Introduction
A definite integral over an interval can often be found without computing a new antiderivative. If the integral values on suitable smaller or overlapping intervals are already known, the value on a target interval may follow by adding or subtracting those known values and accounting for the direction of the limits.
The key is to treat interval endpoints carefully. Adjacent intervals can be joined, a known larger interval can be split, and reversing the limits changes the sign. These simple rules form a reliable way to assemble definite integrals from previously available information.
Orientation and Additivity
For an integrable function \(f\), the definite integral \(\int_a^b f(x)\,dx\) records a signed accumulation from the starting endpoint \(a\) to the ending endpoint \(b\). The order of the endpoints matters: changing their order reverses the sign. If the endpoints agree, the interval has zero length and its integral is zero.
When \(a\), \(b\), and \(c\) are in increasing order, the integral over the full interval is the sum of the integrals over its two adjacent pieces. More generally, additivity holds for any three endpoints when the orientation convention is observed.
Why the Rules Work
For continuous \(f\), define \(F(x)=\int_a^x f(t)\,dt\). The Fundamental Theorem of Calculus gives \(F'(x)=f(x)\), and it gives the endpoint formula \(\int_u^v f(x)\,dx=F(v)-F(u)\) for endpoints in the interval. This formula immediately explains both orientation and additivity.
For three endpoints, the endpoint differences telescope:
The intermediate endpoint contributes once with a positive sign and once with a negative sign, so it cancels. The interval identities also hold for integrable functions more generally; continuity is sufficient for the short proof above. In practical evaluation, these identities let us rearrange known integral values without constructing \(F\).
Worked Example: Join Adjacent Intervals, Then Reverse
Worked Example: Join Adjacent Intervals, Then Reverse
Suppose \(f\) is integrable and the known values are \(\int_{-2}^{1} f(x)\,dx=6\) and \(\int_{1}^{3} f(x)\,dx=-4\). Find \(\int_{3}^{-2} f(x)\,dx\).
The known intervals are adjacent: the first ends at \(1\), where the second begins. Additivity gives the integral in the increasing direction:
The requested integral has its limits reversed. Therefore,
Check: Reversing the result gives \(\int_{-2}^{3} f(x)\,dx=-(-2)=2\), which agrees with the sum \(6+(-4)=2\).
Subtracting to Find a Missing Piece
Additivity can be rearranged. If a known interval is split at an interior point, the integral over either piece can be found by subtracting the value over the other piece from the value over the whole interval. The endpoints must line up: one interval must genuinely be the union of the other two adjacent intervals.
Worked Example: Recover an Integral on a Subinterval
Worked Example: Recover an Integral on a Subinterval
Suppose \(g\) is integrable and \(\int_0^4 g(x)\,dx=9\), while \(\int_1^4 g(x)\,dx=5\). Find \(\int_1^0 g(x)\,dx\).
First use additivity at the point \(1\):
Substitute the known values and solve for the missing piece:
The requested limits are reversed, so
Check: The recovered piece and the known piece reconstruct the whole: \(4+5=9\). Reversing the recovered piece gives \(\int_1^0 g(x)\,dx=-4\), as required.
A Reliable Strategy for Interval Data
When several values are provided, it helps to write down the target interval and identify which known intervals can be joined or subtracted to produce it. Think of each integral as carrying a direction: an interval from left to right is positive in orientation, and the same interval from right to left has the opposite sign.
List the points that divide the relevant intervals, including any endpoints from the known values.
Use additivity to join intervals that meet at an endpoint, or subtract a known piece from a known whole.
First compute the integral with increasing limits when convenient; reverse the sign if the requested limits run in the opposite direction.
Substitute the result into the interval identity that produced it and confirm that the known value is recovered.
Worked Example: Combine Overlapping Data and Reverse
Worked Example: Combine Overlapping Data and Reverse
Suppose \(h\) is integrable and the available values are \(\int_{-1}^{2} h(x)\,dx=7\), \(\int_0^2 h(x)\,dx=3\), and \(\int_0^4 h(x)\,dx=-2\). Find \(\int_4^2 h(x)\,dx\).
The first two values share the interval from \(0\) to \(2\). Subtracting that shared piece from the interval from \(-1\) to \(2\) gives:
For the target, split the known interval from \(0\) to \(4\) at \(2\):
Insert the given values and isolate the unknown integral:
The requested limits are reversed, so \(\int_4^2 h(x)\,dx=-(-5)=5\).
Check: The increasing-direction pieces give \(\int_0^4 h(x)\,dx=3+(-5)=-2\), matching the supplied value. Reversing the computed piece gives \(\int_4^2 h(x)\,dx=5\).
Common Mistake: Forgetting the Orientation Sign
A frequent error is to reverse the limits but keep the same value. For instance, if \(\int_2^5 q(x)\,dx=8\), it is incorrect to conclude that \(\int_5^2 q(x)\,dx=8\). The correct calculation is
This sign is essential even when the integrand is sometimes negative: orientation is a property of the limits, not a claim that the function itself is positive. A useful check is to reverse the limits a second time. The sign should reverse again, returning to the original value.
When This Method Is Useful
Known-integral assembly is especially useful when a problem supplies values on several intervals, asks for an integral with reversed limits, or gives a whole-interval value together with one subinterval value. It also provides a clear way to organize computations: the endpoint identities show exactly which information is being combined and why the result follows.
The method does not create missing information from nothing. If the known intervals do not combine or subtract to produce the target interval, additivity and orientation alone may not determine the answer. Nor should an integral over one interval be treated as if it automatically gives the value on a different interval. The endpoints and the function must match the stated data.
Check Your Understanding
Use interval additivity and orientation to answer these questions. Show the interval identity before doing the arithmetic.
- If \(\int_0^2 f(x)\,dx=5\) and \(\int_2^6 f(x)\,dx=-1\), find \(\int_6^0 f(x)\,dx\).
- If \(\int_{-2}^3 g(x)\,dx=8\) and \(\int_1^3 g(x)\,dx=6\), find \(\int_{-2}^1 g(x)\,dx\).
- If \(\int_1^5 h(x)\,dx=-3\), what is \(\int_5^1 h(x)\,dx\)?
- If \(\int_0^4 k(x)\,dx=2\) and \(\int_0^1 k(x)\,dx=7\), find \(\int_1^4 k(x)\,dx\).
- What identity would you use to check a value you found for an integral over a subinterval?
