Proof Of Symmetry Rules For Integrals

Why Reflection Simplifies an Integral

On an interval extending equally far to the left and right of zero, points occur in pairs: $x$ and $-x$. If a function has the same value at both points, their contributions to the integral reinforce each other. If the values are opposites, the contributions cancel.

This observation gives useful shortcuts, but the shortcuts depend on two conditions: the interval must be symmetric about zero, and the function must have the appropriate symmetry. We will prove the rules by reflecting partitions and their sample points, so the doubling and cancellation happen directly in the Riemann sums.

Even and Odd Functions

Definition: Let $a>0$, and let $f:[-a,a]\to\mathbb{R}$. The function $f$ is even if $f(-x)=f(x)$ for every $x\in[-a,a]$. It is odd if $f(-x)=-f(x)$ for every $x\in[-a,a]$.

For example, $f(x)=x^2$ is even because $(-x)^2=x^2$. The function $g(x)=x^3$ is odd because $(-x)^3=-x^3$. These definitions compare values at reflected inputs; they do not by themselves say anything about whether the function is integrable.

Theorem: Suppose $a>0$ and $f$ is Riemann integrable on $[-a,a]$. If $f$ is even, then $$ \int_{-a}^{a} f(x)\,dx=2\int_{0}^{a}f(x)\,dx. $$ If $f$ is odd, then $$ \int_{-a}^{a} f(x)\,dx=0. $$

The integrability condition ensures that the Riemann sums used below converge to the stated integrals. The restriction of an integrable function to either half-interval is integrable as well. The interval condition matters just as much: the two halves must have equal length, with endpoints $-a$ and $a$.

Proof by Reflecting Tagged Partitions

Begin with any partition of the positive half-interval:

0=x_0<x_1<\cdots<x_n=a.

Choose a tag $\xi_i\in[x_{i-1},x_i]$ for each subinterval, and let $\Delta x_i=x_i-x_{i-1}$. The tagged Riemann sum on $[0,a]$ is

S^+=\sum_{i=1}^{n}f(\xi_i)\Delta x_i.

Reflect each positive subinterval $[x_{i-1},x_i]$ across zero. It becomes $[-x_i,-x_{i-1}]$, which has the same width $\Delta x_i$. Reflect its tag as well: the tag $\xi_i$ becomes $-\xi_i$. Together, the original partition and its reflection form a partition of $[-a,a]$. Its tagged sum is

S=\sum_{i=1}^{n}\left(f(\xi_i)+f(-\xi_i)\right)\Delta x_i.

This formula displays the paired contributions. If $f$ is even, then $f(-\xi_i)=f(\xi_i)$, so $S=2S^+$. If $f$ is odd, then $f(-\xi_i)=-f(\xi_i)$, so $S=0$. To pass from these exact finite-sum identities to integral identities, take a sequence of partitions whose mesh—the largest subinterval width—tends to zero. The reflected partitions have the same mesh. Since $f$ is integrable, their tagged sums converge to the corresponding integrals.

Partition $[0,a]$.
Choose subintervals and one tag in each.

Reflect every subinterval and tag.
The reflected intervals have the same widths, and their tags are the negatives of the original tags.

Pair the terms.
Evenness makes each pair equal; oddness makes each pair sum to zero.

Take the limit.
As the mesh tends to zero, the tagged sums converge to the integrals, giving the theorem.

In the even case, the reflected sum is twice the positive-half sum at every stage. Taking limits gives $\int_{-a}^{a} f(x)\,dx=2\int_0^a f(x)\,dx$. In the odd case, every reflected sum is exactly zero, so its limit is zero. This proves both rules.

Worked Examples

Worked Example: An Even Quadratic

Evaluate $\int_{-2}^{2}x^2\,dx$ using symmetry and Riemann sums. The function is even because $(-x)^2=x^2$, and it is integrable on this interval. The even-function rule gives

\int_{-2}^{2}x^2\,dx=2\int_0^2x^2\,dx.

For the integral on $[0,2]$, divide the interval into $n$ equal pieces. Each width is $\Delta x=2/n$, and choose the right endpoint $x_i=2i/n$ as the tag for the $i$-th piece. The sum is

\begin{aligned} S_n &=\sum_{i=1}^{n}\left(\frac{2i}{n}\right)^2\frac{2}{n}\\ &=\frac{8}{n^3}\sum_{i=1}^{n}i^2\\ &=\frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}\\ &=\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}. \end{aligned}

As $n$ tends to infinity, the last two terms tend to zero, so $\int_0^2x^2\,dx=8/3$. Doubling gives $\int_{-2}^{2}x^2\,dx=16/3$. The reflected sums also check the symmetry step: a tag $x_i$ on the positive side is paired with $-x_i$, and their function values are both $x_i^2$. Thus each pair contributes twice the positive-side term, as required.

