Why Split at the Midpoint?
A definite integral over an interval can sometimes become easier to understand when the interval is divided at its midpoint. The two resulting halves are mirror images: each point in one half has a reflected point in the other. Pairing the integrand values at these points can reveal a simpler expression, even when the original integrand is not itself symmetric.
The key is to compare values at matching reflected points, not merely to compute the two halves independently. This method can turn two complicated-looking contributions into one simpler integral over half the original interval.
The Midpoint and the Reflection
Let \(a<b\), and let \(c\) be the midpoint of \([a,b]\). Thus \(c=(a+b)/2\), so \(b=2c-a\). Reflection across \(c\) sends a point \(u\) to \(2c-u\). In particular, it sends the left endpoint \(a\) to \(b\), and sends \(c\) to itself.
For this formula, it is enough to assume that \(f\) is continuous on \([a,b]\). It also holds for piecewise continuous functions, as in the piecewise example below. The variable \(u\) in the right-hand integral ranges only over the left half; the term \(f(2c-u)\) accounts for the reflected point in the right half.
Why the Pairing Formula Works
First split the integral at \(c\), using additivity of the definite integral. Then rewrite the integral over \([c,b]\) by setting \(x=2c-u\). As \(x\) runs from \(c\) to \(b\), \(u\) runs from \(c\) down to \(a\), and \(dx=-du\). Reversing the bounds removes the minus sign. This gives the following derivation.
$$ \int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx. $$
In the second integral, set \(x=2c-u\). The bounds \(x=c\) and \(x=b=2c-a\) correspond to \(u=c\) and \(u=a\), respectively. Therefore $$ \int_c^b f(x)\,dx =\int_c^a f(2c-u)(-du) =\int_a^c f(2c-u)\,du. $$
Rename the dummy variable \(x\) in the first integral as \(u\), then add the integrands: $$ \int_a^b f(x)\,dx =\int_a^c f(u)\,du+\int_a^c f(2c-u)\,du =\int_a^c \bigl(f(u)+f(2c-u)\bigr)\,du. $$
This is a change of variables applied to just one half of the original integral. It does not require the two reflected values to be equal. Rather, it collects them into a sum that can be simpler than either value on its own.
Worked Examples
Worked Example: Pairing a Polynomial on a Shifted Interval
Evaluate \(\int_1^5 x^2\,dx\) by splitting at the midpoint.
The midpoint is \(c=(1+5)/2=3\), so reflection sends \(u\) to \(2c-u=6-u\). Apply the paired-integrand formula:
Check directly using the antiderivative \(x^3/3\): $$ \int_1^5 x^2\,dx=\left[\frac{x^3}{3}\right]_1^5=\frac{125-1}{3}=\frac{124}{3}. $$ The paired calculation agrees with the direct evaluation.
Worked Example: A Pair That Simplifies to a Constant
Evaluate \(\int_0^4 (x^2-2x+5)\,dx\) by pairing reflected values.
The midpoint is \(c=2\), so the reflected point is \(4-u\). Compute both values before adding them:
Thus the integral is $$ \begin{aligned} \int_0^4 (x^2-2x+5)\,dx &=\int_0^2 (2u^2-8u+18)\,du\\ &=\left[\frac{2}{3}u^3-4u^2+18u\right]_0^2\\ &=\frac{16}{3}-16+36 =\frac{76}{3}. \end{aligned} $$ Check directly using the antiderivative \(x^3/3-x^2+5x\): $$ \left[\frac{x^3}{3}-x^2+5x\right]_0^4 =\frac{64}{3}-16+20 =\frac{76}{3}. $$ The paired values produce a quadratic integrand on only half the interval, and the result matches the direct calculation.
Worked Example: Equal Half-Integrals Without Matching Reflected Values
On \([0,2]\), define \(f(x)=0\) for \(0\le x\le1\) and \(f(x)=\frac32-x\) for \(1<x\le2\). Compare the integrals over the two halves and the reflected values.
