How to integrate even and odd functions on symmetric intervals

If your integration bounds are the same distance from zero, take a look at the function before you integrate. Matching halves can save you half the work, and opposite halves can cancel. Let’s see how to tell which rule applies.

Check the bounds and the function

Start with an interval [-a,a], where a>0, and a function that is integrable on that interval. Substitute -x into the function. The two tests are

f(-x)=f(x)evenf(-x)=-f(x)odd

An even graph has mirror symmetry across the vertical axis. An odd graph has symmetry about the origin, so the signed contributions on the two sides are opposites.

Even and odd symmetry on a symmetric intervalThe even graph has two matching regions above the axis. On the odd graph the region on the left is below the axis and has the same size as the region above the axis on the right.Even−11Odd−11
The even example doubles one side. The signed areas in the odd example cancel.

Use the symmetry rule

For an even function, integrate from zero to the upper bound, then double the result.

∫-aaf(x)dx=2∫0af(x)dx

For an odd function, the signed integrals on the two halves add to zero.

∫-aaf(x)dx=0

Evaluate the integral of an even function

Let f(x)=3x2+2. Squaring -x gives x2, so f(-x)=f(x). The bounds are symmetric, and we can use the even rule. An antiderivative is x3+2x, which is zero at the lower bound of the half interval.

∫-22(3x2+2)dx=2∫02(3x2+2)dx=2(8+4)=24

Evaluate the integral of an odd function

For f(x)=3x7+4sinx, replacing x with -x changes the sign of both terms.

f(-x)=-3x7-4sinx=-f(x)∫-77(3x7+4sinx)dx=0

The cancellation uses both the odd symmetry of the function and the symmetric bounds. If the bounds are uneven, evaluate over the given interval or apply symmetry to a symmetric part of it.

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