Worked Example: An Odd Polynomial

Evaluate $\int_{-1}^{1}(x^3-2x)\,dx$. Let $f(x)=x^3-2x$. Then

$$ f(-x)=(-x)^3-2(-x)=-x^3+2x=-(x^3-2x)=-f(x), $$

so $f$ is odd. The odd-function rule immediately gives $\int_{-1}^{1}f(x)\,dx=0$. To see the cancellation in a concrete sequence of sums, divide $[0,1]$ into $n$ equal pieces, with right-endpoint tags $i/n$. The positive-side sum is

\begin{aligned} S_n^+ &=\sum_{i=1}^{n}\left(\left(\frac{i}{n}\right)^3-2\frac{i}{n}\right)\frac{1}{n}\\ &=\frac{1}{n^4}\sum_{i=1}^{n}i^3-\frac{2}{n^2}\sum_{i=1}^{n}i\\ &=\frac{1}{n^4}\cdot\frac{n^2(n+1)^2}{4} -\frac{2}{n^2}\cdot\frac{n(n+1)}{2}\\ &=\frac{(n+1)^2}{4n^2}-\frac{n+1}{n}. \end{aligned}

Taking the limit gives $1/4-1=-3/4$ for the positive-side integral. On the reflected interval, each tag $i/n$ is paired with $-i/n$, and the function values are opposites. The negative-side sum is therefore the negative of the positive-side sum, with limit $3/4$. The two contributions add to $-3/4+3/4=0$, confirming the theorem.

Worked Example: A Function with Even and Odd Parts

Evaluate $\int_{-1}^{1}(x^2+3x+1)\,dx$ by pairing reflected points. Write $g(x)=x^2+3x+1$. For each $x$,

$$ g(x)+g(-x) =(x^2+3x+1)+(x^2-3x+1) =2x^2+2. $$

Thus, on a reflected partition, the paired contribution at tag $x$ is $(2x^2+2)\Delta x$. Equivalently, the even terms $x^2+1$ contribute twice their positive-half integral, while the odd term $3x$ cancels. Compute the positive-half integral of $x^2+1$ by right-endpoint sums on $[0,1]$:

\begin{aligned} T_n &=\sum_{i=1}^{n}\left(\left(\frac{i}{n}\right)^2+1\right)\frac{1}{n}\\ &=\frac{1}{n^3}\sum_{i=1}^{n}i^2+1\\ &=\frac{(n+1)(2n+1)}{6n^2}+1 \longrightarrow \frac{1}{3}+1=\frac{4}{3}. \end{aligned}

Doubling gives $2(4/3)=8/3$, and the $3x$ contributions cancel pair by pair. Therefore the integral is $8/3$. As a direct check of the pairing, the limit of the full symmetric sums equals $\int_0^1(2x^2+2)\,dx=2(4/3)=8/3$, the same result.

Conditions and a Common Mistake

The rules concern integration over $[-a,a]$, not every interval whose endpoints happen to include both negative and positive numbers. The distances from zero to the two endpoints must agree. Also, a function's parity must be checked from its definition; a graph that appears symmetric over only part of its domain is not enough to apply the theorem to a larger interval.

Common Mistake: Applying Odd Symmetry on the Wrong Interval

A tempting but incorrect calculation is

\int_{-1}^{2}x\,dx=0 \qquad\text{because }x\text{ is odd}.

The odd-function rule does not apply: $[-1,2]$ is not symmetric about zero. The contributions can instead be separated at zero. The signed area under $y=x$ on $[-1,0]$ is a triangle below the axis, with integral $-1/2$; on $[0,2]$, it is a triangle above the axis, with integral $2$. Therefore

\int_{-1}^{2}x\,dx=-\frac{1}{2}+2=\frac{3}{2},

not zero. The correction is to check the interval before using a symmetry rule. On the genuinely symmetric interval $[-1,1]$, the reflected contributions of $x$ cancel and the integral is zero.

Why the Proof Is Useful

The proof explains more than a shortcut. It identifies exactly what makes the result work: equal widths under reflection and a precise relationship between the two function values. This viewpoint is useful when a function is not entirely even or odd, as in the third example, because the paired sum $f(x)+f(-x)$ reveals which contributions remain.

When applying either rule, check the interval, verify evenness or oddness using the definition, and confirm integrability. Once these requirements hold, the result follows from pairing the Riemann-sum terms—not from an informal impression that areas look alike or cancel.

Check Your Understanding

Use the reflected-partition argument or the stated symmetry rules to answer these questions.

If $f$ is integrable and even on $[-3,3]$, how can $\int_{-3}^{3}f(x)\,dx$ be written using an integral on $[0,3]$?
Why do the two tagged contributions from $x$ and $-x$ cancel when $f$ is odd?
For an odd integrable function, what is its integral over $[-a,a]$, and which interval condition is essential?
Is the even-function rule applicable to $\int_{-2}^{3}f(x)\,dx$ just because $f$ is even? Explain.
If $g(x)=x^2+x$, what is $g(x)+g(-x)$, and which part of $g$ cancels under reflection?

Proof Of Symmetry Rules For Integrals

What You'll Learn

Why Reflection Simplifies an Integral

Even and Odd Functions

Proof by Reflecting Tagged Partitions