Why Split a Definite Integral?
A definite integral over one interval can be evaluated by breaking that interval into smaller pieces. A split can make a calculation easier to organize, highlight what happens on different parts of the interval, or let us examine positive and negative contributions separately. The crucial point is that the pieces fit together: when their endpoint differences are added, the contributions at internal split points cancel.
Suppose, for example, that an interval runs from \(a\) to \(b\), and we choose a point \(c\) between them. We can evaluate the integral from \(a\) to \(c\), then the integral from \(c\) to \(b\), and add the results. The Fundamental Theorem of Calculus gives each piece as a difference of antiderivative values. The value at \(c\) appears once with a positive sign and once with a negative sign.
Partitions and the Integral on Each Piece
A partition of \([a,b]\) is a finite ordered list of points beginning at \(a\) and ending at \(b\). Write it as
The partition divides the original interval into the subintervals \([x_0,x_1]\), \([x_1,x_2]\), and so on, up to \([x_{n-1},x_n]\). If \(f\) is continuous on \([a,b]\), then the Fundamental Theorem of Calculus applies on each of these closed subintervals. In particular, if \(F\) is an antiderivative of \(f\) on \([a,b]\), then \(F'(x)=f(x)\), and on every subinterval \([x_{i-1},x_i]\) we have
Notice the endpoint order: the value at the right endpoint is written first, and the value at the left endpoint is subtracted. The same rule applies separately to every piece.
Why the Intermediate Values Cancel
Apply the Fundamental Theorem to each subinterval and add the results. Written out, the endpoint differences are
Each internal value has two appearances. For instance, \(F(x_1)\) is added in the first difference and subtracted in the second. The same pattern holds for every split point. After cancellation, only the value at the final endpoint and the negative of the value at the initial endpoint remain.
This proves the stated evaluation formula: each subinterval integral is the corresponding endpoint difference by the Fundamental Theorem, and those differences telescope when added. Splitting does not change the value of the integral. It only expresses the same total as a sum of integrals over adjoining pieces.
For two subintervals, the cancellation can be seen in its simplest form:
Worked Examples
Worked Example: Splitting a Polynomial Integral at One Point
Evaluate \(\int_0^2 (3x^2+2x)\,dx\) by splitting the interval at \(x=1\).
An antiderivative is \(F(x)=x^3+x^2\), since \(F'(x)=3x^2+2x\). The partition is \(0<1<2\), so evaluate the two pieces:
Add the two results: \(2+10=12\). At the split, \(F(1)=2\) is added in the first endpoint difference and subtracted in the second. As a check, evaluating over the original interval gives
which agrees with the sum of the pieces. The proposed antiderivative also checks because \(\frac{d}{dx}(x^3+x^2)=3x^2+2x\).
Worked Example: Splitting a Trigonometric Integral
Evaluate \(\int_0^\pi \sin x\,dx\) by splitting at \(x=\frac{\pi}{3}\).
Use \(F(x)=-\cos x\), because \(F'(x)=\sin x\). The first piece is
The second piece is
The sum is \(\frac{1}{2}+\frac{3}{2}=2\). The split value \(F(\frac{\pi}{3})=-\frac{1}{2}\) appears positively in the first difference and negatively in the second. Check the combined result directly: \(F(\pi)-F(0)=1-(-1)=2\). Also, differentiating \(F(x)=-\cos x\) gives \(\sin x\), as required.
Worked Example: Using Several Split Points
Evaluate \(\int_{-2}^{2}(x^2-1)\,dx\) by splitting at \(x=-1\) and \(x=1\).
An antiderivative is \(F(x)=\frac{x^3}{3}-x\), since \(F'(x)=x^2-1\). The partition is \(-2<-1<1<2\). First calculate the endpoint values:
Now evaluate each piece, always subtracting the value at the left endpoint from the value at the right endpoint:
Adding the three values gives \(\frac{4}{3}-\frac{4}{3}+\frac{4}{3}=\frac{4}{3}\). The middle piece is negative, and the outer pieces are positive; the definite integral records their signed total. For a final check, the single endpoint difference is
This agrees with the sum of the pieces, and differentiating \(F(x)=\frac{x^3}{3}-x\) returns \(x^2-1\).
Choosing Useful Split Points
The partition points can be chosen to suit the calculation, provided they are in order between the endpoints. There is no requirement that the subintervals have equal length. A split can be useful when you want to report contributions separately, inspect where the integrand changes sign, or organize a long calculation into manageable pieces.
For example, in the third worked example, the points \(-1\) and \(1\) are zeros of \(x^2-1\). Splitting there makes it easy to see the contribution from each region. The sum still gives the signed integral over the whole interval; splitting does not turn a signed integral into a sum of absolute areas.
More generally, if a partition has \(n\) subintervals, apply the endpoint-difference formula to each one and add. Internal values cancel in pairs, regardless of how many split points were chosen. It is not necessary to expand every difference fully when the cancellation is already clear, but writing the terms out is a helpful way to check signs.
A Common Endpoint Error
A frequent mistake is to reverse the endpoints in one subinterval while keeping the usual order in the others. In the first worked example, the correct contributions are \(F(1)-F(0)=2-0\) and \(F(2)-F(1)=12-2\). The second interval starts at \(1\) and ends at \(2\), so its contribution is the value at \(2\) minus the value at \(1\).
This error also prevents the intermediate value from cancelling: the first piece contains \(+F(1)\), and the correctly ordered second piece contains \(-F(1)\). A useful check is to line up the endpoint differences in order. Every internal endpoint should occur once with each sign.
Summary
To evaluate an integral across several intervals, choose an ordered partition, apply the Fundamental Theorem of Calculus on each subinterval, and add the endpoint differences. The method requires the theorem’s hypotheses on the intervals where it is applied; in particular, continuity of \(f\) on the original closed interval ensures the usual version applies to every piece.
The intermediate terms cancel because each split point is the right endpoint of one piece and the left endpoint of the next. Keep the endpoint order consistent—right endpoint minus left endpoint—and the sum of the pieces will agree with the integral over the full interval.
Check Your Understanding
Use the endpoint-difference rule and explain how the split values behave when you combine the pieces.
- For a partition \(a<c<b\), write \(\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\) in terms of an antiderivative \(F\), then simplify.
- Evaluate \(\int_0^3 2x\,dx\) by splitting at \(x=1\). Show both subinterval contributions and their sum.
- For the partition \(-1<0<2\), which signs should the value \(F(0)\) have in the two adjacent endpoint differences?
- Explain why a split point can be chosen where the integrand changes sign without changing the value of the integral over the full interval.
- What is wrong with using \(F(c)-F(a)\) for the interval \([c,b]\)? State the correct endpoint order.
