Why Split an Integral?
A definite integral records accumulated signed change across an interval. Sometimes the whole interval is difficult to interpret at once, while its separate pieces are clear. A graph may cross the horizontal axis, or a rate table may give accumulated change over successive time periods. Additivity lets us calculate or interpret each piece and then combine the results.
The split point must lie between the endpoints. If \(a<c<b\), the interval from \(a\) to \(b\) consists of the adjacent intervals from \(a\) to \(c\) and from \(c\) to \(b\). The shared endpoint \(c\) is a boundary, not an extra interval of positive width.
The equality says that the accumulation over the whole interval is the sum of the accumulations over its two pieces. It does not say that each piece must be positive. When \(f\) is below the horizontal axis, its contribution to the integral is negative.
Why the Pieces Add
The definite integral can be defined as the limit of Riemann sums. A partition divides an interval into smaller subintervals; on each subinterval, a sample value of \(f\) is multiplied by that subinterval’s width, and the products are added. When the largest width tends to zero, these sums approach the integral, provided \(f\) is Riemann integrable.
For a partition that includes \(c\), every subinterval lies either in \([a,c]\) or in \([c,b]\). Thus its Riemann sum separates into the terms from the first interval and the terms from the second. This separation remains true as the widths shrink, so the limits separate in the same way.
Choose partitions of \([a,c]\) and \([c,b]\). Joining them gives a partition of \([a,b]\) that contains \(c\).
The Riemann sum on the joined partition is exactly the sum of the Riemann sum on \([a,c]\) and the Riemann sum on \([c,b]\).
As the widths shrink, the two sums approach their respective integrals. Their sum approaches the integral over \([a,b]\), giving the additivity equation.
More explicitly, let \(P_1\) be a partition of \([a,c]\) and \(P_2\) a partition of \([c,b]\). Use the same sample values as in the two separate sums. Their union is a partition of \([a,b]\), and the terms divide into two groups, so
Because \(f\) is Riemann integrable on \([a,b]\), its restrictions to the two subintervals are integrable as well. As the largest subinterval widths tend to zero, each sum on the right tends to its definite integral. The joined sum on the left tends to the integral on \([a,b]\). Taking these limits proves the stated property. The proof depends on the fact that the intervals meet at \(c\) and together cover \([a,b]\) without a gap or overlap of positive width.
Using Signed Areas
For a graph, a definite integral can be interpreted as signed area: regions above the horizontal axis contribute positively, and regions below contribute negatively. Additivity is especially useful when a convenient split point separates these regions or divides a graph into simple shapes.
Worked Example: Combine Areas on Adjacent Intervals
Suppose a graph has a triangular region above the axis on \([0,1]\), with base \(1\) and height \(4\). On \([1,2]\), it encloses a rectangle below the axis with width \(1\) and height \(2\). On \([2,3]\), it encloses a triangle below the axis with base \(1\) and height \(2\). Find \(\int_0^3 f(x)\,dx\), splitting at \(c=2\).
First calculate the signed areas on \([0,2]\). The region on \([0,1]\) is above the axis, so its contribution is
The rectangle on \([1,2]\) is below the axis, so its contribution is
Therefore, \(\int_0^2 f(x)\,dx=2+(-2)=0\). The triangle on \([2,3]\) is below the axis, and contributes
By additivity at \(c=2\),
Check: Adding all three signed regions directly gives \(2+(-2)+(-1)=-1\), the same result.
The split point does not have to be where the graph crosses the axis. It can be any interior point. A useful choice is one that makes the pieces easier to calculate or interpret.
Worked Example: Split a Linear Graph at an Interior Point
Let the graph of \(f\) be the line segment joining \((0,-2)\) and \((4,2)\). Find \(\int_0^4 f(x)\,dx\) by splitting at \(c=1\). Use signed geometric areas.
On \([0,1]\), the line rises from \(-2\) to \(-1\). The trapezoidal region is below the axis, with width \(1\) and vertical side lengths \(2\) and \(1\). Its signed area is
On \([1,2]\), the line rises from \(-1\) to \(0\). This is a triangle below the axis with base \(1\) and height \(1\), so its signed area is
On \([2,4]\), the line rises from \(0\) to \(2\). This triangle is above the axis, with base \(2\) and height \(2\), so its area is
The integral from \(1\) to \(4\) is the sum of the last two contributions, \(-\frac{1}{2}+2=\frac{3}{2}\). Additivity at \(c=1\) now gives
Check: Across the full interval, the negative triangle from \(0\) to \(2\) has area \(-\frac{1}{2}(2)(2)=-2\), and the positive triangle from \(2\) to \(4\) has area \(\frac{1}{2}(2)(2)=2\). Their sum is \(-2+2=0\), agreeing with the split calculation.
Combining Accumulated Change
The same property applies when \(f\) represents a rate. For instance, if \(r(t)\) is a rate of change, then \(\int_a^b r(t)\,dt\) is the net accumulated change over the time interval, with units equal to rate units multiplied by time units. Additivity says that the net change over consecutive periods is the sum of the net changes in those periods.
Worked Example: Combine Net Changes from a Rate Table
A rate \(r(t)\), measured in liters per minute, has an accumulated change of \(18\) liters from \(t=0\) to \(t=5\) minutes and an accumulated change of \(-7\) liters from \(t=5\) to \(t=8\) minutes. Find the accumulated change from \(t=0\) to \(t=8\).
The periods \([0,5]\) and \([5,8]\) are adjacent, meeting at \(t=5\). Apply additivity:
Substitute the given accumulated changes, keeping their signs:
Check: The first period contributes \(18\) liters and the second removes \(7\) liters, so the total net change is \(18-7=11\) liters.
Common Pitfall: Adding Unsigned Areas
A geometric region’s ordinary area is nonnegative, but a definite integral keeps track of whether the graph is above or below the axis. Forgetting the sign changes the result, even though the interval split itself is correct.
Worked Example: Correct an Unsigned-Area Calculation
Suppose an interval is divided at \(c\). The graph encloses area \(6\) above the axis on \([a,c]\) and area \(4\) below the axis on \([c,b]\). A common incorrect calculation is
This treats the region below the axis as a positive contribution. Its integral is instead \(-4\), so the corrected calculation is
Check: The signed total is positive because the area above the axis exceeds the area below by \(6-4=2\).
Another useful check is to confirm that the intervals really are adjacent: the right endpoint of the first must equal the left endpoint of the second. If there is a gap, an integral over that gap is missing. If the intervals overlap over a positive-width region, that region is counted more than once.
Choosing a Split Point
Any interior point \(c\) gives a valid split when \(f\) is integrable on the interval. In practice, choose a point that simplifies the interpretation. It might be where the graph changes sign, where a piecewise rule changes, or where available rate data mark the end of one time period. For several consecutive intervals, apply additivity repeatedly; each shared endpoint joins one piece to the next.
Check Your Understanding
Use additivity and keep track of signed contributions in each question.
- If \(\int_1^4 f(x)\,dx=7\) and \(\int_4^6 f(x)\,dx=-3\), what is \(\int_1^6 f(x)\,dx\)?
- A graph has area \(5\) above the axis on \([0,2]\) and area \(8\) below the axis on \([2,5]\). What is its integral from \(0\) to \(5\)?
- In the Riemann-sum argument, why must the joined partition include the split point?
- A rate has net accumulated changes of \(-12\) meters on one time interval and \(9\) meters on the next. What is the combined net change?
- What error occurs if two intervals overlap over a region of positive width and their integrals are added as if they were adjacent?
