Why Split an Integral?
A definite integral can describe a total accumulated quantity over an interval. Sometimes the full interval is easier to understand as two adjacent pieces. A useful point might be where a formula changes, where data are reported, or simply where an antiderivative gives convenient values.
The key fact is that the integral over the whole interval is the sum of the integrals over its adjacent parts. This is not a claim that the function must be positive, or that each part must represent a geometric area. It is a property of the definite integral that follows from how the integral is evaluated.
For AP Calculus problems, continuity of \(f\) on \([a,c]\) is a common sufficient condition for the integrals in this identity to exist. The point \(b\) is the shared endpoint: the two smaller intervals meet there, with no gap between them.
Deriving Additivity from Endpoint Values
For a continuous function \(f\) on \([a,c]\), the Fundamental Theorem of Calculus gives a way to evaluate each definite integral. If \(F\) is an antiderivative of \(f\) on that interval, then the integral from one endpoint to another is the difference of the values of \(F\) at those endpoints.
Apply this result first to \([a,b]\), then to \([b,c]\). The value \(F(b)\) appears once with a positive sign and once with a negative sign, so it cancels. The result is precisely the endpoint difference for the whole interval.
By the Fundamental Theorem of Calculus, \(\int_a^b f(x)\,dx=F(b)-F(a)\).
On the adjacent interval, \(\int_b^c f(x)\,dx=F(c)-F(b)\).
The terms \(F(b)\) cancel, leaving \(F(c)-F(a)\).
The Fundamental Theorem identifies \(F(c)-F(a)\) as \(\int_a^c f(x)\,dx\).
Written as a proof, the calculation is:
The cancellation explains why the intervals must be adjacent in this form: the end of the first piece must be the start of the second. The equality does not depend on which antiderivative is chosen. If a constant is added to \(F\), it appears in endpoint differences with opposite signs and cancels as well.
Choosing a Splitting Point
The identity allows a definite integral to be divided wherever the interval can be divided at a point \(b\) between the endpoints. A good choice of \(b\) can make the arithmetic simpler or align the calculation with information given in a problem. The split does not change the value of the full integral; it only expresses that value as a sum.
A split can also be used in reverse. If the integral over the full interval and the integral over one subinterval are known, subtraction can recover the integral over the remaining adjacent interval. In either direction, first confirm that the pieces fit together and cover the interval exactly.
Worked Example: Split a Polynomial Integral at an Interior Point
Evaluate \(\int_0^3 (x^2-2x)\,dx\) by splitting at \(x=1\), then check the result using the full interval.
An antiderivative is \(F(x)=\frac{x^3}{3}-x^2\), because \(F'(x)=x^2-2x\).
First evaluate the part from \(0\) to \(1\):
Next evaluate the part from \(1\) to \(3\):
Add the two adjacent-interval integrals:
For the check, evaluate directly over the full interval:
The split sum and the full-interval calculation both give \(0\), as required.
Using a Split to Find an Unknown Integral
Additivity is useful even when a function or its antiderivative is not given explicitly. If the integral over the whole interval and some subinterval integrals are supplied, the identity relates those values. The unknown integral can then be found by ordinary algebra.
Worked Example: Recover an Integral from Two Adjacent Parts
Suppose \(f\) is integrable on \([0,4]\), and \(\int_0^4 f(x)\,dx=7\), \(\int_0^1 f(x)\,dx=-2\), and \(\int_1^3 f(x)\,dx=5\). Find \(\int_3^4 f(x)\,dx\).
The intervals \([0,1]\), \([1,3]\), and \([3,4]\) meet end to end and cover \([0,4]\). Applying additivity twice gives
Let \(I=\int_3^4 f(x)\,dx\). Substitute the given values:
Since \((-2)+5=3\), it follows that \(7=3+I\), so \(I=4\).
Check by adding the three pieces: \(-2+5+4=7\), which agrees with the given integral over \([0,4]\). Therefore, \(\int_3^4 f(x)\,dx=4\).
The subinterval values in this example are signed integral values. Additivity applies to those values as given; it does not require replacing a negative value with its absolute value.
Splitting Accumulated Quantities
An integral often represents an accumulated quantity over time or over another measured variable. If the integrand is a rate, the integral over an interval records the accumulation across that interval. When a time interval is split at an intermediate time, the accumulation over the full time period is the sum of the accumulations on the two adjacent periods.
Units provide a useful check. For example, if \(r(t)\) is measured in liters per minute and \(t\) in minutes, then \(\int_a^b r(t)\,dt\) is measured in liters. Each subinterval integral has the same units, so their sum can be compared with the full-interval integral.
Worked Example: Add Accumulation on Two Time Intervals
A rate \(r(t)\) is measured in liters per minute. Suppose the accumulated amount is \(12\) liters from minute \(0\) to minute \(2\), and \(9\) liters from minute \(2\) to minute \(5\). Find the accumulation from minute \(0\) to minute \(5\).
The intervals \([0,2]\) and \([2,5]\) are adjacent and together make \([0,5]\). Thus,
The units agree because liters are being added to liters. As a check, the two stated subinterval accumulations sum to \(21\) liters, the value assigned to the entire interval by additivity.
Common Mistakes and Checks
A common error is to calculate the pieces correctly but fail to include every part of the original interval. For example, the expression \(\int_0^1 f(x)\,dx+\int_2^3 f(x)\,dx\) does not, in general, equal \(\int_0^3 f(x)\,dx\). The interval from \(1\) to \(2\) is missing. The correct decomposition is
Another frequent mistake is to lose a sign while evaluating an endpoint difference. For the second subinterval, the order is \(F(c)-F(b)\), not \(F(b)-F(c)\). A quick check is to write the endpoint subtraction before substituting numbers, then verify that the intermediate values at shared endpoints cancel when the pieces are added.
Worked Example: Correct an Endpoint-Subtraction Error
Consider \(\int_0^2 x\,dx\), split at \(1\). An incorrect calculation might write
The endpoint order is reversed. The correct calculation is
The first part is \(\int_0^1 x\,dx=F(1)-F(0)=\frac{1}{2}-0=\frac{1}{2}\). Adding the corrected pieces gives \(\frac{1}{2}+\frac{3}{2}=2\). Direct evaluation confirms this: \(\int_0^2 x\,dx=F(2)-F(0)=2-0=2\).
Why the Identity Matters
Additivity makes definite integrals flexible. It lets you divide an interval to match the information in a problem, combine known portions into a total, or isolate an unknown portion. The underlying reason remains simple: when endpoint differences are added, each shared endpoint cancels.
Before using the identity, check three things: the subintervals are adjacent, their union is the entire interval of interest, and the integral exists on those intervals. Then keep the order of endpoint subtraction consistent. These checks make the split reliable whether the integral is evaluated from an antiderivative or supplied as accumulated data.
Check Your Understanding
Use additivity and the endpoint-difference rule where appropriate.
- Write the additivity identity for an interval \([p,r]\) split at \(q\), where \(p\le q\le r\).
- If \(\int_0^5 g(x)\,dx=11\) and \(\int_0^2 g(x)\,dx=4\), find \(\int_2^5 g(x)\,dx\).
- For \(f(x)=2x\), split \(\int_0^3 f(x)\,dx\) at \(x=1\), evaluate both parts, and verify their sum against the full integral.
- Why does the shared endpoint value cancel when the two subinterval integrals are evaluated using an antiderivative?
- Does \(\int_0^3 f(x)\,dx=\int_0^1 f(x)\,dx+\int_2^3 f(x)\,dx\) follow from additivity alone? Explain what interval is unaccounted for.