The left-half integral is $$ \int_0^1 f(x)\,dx=\int_0^1 0\,dx=0. $$ The right-half integral is $$ \begin{aligned} \int_1^2 f(x)\,dx &=\int_1^2 \left(\frac32-x\right)\,dx\\ &=\left[\frac32x-\frac{x^2}{2}\right]_1^2\\ &=(3-2)-\left(\frac32-\frac12\right) =1-1 =0. \end{aligned} $$ The half-integrals are equal, but the reflected values do not match almost everywhere. Here the midpoint is \(1\), and for \(0\le u<1\), $$ f(u)=0,\qquad f(2-u)=\frac32-(2-u)=u-\frac12. $$ These values are unequal except at \(u=\frac12\). Pairing explains why the full integral is nevertheless zero: $$ \int_0^2 f(x)\,dx =\int_0^1 \bigl(0+u-\tfrac12\bigr)\,du =\left[\frac{u^2}{2}-\frac{u}{2}\right]_0^1 =0. $$ Thus matching reflected values guarantee equal half-integrals, but equality of the half-integrals can also occur without pointwise matching. In this example, the half-integrals are equal, not unequal.
For contrast, if \(g(x)=0\) on \([0,1]\) and \(g(x)=1\) on \((1,2]\), the left-half integral is \(0\) and the right-half integral is \(1\). The reflected values differ, and so do the half-integrals. A mismatch by itself does not determine whether the half-integrals are equal; their integrals must be compared.
What the Pairing Does and Does Not Tell You
The paired formula always gives the total integral as an integral of a sum over the left half. Further conclusions require additional information about that sum. For instance, if \(f(u)+f(2c-u)=K\) throughout the left half for a constant \(K\), then the total integral is \(K(c-a)\), since the length of \([a,c]\) is \(c-a\).
There is also a useful distinction between pointwise equality and equality of integrals. If \(f(u)=f(2c-u)\) for every \(u\in[a,c]\), then the two half-integrals are equal: the reflected integrands are the same at every corresponding point, so their integrals agree. But this is a sufficient condition, not a necessary one. Different reflected values can still have the same integral, as the preceding example demonstrates.
In the same way, reflected values that differ do not automatically imply unequal half-integrals. The sign and size of their differences over the interval matter. A positive contribution on one part can be balanced by a negative contribution elsewhere. The integral records the net accumulated value, not whether the integrands agree point by point.
A Reliable Procedure
When a midpoint split looks useful, work through the steps in order.
For bounds \(a\) and \(b\), calculate \(c=(a+b)/2\) and check that it lies halfway between them.
The reflected input is \(2c-u\). Check that it sends \(a\) to \(b\) and \(c\) to itself.
Calculate \(f(u)+f(2c-u)\), simplifying each term before drawing a conclusion.
Evaluate the paired expression from \(a\) to \(c\), then check the answer by another valid evaluation when practical.
Common Pitfall: Confusing Matching with Equal Integrals
A common error is to say that equal half-integrals require \(f(u)=f(2c-u)\) at every point. That conclusion is too strong. Pointwise matching is enough to guarantee equality of the integrals, but it is not required: unequal values can balance after integration. The piecewise example above gives unequal reflected values almost everywhere and two half-integrals that are both zero.
A different mistake is to infer the total integral from only one half without checking the reflected values. For example, the rule \(\int_a^b f(x)\,dx=2\int_a^c f(u)\,du\) is justified when the corresponding reflected values match, but it cannot be used merely because the interval has a midpoint. In general, use the paired formula and retain both terms unless a stated condition allows simplification.
Why Midpoint Pairing Is Useful
Splitting at the symmetry point provides a systematic way to exploit relationships between opposite sides of an interval. The original integrand need not be symmetric: sometimes the sum of the reflected values is simpler, as in the polynomial example. In other cases, the pairing clarifies why positive and negative contributions cancel or why two half-integrals agree.
The method is also a useful check on reasoning. A correct reflection must use \(2c-u\), and a correct change of variables must account for the reversed bounds. Writing both reflected values explicitly helps prevent errors in signs, bounds, or claims about symmetry.
Check Your Understanding
For each question, identify the midpoint and keep pointwise statements distinct from statements about integrals.
- What is the midpoint of \([2,8]\), and what expression gives the reflection of \(u\) across it?
- State the paired-integrand formula for \(\int_a^b f(x)\,dx\) when \(c=(a+b)/2\).
- If \(f(u)+f(2c-u)=5\) for every \(u\in[a,c]\), what is \(\int_a^b f(x)\,dx\) in terms of \(a\) and \(c\)?
- Does pointwise equality \(f(u)=f(2c-u)\) guarantee equal half-integrals? Does equal half-integrals guarantee pointwise equality?
- For the piecewise function in the third worked example, what are the two half-integrals, and how do the reflected values compare?
