AP Syllabus focus:
‘Apply the property that the integral over [a, c] equals the sum of integrals over [a, b] and [b, c], and understand the effect of reversing limits of integration.’
Understanding how definite integrals behave across adjacent intervals allows you to break complex integration domains into manageable parts while preserving the accumulated quantity’s meaning and structure.
Integrals Over Adjacent Intervals
The idea of integrating over adjacent intervals is central to AP Calculus AB because it reinforces how definite integrals measure accumulated change consistently across connected segments of an interval. When a function is integrable on an interval, splitting that interval at any intermediate point produces integrals that add together cleanly.

A compact summary of definite integral rules, including additivity over adjacent intervals and the effect of reversing limits. The diagram also contains additional rules such as dominance and comparison, which extend beyond the current topic but remain consistent with its conceptual framework. Students may focus on the expressions and . Source.
The Additive Property of Definite Integrals
A key relationship for adjacent intervals is that a definite integral over a broad interval can be decomposed into a sum of integrals over smaller intervals. This reinforces the idea that accumulation over time or space is naturally segmented while retaining consistency across the entire domain.
Adjacent Interval Property: For any integrable function on , choosing a point between and allows the integral over to be expressed as the sum of the integrals over and .
This property reflects the linear accumulation inherent in definite integrals. Because integrals measure net signed area, splitting the domain does not alter the underlying meaning; it simply reorganizes the computation.

A graph of showing adjacent subintervals with positive and negative shaded areas, illustrating how definite integrals combine signed contributions. The figure emphasizes the net nature of accumulation across segments. Some labels identifying areas as , , and extend beyond this topic but remain consistent with its concepts. Source.
This makes the property particularly useful in conceptual reasoning, analytical proofs, and practical applications such as constructing piecewise rate models.
Formal Expression of the Adjacent Interval Property
The additive relationship is typically written using standard integral notation. Students must be fluent with this structure, as it appears both in conceptual questions and in computational settings.
= Numbers defining adjacent interval endpoints (units depend on context)
= Rate or function being integrated (units vary; often units of quantity per unit of )
This equation highlights a core structural feature: the limits must be in increasing order from left to right. The meaning of each sub-interval integral remains tied to the original function, making the overall accumulated change the sum of its parts.
A meaningful consequence of this structure is that students can decompose integrals surrounding discontinuities, changing behaviors, or piecewise definitions. Doing so ensures that complex functions are understood locally while still contributing to a global interpretation.
Reversing the Limits of Integration
The syllabus emphasizes that understanding the effect of reversing limits is essential. When the direction of integration reverses, the sign of the definite integral changes. This reflects the orientation of accumulation, meaning moving backward along the axis produces the opposite signed area.

A diagram of showing positive and negative contributions to a definite integral, clarifying how reversing orientation changes the sign of . The image includes adjacent regions of opposite sign, reinforcing the interpretation of the integral as net accumulation. Some surrounding formulas on the page extend beyond this topic but align conceptually. Source.
The Combined Effect of Adjacent Intervals and Reversed Limits
Together, the adjacent interval property and the concept of reversed limits create a coherent framework for working with definite integrals in flexible ways. Students should be comfortable integrating across multiple connected intervals, reassembling them in any order, and interpreting the associated signs.
Important ideas include:
Breaking complex integrals into simpler, adjoining parts without changing overall accumulated value.
Reconstructing integrals by combining smaller integrals in an order consistent with the orientation of the axis.
Avoiding errors by checking that limits progress from lower to higher values unless intentionally using reversed limits.
Maintaining correct signs when integrals are recombined after manipulation.
These principles reveal that definite integrals encode direction-sensitive accumulation, not just total area. When used together, they provide a robust set of tools for interpreting and manipulating integrals in analytic and applied contexts.
Practical Structure for Working With Adjacent Intervals
To apply these ideas effectively, students should follow an organized thought process:
Identify the interval being integrated and determine whether it can or should be split into adjacent segments.
Choose interior points deliberately, often matching points where the function changes behavior or where data is provided.
Check orientation to ensure that all integral bounds move in the correct direction.
Apply the additive property to sum integrals over the chosen segments.
Confirm sign consistency when reversing limits or rearranging expressions.
These steps reinforce conceptual clarity and help prevent common errors involving interval direction, sign interpretation, and domain partitioning. Integrals over adjacent intervals are a fundamental part of the AP Calculus AB framework, supporting deeper understanding of accumulation and integral structure across diverse contexts.
FAQ
When a function changes definition across different segments of an interval, splitting the integral at each transition point ensures that the correct expression is used on each part.
This avoids mixing formulas and ensures each segment is integrated under the correct rule.
It also allows you to identify where the signed area may switch behaviour, particularly when the graph crosses the axis or changes shape.
The property only works when the chosen point lies between the original bounds, because the definition relies on forming subintervals that exactly reconstruct the full domain.
If the point lies outside the interval, the resulting integrals no longer represent a partition of the original region and therefore cannot recombine in a meaningful or consistent way.
Yes, provided the function is integrable on the overall interval. A jump discontinuity does not break integrability as long as there are only finitely many such points.
In this case, splitting the integral at the discontinuity is often helpful because it isolates the change in behaviour and allows each side to be evaluated separately.
Signed area accumulates continuously along the horizontal axis, so dividing the domain simply separates the accumulation into smaller segments.
Because the sign is based on whether the function is above or below the axis, each interval contributes positively or negatively to the total. The sum of the pieces still represents the full net area.
Reversing limits flips the orientation of travel along the axis. Since the definite integral depends on direction, this introduces a negative sign.
When combining adjacent integrals, keeping track of orientation prevents sign errors, ensures consistent direction of accumulation, and preserves the correct total when recombining intervals.
Practice Questions
Question 1 (1–3 marks)
A function f is continuous on the interval [2, 10]. It is known that the integral from 2 to 6 of f(x) dx is 7, and the integral from 6 to 10 of f(x) dx is −3.
Use the property of integrals over adjacent intervals to find the value of the integral from 2 to 10 of f(x) dx.
Question 1
• Correctly adds the two integrals to obtain the integral from 2 to 10: 7 + (−3) = 4. (1 mark)
Total: 1 mark
Question 2 (4–6 marks)
A function g is continuous on the interval [−4, 5]. The graph of g is not given, but the following information is known:
• The integral from −4 to −1 of g(x) dx is 5.
• The integral from −1 to 3 of g(x) dx is −2.
• The integral from 3 to 5 of g(x) dx is k, where k is a constant.
(a) Use the property of integrals over adjacent intervals to express the integral from −4 to 5 of g(x) dx in terms of k.
(b) Using the fact that reversing the limits of integration changes the sign, write an expression for the integral from 5 to −4 of g(x) dx.
(c) Hence determine the value of the integral from 5 to −4 of g(x) dx in terms of k.
Question 2
(a)
• Uses additivity to state that the integral from −4 to 5 of g(x) dx equals 5 + (−2) + k. (1 mark)
• Simplifies to 3 + k. (1 mark)
(b)
• States that reversing limits introduces a negative sign: integral from 5 to −4 of g(x) dx = − integral from −4 to 5 of g(x) dx. (1 mark)
(c)
• Substitutes the expression from part (a). (1 mark)
• Writes final expression: integral from 5 to −4 of g(x) dx = −(3 + k). (1 mark)
Total: 5 marks
