Why the Center Matters
A function can have a clear symmetry even when its center is not zero. For example, the graph of \(f(x)=(x-3)^2\) is symmetric about the vertical line \(x=3\), not about the \(y\)-axis. If an integral runs over an interval centered at \(3\), that symmetry can simplify the calculation just as symmetry about zero simplifies integrals on intervals such as \([-a,a]\).
The key is to measure positions relative to the center. A point \(c+t\) is \(t\) units to the right of \(c\), while \(c-t\) is \(t\) units to its left. Comparing the function at these two points gives a precise test for symmetry about \(x=c\).
Symmetry About \(x=c\)
These definitions compare values at equal distances from \(c\). In the symmetric case, the values agree; in the antisymmetric case, they are opposites. When \(c=0\), these are the usual definitions of an even and an odd function. For a general \(c\), symmetry about \(x=c\) is not the same as evenness or oddness about zero.
Equivalently, let \(g(t)=f(c+t)\). Then \(f\) is symmetric about \(x=c\) exactly when \(g\) is even, and antisymmetric about \(x=c\) exactly when \(g\) is odd. This translates the question into the familiar parity test, but applies it to the shifted function \(g\).
The Centered-Interval Integral Rules
Suppose \(a>0\), \(f\) is integrable on \([c-a,c+a]\), and its symmetry condition holds throughout that interval. The translated function \(g(t)=f(c+t)\) is integrable on \([-a,a]\), and translating the integration variable gives
Thus the centered-interval rules for even and odd functions apply to \(g\). In particular, an antisymmetric function has zero integral over the centered interval. A symmetric function has an integral equal to twice the integral over either half of the interval.
- if \(f(c+t)=f(c-t)\), then \(\displaystyle\int_{c-a}^{c+a}f(x)\,dx=2\int_c^{c+a}f(x)\,dx\);
- if \(f(c+t)=-f(c-t)\), then \(\displaystyle\int_{c-a}^{c+a}f(x)\,dx=0\).
Continuity on the interval is a sufficient condition for integrability, though the theorem only requires integrability. The symmetry condition needs to hold across the interval: for each \(t\) with \(0\leq t\leq a\), it relates the two values \(f(c+t)\) and \(f(c-t)\).
Why the Rules Work
Here is a direct justification using the centered coordinate \(t\). Put \(g(t)=f(c+t)\). Translating the integral gives \(\int_{-a}^{a}g(t)\,dt\). Split that integral at zero. In the integral over \([-a,0]\), write the points as \(-t\) with \(t\) ranging from \(0\) to \(a\); equivalently, use the standard reflected-interval identity \(\int_{-a}^{0}g(t)\,dt=\int_0^a g(-t)\,dt\). Therefore
If \(g\) is even, then \(g(-t)=g(t)\), so the right side is \(2\int_0^a g(t)\,dt\). If \(g\) is odd, then \(g(-t)=-g(t)\), so the two terms cancel. Translating back, the half-interval \([0,a]\) corresponds to \([c,c+a]\), giving the stated formulas. The cancellation is a consequence of pairing values at equal distances from the center.
Worked Example: A Symmetric Function About \(x=3\)
Worked Example: A Symmetric Function About \(x=3\)
Evaluate \(\displaystyle\int_1^5\big((x-3)^2+1\big)\,dx\) by using symmetry.
The interval \([1,5]\) has center \(c=(1+5)/2=3\) and half-width \(a=2\). At equal distances from 3, the squared terms agree:
So \(f\) is symmetric about \(x=3\). The symmetric-function rule gives
For a check, an antiderivative in \(x\) is \(\frac{(x-3)^3}{3}+x\). Its endpoint difference is \(\left(\frac{8}{3}+5\right)-\left(-\frac{8}{3}+1\right)=\frac{28}{3}\), as required.
Worked Example: An Antisymmetric Function About \(x=-1\)
Worked Example: An Antisymmetric Function About \(x=-1\)
Evaluate \(\displaystyle\int_{-3}^{1}\big((x+1)^3-2(x+1)\big)\,dx\).
