Why Adjacent Intervals Can Cancel
A definite integral records a signed total, not merely the total geometric area between a graph and the horizontal axis. Contributions above the axis count positively, and contributions below it count negatively. Consequently, a function can have nonzero contributions on two adjoining intervals while its integral over their union is zero.
The key tools are additivity and orientation. Additivity lets us split an interval at a shared endpoint. Orientation tells us that reversing the direction of integration changes the sign. Together, these facts make it possible to identify cancellation without assuming that the function has one sign or that its graph has a particular shape.
Additivity and Orientation
For a continuous function \(f\) on an interval containing \(a\), \(b\), and \(c\), with \(a\leq b\leq c\), additivity gives
Thus, the integral over two adjacent intervals is the sum of the two individual integrals. The shared endpoint \(b\) does not itself contribute a separate term.
The orientation rule is
It follows that traversing the same interval in opposite directions cancels exactly:
For adjacent intervals, however, the intervals are not the same interval traversed in opposite directions. Cancellation occurs when their integral values are equal and opposite.
The function need not be zero anywhere, and its values on one interval need not be pointwise opposite to its values on the other. What must cancel is the signed total.
A General Cancellation Test
Let \(A=\int_a^b f(x)\,dx\) and \(B=\int_b^c f(x)\,dx\). Additivity says that the integral across the combined interval is \(A+B\). Therefore, the complete test is to determine whether \(A+B=0\). In practice, a useful route is to compute each contribution separately, or to relate one interval to the other by a change of variable.
Write the integral over the combined interval as the sum of the adjacent integrals.
Evaluate each integral as a signed quantity, retaining contributions from regions where the function is negative.
If the two values add to zero, the integral over the combined interval is zero.
A reflection relation can sometimes provide the comparison directly. For instance, if the adjacent intervals have equal length \(L\), and \(f(b+t)=-f(b-t)\) for \(0\leq t\leq L\), the contributions cancel. This relation pairs points equally far from the shared endpoint. The change of variable \(x=b+t\) on the right interval and \(x=b-t\) on the left interval shows why the paired contributions have opposite signs.
Why Reflection Gives Cancellation
Here is a complete justification under the stated conditions. Assume \(f\) is continuous on \([b-L,b+L]\) and satisfies \(f(b+t)=-f(b-t)\) for \(0\leq t\leq L\). Substitute \(x=b+t\) in the right-hand integral. Then substitute \(x=b-t\) in the left-hand integral. The second substitution reverses the bounds, which contributes a minus sign:
Adding these equations and using the assumed relation gives
The cancellation is between the two interval totals. The proof also shows a sufficient condition for cancellation, not the only possible one: two integrals can be equal and opposite even if this point-by-point reflection relation does not hold.
Worked Example: Cancellation Across a Shared Reflection Point
Let \(f(x)=x-2\) on \([0,4]\). Split the interval at \(2\), so the adjacent intervals are \([0,2]\) and \([2,4]\).
On the first interval, an antiderivative is \(F(x)=\frac{x^2}{2}-2x\). Therefore,
On the second interval,
Additivity now gives
Check: using the same antiderivative directly gives \(F(4)-F(0)=(8-8)-0=0\), agreeing with the split calculation. The integrand is negative to the left of \(2\) and positive to the right, so its signed contributions cancel.
Cancellation Does Not Require a Constant Sign
On an interval where the function changes sign, its integral is the net result of positive and negative contributions within that interval. It is still valid to compare this net result with the integral over an adjacent interval. The relevant quantities are the two signed integrals, not the unsigned areas and not the values at just the endpoints.
Worked Example: Both Adjacent Contributions Change Sign
Define \(f\) on \([0,4]\) by
The value assigned at the single shared point does not affect either integral. On \([0,2]\), \(3x-2\) changes sign at \(x=\frac{2}{3}\). Its integral is
On \([2,4]\), \(8-3x\) changes sign at \(x=\frac{8}{3}\). Its integral is
Thus,
Check: applying the piecewise antiderivative calculation interval by interval gives the same total, zero. Each integral includes both positive and negative contributions; it is their net values, \(2\) and \(-2\), that cancel.
Equal Totals Do Not Require Pointwise Opposites
A reflection relation is convenient when it is present, but cancellation can arise for a more general reason: two different function shapes can have equal and opposite signed integrals. An integral compresses all the values on an interval into one number. Matching those numbers does not require matching the function values point by point.
Worked Example: Opposite Integrals Without Pointwise Matching
Define \(f\) on \([0,2]\) by
The first contribution is
For the second contribution, set \(u=x-1\), so that \(du=dx\), and the bounds \(x=1,2\) become \(u=0,1\). Then
Therefore,
Check: the two displayed integrals are equal and opposite. But the first interval has a quadratic function and the second a linear one, so the cancellation does not depend on pointwise opposite function values under reflection.
Common Mistake: Assuming the Shared Endpoint Cancels
A common error is to treat the shared endpoint as if it were a term that appears with opposite signs in the two integrals. For example, the following calculation is not a valid consequence of additivity:
The shared endpoint sets the boundary between the intervals; it does not determine either integral's value. For the concrete function \(f(x)=1\), both adjacent integrals are \(2\), so
not zero. The correction is to calculate or otherwise compare the two signed contributions. They cancel only if one is the negative of the other. Reversing the bounds on one interval would change its sign, but that would describe integration in the opposite direction, not cancellation caused merely by adjacency.
Why This View Is Useful
When evaluating a definite integral, a split can reveal structure that is hidden in the unsplit expression. It can make it clear that a negative contribution offsets a positive one, or that two adjacent intervals contribute equal and opposite totals even when the function changes sign within each. Conversely, additivity prevents an unjustified conclusion: adjacency alone never guarantees a zero total.
For any split point \(b\), keep the exact relationship in view:
Then ask whether the two terms on the right actually sum to zero. If they do, cancellation is established. If they do not, the combined integral is their sum, regardless of where the intervals meet.
Check Your Understanding
Use additivity and signed integral values to answer these questions.
- If \(\int_a^b f(x)\,dx=5\) and \(\int_b^c f(x)\,dx=-5\), what is \(\int_a^c f(x)\,dx\)?
- Does a shared endpoint, by itself, imply that integrals over adjacent intervals cancel? Explain.
- Can two adjacent interval integrals cancel if the function changes sign on both intervals? What quantities should be compared?
- If \(f(b+t)=-f(b-t)\) for \(0\leq t\leq L\), what is the integral over \([b-L,b+L]\), assuming \(f\) is continuous there?
- What sign change occurs when the bounds of a definite integral are reversed?
