Why Adjacent Intervals Can Be Combined
A rate can change over time: it may follow one formula in the morning and another in the afternoon, or it may become positive during one interval and negative during another. Even when the rate changes, the net change over the full period is the sum of the net changes on the pieces.
This works because the definite integral is additive across adjacent intervals. The pieces must meet at their shared endpoint, and the integral must exist on the full interval. The rate does not have to be positive, or even use the same formula throughout.
For an accumulation problem, \(f(x)\) is often a rate and \(x\) represents time. The integral over an interval gives the net change during that interval. Additivity lets us combine those changes, even if the rate has different formulas on different pieces.
The Additivity Rule
Let \(f\) be integrable on \([a,b]\), and choose a point \(c\) between \(a\) and \(b\). Then the interval \([a,b]\) is exactly the union of the adjacent intervals \([a,c]\) and \([c,b]\). The additivity rule is
The same rule can be applied repeatedly. If \(a=x_0<x_1<\cdots<x_n=b\), then
In AP applications, rates are commonly continuous on each piece and may have jumps at finitely many boundaries. Such piecewise continuous functions are integrable on a closed, bounded interval, so the rule applies. A jump at a shared endpoint does not prevent additivity.
Why the Rule Works
The definite integral is the limit of sums of function values times small interval widths. Divide \([a,c]\) into small pieces and divide \([c,b]\) into small pieces. Together, those pieces form a partition of \([a,b]\). The sum over the whole interval is therefore the first subinterval sum plus the second subinterval sum.
Choose partitions of \([a,c]\) and \([c,b]\). Their combined partition covers \([a,b]\) without a gap or overlap of positive length.
The sum for the combined partition is exactly the sum on \([a,c]\) plus the sum on \([c,b]\), because the terms come from those two pieces.
Since \(f\) is integrable on \([a,b]\), the integral limits exist. Taking the limits of the sums gives \(\int_a^b f=\int_a^c f+\int_c^b f\).
This also explains why the rule does not depend on whether \(f\) is positive. The sums include signed function values. Terms from a region where \(f\) is negative are negative, and they remain so when the sums are combined.
Using Additivity with a Formula
Worked Example: Split an Integral at a Chosen Point
Find \(\int_0^4 (3t-2)\,dt\) by splitting at \(t=2\), and verify the result by evaluating the integral over the whole interval.
First calculate the contribution from \([0,2]\). An antiderivative of \(3t-2\) is \(\frac{3}{2}t^2-2t\), so
Now calculate the contribution from \([2,4]\):
Add the two adjacent contributions:
For the check, evaluate directly over \([0,4]\):
The split and direct calculations agree. Notice that the first subinterval’s net change is \(2\), not necessarily a description of every value the rate takes there; it is the signed integral over that whole piece.
Piecewise Rates and Net Change
When a rate is defined by different formulas on adjacent intervals, calculate each integral using the formula valid on that interval. Then add the resulting signed changes. The value at a single boundary does not affect a definite integral, so a rate may jump there without changing this procedure.
If \(R(t)\) is a rate of change of a quantity \(Q(t)\), the net change in \(Q\) from \(t=a\) to \(t=b\) is \(\int_a^b R(t)\,dt\), provided the rate is integrable. If \(Q(a)\) is known, then \(Q(b)=Q(a)+\int_a^b R(t)\,dt\). Additivity means the accumulated change can be found piece by piece.
Worked Example: Combine Rates with Different Signs
A tank initially contains \(20\) liters of water. During the first \(3\) minutes, water enters at \(4\) liters per minute. During the next \(2\) minutes, water leaves at \(2\) liters per minute. During the final \(2\) minutes, water enters at \(3\) liters per minute. Find the net change over all \(7\) minutes and the final amount.
Let \(R(t)\) denote the rate in liters per minute. Use \(t\) in minutes. The contributions on the three adjacent intervals are
The negative sign in the second contribution records that water leaves the tank. By additivity, the net change over the full interval is
Add this net change to the initial amount:
Check by tracking the amount at each boundary: after \(3\) minutes it is \(20+12=32\) liters; after the next \(2\) minutes it is \(32-4=28\) liters; after the final \(2\) minutes it is \(28+6=34\) liters. This agrees with the integral calculation.
Piecewise Formulas Can Be Integrated Separately
Suppose a function is defined by one formula on \([a,c]\) and another on \([c,b]\). The integral over the full interval is the sum of the integral of the first formula over \([a,c]\) and the integral of the second formula over \([c,b]\). Use the correct formula on each interval; do not extend one piece’s formula into another piece unless the definition permits it.
Worked Example: Integrate a Piecewise Rate
A quantity changes at the rate
Find the net change from \(t=0\) to \(t=4\). First integrate the constant rate on \([0,1]\):
On \([1,3]\), use \(2-t\), whose antiderivative is \(2t-\frac{t^2}{2}\):
On \([3,4]\), the rate is \(-1\):
Combining the adjacent intervals gives
For a direct check, an antiderivative of the piecewise rate can be chosen separately on each interval, with constants selected to keep accumulated change continuous. Starting at \(A(0)=0\), the accumulated change is \(A(1)=2\). From \(1\) to \(3\) the additional change is \(0\), so \(A(3)=2\). From \(3\) to \(4\) the additional change is \(-1\), so \(A(4)=1\). Thus the full net change is \(A(4)-A(0)=1-0=1\), as found by additivity.
Signs Must Be Preserved
A definite integral records net change, so its contributions are signed. If the rate is positive on one interval, its integral contributes positively. If the rate is negative on another, its integral contributes negatively. Adding the signed values accounts for increases and decreases in the same calculation.
A common error is to turn every subinterval result into a positive number before adding. For example, suppose a rate contributes \(8\) units on the first interval and \(-5\) units on the next. The net change is \(8+(-5)=3\) units, not \(8+5=13\) units. The incorrect calculation treats the decrease as though it were another increase. Keep the sign attached to each integral until the final sum is complete.
Another error is to split at a point but use the wrong interval endpoints. For a split at \(c\), the pieces must be \([a,c]\) and \([c,b]\): together they cover the original interval exactly. Leaving out a piece omits some accumulated change; overlapping intervals can count some change twice.
Check Your Understanding
Use additivity and preserve the sign of each contribution in your answers.
- If \(f\) is integrable on \([0,6]\), write \(\int_0^6 f(x)\,dx\) as a sum of integrals split at \(x=2\) and \(x=5\).
- Evaluate \(\int_0^5 3\,dt+\int_5^7(-2)\,dt\). What is the combined integral over the adjacent intervals?
- A rate contributes \(-4\) units on \([1,3]\) and \(9\) units on \([3,6]\). Find the net change from \(1\) to \(6\).
- For a piecewise continuous rate with a formula change at \(t=4\), explain why the integral over \([0,7]\) can be split at \(4\).
- A quantity starts at \(12\) units and changes by \(-3\) units on the first interval and \(5\) units on the next. Find its final amount.