The interval has center \((-3+1)/2=-1\) and half-width \(2\). Set \(g(t)=f(-1+t)\). Then
The integrand is antisymmetric about \(x=-1\). The interval is centered there, so the integral is zero.
Check by direct evaluation. An antiderivative is \(F(x)=\frac{(x+1)^4}{4}-(x+1)^2\). At \(x=1\), \(F(1)=16/4-4=0\); at \(x=-3\), \(F(-3)=16/4-4=0\). Thus \(F(1)-F(-3)=0\), confirming the result.
Separating the Symmetric and Antisymmetric Parts
An integrand need not be wholly symmetric or wholly antisymmetric. In the centered coordinate, an expression can contain both even and odd powers of \(t=x-c\). The even terms contribute to the integral over a centered interval; the odd terms cancel in pairs. This is the same even-and-odd decomposition used for functions centered at zero, applied after expressing the function relative to \(c\).
Worked Example: A Mixture of Both Types
Worked Example: A Mixture of Both Types
Evaluate \(\displaystyle\int_{-1}^{3}\big((x-1)^4+(x-1)^3+3(x-1)^2\big)\,dx\).
The interval has center \(c=1\) and half-width \(a=2\). With \(t=x-1\), the integrand becomes \(t^4+t^3+3t^2\). The terms \(t^4\) and \(3t^2\) are even, while \(t^3\) is odd. Thus the odd term contributes zero, and the integral is
Check with an antiderivative in \(x\): \(F(x)=\frac{(x-1)^5}{5}+\frac{(x-1)^4}{4}+(x-1)^3\). At \(x=3\), \(F(3)=32/5+4+8=92/5\). At \(x=-1\), \(F(-1)=-32/5+4-8=-52/5\). Therefore \(F(3)-F(-1)=92/5-(-52/5)=144/5\).
A Common Mistake: Testing Parity Before Shifting
A frequent error is to look at an expression in \(x-c\) and apply the even-or-odd test as though the center were zero. Consider \(f(x)=(x-2)^3+1\) on \([0,4]\). The interval is centered at \(2\), and in the centered coordinate \(t=x-2\), the function is \(t^3+1\). It is not antisymmetric about \(x=2\): its values at equal offsets add to \(2\), not to \(0\).
The incorrect calculation treats the whole expression as odd because it contains a cubic term, then concludes that the integral is zero. The constant \(1\) does not cancel. The correct calculation is
The practical correction is to identify the center of the interval first, then test the shifted function \(g(t)=f(c+t)\). A power of \(x-c\) has parity as a function of the centered coordinate \(t\); that does not mean the full integrand has that parity if other terms are present.
When the Shortcut Applies
Before using a symmetry rule, check each part of the setup:
- The interval must be centered at \(c\), so its endpoints have the form \(c-a\) and \(c+a\).
- The function must be integrable over the interval.
- The symmetry relation must hold for corresponding points on both sides of \(c\).
- If the integrand contains both symmetric and antisymmetric terms, only the antisymmetric part cancels.
The symmetry rule is most useful when a direct antiderivative would be lengthy or when an integrand separates naturally into powers of \(x-c\). It can reduce the amount of calculation, but it does not make an asymmetric interval symmetric and does not justify cancellation for a function that fails the relevant test.
Check Your Understanding
Use the center of each interval and the symmetry definitions to answer these questions.
- Is \(f(x)=(x-4)^6\) symmetric, antisymmetric, or neither about \(x=4\)?
- Evaluate \(\displaystyle\int_{-2}^{4}(x-1)^5\,dx\) using symmetry.
- For \(f(x)=(x-2)^2+(x-2)^3\), identify the symmetric and antisymmetric parts about \(x=2\).
- Why is \([1,7]\) a suitable interval for applying symmetry about \(x=4\)?
- What condition must be checked before concluding that an integral over \([c-a,c+a]\) is zero?
