, it may be valid but unnecessary.
Check Your Understanding
Use interval additivity and signed contributions in your answers.
- For \(h(x)=x-2\) on \([0,4]\), where is the root, and what geometric feature makes it a natural split point?
- Why does a root of a continuous function separate sign regions, while a root alone does not always imply a sign change?
- A bounded piecewise-constant function has one jump on a closed interval. What condition allows its integral to be split at the jump?
- What is the difference between total geometric area and the signed integral when a graph crosses the horizontal axis?
- Give one situation in which splitting at a change of formula would simplify the interpretation of a definite integral.
