Symmetry in an Accumulation
A rate can have a symmetry that is reflected, in a precise way, in the accumulation measured from zero. The direction of that reflection depends on whether the rate itself is even or odd. The key point is that an integral with its bounds reversed changes sign.
Throughout, let \(f\) be integrable on an interval containing zero, and define its accumulation from zero by \(F(x)=\int_0^x f(t)\,dt\) wherever this integral is defined. The variable \(t\) is the integration variable; \(x\) is the input at which the accumulated amount is evaluated.
For the accumulation \(F\) to have either parity on an interval about zero, its domain must also contain both \(x\) and \(-x\). The theorem below applies on any symmetric interval about zero where the relevant integrals exist. In particular, it applies when \(f\) is continuous or piecewise continuous there. The theorem is about the accumulation basepoint being zero; a different basepoint need not produce the same symmetry.
The Reflection Identity for Definite Integrals
We first establish the integral fact that connects the values of \(f\) on opposite sides of zero. For \(x\geq 0\), reflecting the interval \([-x,0]\) across zero gives \([0,x]\). The values of \(f\) on the first interval therefore correspond to the values of \(f(-u)\) on the second:
Here \(u\) is just a variable of integration. To see why the identity holds, divide \([0,x]\) into subintervals with partition points \(0=u_0<u_1<\cdots<u_n=x\). Reflecting those subintervals gives a partition of \([-x,0]\) with endpoints \(-u_n,\ldots,-u_1,-u_0\). The subinterval corresponding to \([u_{i-1},u_i]\) has the same width, and the sampled function value \(f(-u_i)\) corresponds to the value of \(f\) at the reflected point \(-u_i\). Thus the corresponding Riemann sums agree. Taking their limits gives the identity whenever the integrals exist.
There is a second sign to keep track of. An integral whose upper bound is below its lower bound is oriented in the reverse direction:
Parity Theorem for Accumulations
The theorem says that an even rate produces an odd accumulation, while an odd rate produces an even accumulation. The proof uses reflection of the interval and then the defining parity relation for \(f\).
Take \(x\geq0\), so that the intervals from \(-x\) to zero and from zero to \(x\) are both within the domain.
By integral orientation, \(F(-x)=-\int_{-x}^{0}f(t)\,dt\).
The reflection identity gives \(F(-x)=-\int_0^x f(-u)\,du\).
If \(f\) is even, replace \(f(-u)\) by \(f(u)\), obtaining \(-F(x)\). If \(f\) is odd, replace it by \(-f(u)\), obtaining \(F(x)\).
In full, if \(f\) is even, then for \(x\geq0\),
If \(f\) is odd, the corresponding calculation is
These equations prove the claims for every \(x\geq0\). They prove them for negative inputs as well: if \(x<0\), apply the established equation to the nonnegative input \(-x\). For example, in the even-integrand case the relation gives \(F(x)=-F(-x)\), which is the same as \(F(-x)=-F(x)\). In the odd-integrand case it gives \(F(x)=F(-x)\). This completes the proof.
Worked Examples
Worked Example: A Constant Even Rate
Let \(f(t)=2\) on \([-3,3]\), and let \(F(x)=\int_0^x 2\,dt\). Since \(f(-t)=2=f(t)\), the rate is even. For \(x\geq0\), the rate is a rectangle of height \(2\) and width \(x\), so \(F(x)=2x\).
For a negative input, the bounds are reversed. For example,
Thus \(F(x)=2x\) on the full interval, and \(F(-x)=-2x=-F(x)\). The formula agrees with the theorem: the accumulation is odd. As a direct check at \(x=1\), \(F(1)=2\) and \(F(-1)=-2\), so the two values are negatives of each other.
Worked Example: An Odd Sign-Changing Rate
Define \(f(t)=-1\) for \(t<0\), \(f(0)=0\), and \(f(t)=1\) for \(t>0\), on \([-2,2]\). For every \(t\), \(f(-t)=-f(t)\), so \(f\) is odd. Its accumulation is \(F(x)=\int_0^x f(t)\,dt\).
For \(x>0\), the rate is \(1\) throughout \((0,x]\). The accumulated area is a rectangle with width \(x\) and height \(1\), giving \(F(x)=x\). For \(x<0\), the rate is \(-1\) on \([x,0)\), and reversing the bounds gives
Therefore \(F(x)=|x|\) for all \(x\), including \(F(0)=0\). For \(x\geq0\), \(F(-x)=|-x|=x=|x|=F(x)\); the accumulation is even. For instance, \(F(1)=1\) and \(F(-1)=1\), as required. The value assigned at the single point \(t=0\) does not change these area calculations.
Worked Example: A Piecewise-Constant Even Rate
Let \(f(t)=1\) when \(|t|\leq1\), and \(f(t)=3\) when \(1<|t|\leq2\). The value depends only on \(|t|\), so \(f(-t)=f(t)\): the rate is even. Define \(F(x)=\int_0^x f(t)\,dt\) for \(-2\leq x\leq2\).
For \(0\leq x\leq1\), the rectangle has height \(1\) and width \(x\), so \(F(x)=x\). For \(1\leq x\leq2\), split the accumulation at \(1\). The first rectangle has area \(1\cdot1=1\), and the second has area \(3(x-1)\). Hence
Check at \(x=1.5\): \(F(1.5)=1+3(1.5-1)=1+1.5=2.5\). Directly on the negative side, the rectangle from \(-1.5\) to \(-1\) has area \(3(0.5)=1.5\), and the rectangle from \(-1\) to \(0\) has area \(1(1)=1\). Reversing the bounds gives
The general parity result gives \(F(-x)=-F(x)\) throughout the interval, so \(F\) is odd. The calculation at \(x=1.5\) checks the result using the actual accumulated areas on both sides.
Why the Basepoint and the Sign Matter
The basepoint zero is central because reflection across zero swaps the two intervals that meet there: \([0,x]\) and \([-x,0]\). The defining parity of \(f\) compares the rates at the reflected points, so their accumulations are directly related. If accumulation instead starts at another point, the same reflection argument does not directly compare the two values; the theorem as stated makes no claim about that situation.
Also distinguish the signed integral from the unsigned area. Reversing the bounds changes the sign of the integral even when the rate is positive. This is precisely why an even rate gives an odd accumulation: the reflected regions have equal signed contributions when both are measured in the increasing direction, but \(F(-x)\) measures its interval in the decreasing direction.
For a concrete wrong calculation, take the even rate \(f(t)=2\) and \(x=1\). Ignoring orientation would give \(F(-1)=\int_{-1}^0 2\,dt=2\), falsely suggesting \(F(-1)=F(1)\). The correction is
Therefore \(F(-1)=-F(1)\), as the even-integrand theorem requires.
Check Your Understanding
Use the definition of accumulation, the reflection identity, and integral orientation to answer these questions.
- State the definition of an even function and the definition of an odd function on a domain symmetric about zero.
- If \(f\) is even and \(F(x)=\int_0^x f(t)\,dt\), what equation relates \(F(-x)\) and \(F(x)\)?
- For an odd integrand, why do the two minus signs in the reflected-integral calculation result in \(F(-x)=F(x)\)?
- For the even rate in the third worked example, calculate \(F(2)\) from the rectangle areas, then use parity to find \(F(-2)\).
- What sign error occurs if the bounds in \(\int_0^{-x}f(t)\,dt\) are treated as though they were in increasing order?
