Why Split an Integral?
A definite integral records net accumulation over an interval. If a quantity accumulates one way on part of an interval and another way on the rest, it can be useful to calculate the contributions separately. The same is true when a function is defined by different formulas on different subintervals.
Interval additivity says that dividing an interval at a point does not change the total: the integral over the whole interval is the sum of the integrals over its consecutive parts. The contributions are signed. A portion where the function is negative contributes a negative amount, so it must not be replaced by its geometric area.
The sign of an integral depends both on the values of the function and on the direction in which the interval is traversed. Additivity respects both.
The Additivity Rule
Let \(f\) be Riemann integrable on an interval containing \(a\), \(b\), and an interior point \(c\). If \(a\leq c\leq b\), the interval from \(a\) to \(b\) consists of the consecutive intervals from \(a\) to \(c\) and from \(c\) to \(b\). The integral over the whole interval is the sum of the integrals over those pieces.
The reason for the rule is that net accumulation over consecutive portions adds: the accumulation from \(a\) to \(c\), followed by the accumulation from \(c\) to \(b\), accounts for exactly the same traversal as going from \(a\) to \(b\). The shared endpoint \(c\) does not create an extra contribution; a single point has no interval length. In particular, changing a bounded integrable function's value at that one point does not alter either piece's integral.
The formula also works when one of the pieces has zero length. If \(c=a\), the first term is \(\int_a^a f(x)\,dx=0\), leaving the original integral. If \(c=b\), the second term is zero. These cases agree with the convention for zero-length intervals.
Why the Reverse-Orientation Formula Works
The reverse case follows from the increasing-orientation additivity rule and the sign change when bounds are reversed. Here is the full algebra when \(b\leq c\leq a\). Additivity in increasing order gives \(\int_b^a f(x)\,dx=\int_b^c f(x)\,dx+\int_c^a f(x)\,dx\). Reverse the bounds of the whole interval and of each piece:
The last equality uses commutativity of addition. Thus the same split formula applies in either direction, provided the pieces keep their own orientation. A common error is to reverse the whole interval but leave the two piece integrals in the forward direction.
Worked Examples
Worked Example: Splitting at a Sign Change and Handling an Endpoint Value
Define \(f\) on \([0,2]\) by \(f(x)=-2\) for \(0\leq x<1\), \(f(1)=3\), and \(f(x)=4\) for \(1<x\leq2\). Find \(\int_0^2 f(x)\,dx\) by splitting at \(c=1\).
On the first piece, \(f(x)=-2\) for \(0\leq x<1\), while \(f(1)=3\). The value at the single endpoint \(1\) does not affect the Riemann integral, so this piece contributes the constant value \(-2\) over a length of \(1-0=1\):
On the second piece, \(f(x)=4\) for \(1<x\leq2\). Its value at the left endpoint \(1\) is \(3\), not \(4\), but that single-point difference likewise does not affect the integral. The constant value \(4\) over a length of \(2-1=1\) gives
Additivity now gives
Check: the signed contributions are \(-2\) and \(+4\), whose sum is \(2\). The endpoint value \(f(1)=3\) has not been treated as if it held across an interval; it contributes no additional area.
Worked Example: Cancellation Across Two Constant Pieces
Let \(g(x)=-3\) for \(0\leq x<2\) and \(g(x)=2\) for \(2\leq x\leq5\). Split \(\int_0^5 g(x)\,dx\) at \(c=2\).
The first piece has length \(2-0=2\) and constant value \(-3\), apart from a possible endpoint value at \(2\), which does not affect its integral:
The second piece has length \(5-2=3\) and constant value \(2\):
Therefore,
Check: the negative and positive signed contributions have equal magnitudes, so their net accumulation is zero. The geometric area is not zero; it is \(6+6=12\). This distinction is why the signs must be retained when using additivity.
Worked Example: Splitting an Integral Traversed in Reverse
Use the function \(g\) from the previous example to compute \(\int_5^0 g(x)\,dx\), splitting the reverse traversal at \(c=2\).
The oriented split is from \(5\) to \(2\), followed by \(2\) to \(0\):
The forward integral over \([2,5]\) is \(6\), so reversing its bounds gives \(\int_5^2 g(x)\,dx=-6\). The forward integral over \([0,2]\) is \(-6\), so reversing its bounds gives \(\int_2^0 g(x)\,dx=6\). Hence
Check: the whole forward integral from \(0\) to \(5\) is \(0\), and reversing its bounds gives \(-0=0\), agreeing with the split calculation. The individual pieces do change sign when reversed, even though their sum here remains zero.
How to Use Additivity Reliably
When splitting an integral, keep track of the order of the endpoints and the contribution from each piece. A short procedure prevents most sign errors:
Confirm that the function is integrable over the interval containing the whole traversal and the split point.
For a traversal from \(a\) to \(b\) through \(c\), write \(\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\), without rearranging endpoints.
Account for where the function is positive or negative, and reverse the sign of a piece if its bounds run backward.
Compare the result with the signed contributions over the full interval, or reverse all bounds and verify that the total changes sign.
Common Mistake: Ignoring the Direction of a Piece
Suppose \(\int_0^2 f(x)\,dx=5\), and split the reverse traversal at \(c=1\). A mistaken calculation might write \(\int_2^0 f(x)\,dx=\int_0^1 f(x)\,dx+\int_1^2 f(x)\,dx=5\). This incorrectly copies the forward pieces into the reverse traversal.
The correct ordered split is
The correction reverses each piece, not just the whole interval. Another frequent error is to replace a negative contribution by its positive geometric area. An integral is signed net accumulation; taking absolute values changes the quantity being computed.
Why Interval Additivity Matters
Additivity allows an integral to be evaluated or interpreted piece by piece. This is especially useful for piecewise-defined functions, functions that change sign, and accumulated quantities whose rates change at a known time or location. The pieces need not contribute with the same sign, and the interval need not be traversed from left to right.
The rule does not say that each piece is positive, nor that splitting changes the integrand. It says that the oriented total equals the sum of the oriented contributions over consecutive subintervals. Keep the order of the endpoints visible, and signed contributions will combine consistently.
Check Your Understanding
Use additivity and orientation carefully in each question.
- If \(\int_0^3 f(x)\,dx=7\) and \(\int_0^1 f(x)\,dx=-2\), what is \(\int_1^3 f(x)\,dx\)?
- If \(\int_2^6 h(x)\,dx=-4\), what is \(\int_6^2 h(x)\,dx\)?
- For a split at \(c\) with \(a\leq c\leq b\), write the additivity formula for \(\int_a^b f(x)\,dx\).
- A bounded integrable function is changed at the single split point \(c\). Does this change either piece's integral? Explain briefly.
- Why can two nonzero signed integrals over adjacent intervals sum to zero?
