Introduction
A definite integral over an interval from \(-a\) to \(a\) includes both a negative and a positive half. When the function is even, its values on those halves match under reflection across the vertical axis. The two halves therefore contribute equally, so one half can be integrated and then doubled.
This can make a calculation shorter, but the symmetry condition matters: the rule applies only when the function is even and the interval is symmetric about zero. We will state those conditions precisely, see why the reduction is natural, and use it in several evaluations.
Even Functions
A function is even when reflecting its input across zero leaves its output unchanged. The domain must contain both \(x\) and \(-x\) whenever the condition is used.
For example, \(f(x)=x^2\) is even because \((-x)^2=x^2\). The function \(g(x)=|x|\) is also even because \(|-x|=|x|\). By contrast, \(h(x)=x\) is not even: for nonzero \(x\), \(h(-x)=-x\), which is not \(h(x)=x\).
The definition concerns the function values, not whether the function is positive. An even function may be positive, negative, or change sign. The essential feature is that values at opposite inputs agree.
The Symmetry Rule for Definite Integrals
Suppose \(a>0\), \(f\) is even on \([-a,a]\), and \(f\) is integrable there. Then the integral over the full symmetric interval is twice the integral over its positive half.
For the usual definite integrals in this course, continuity on \([-a,a]\) is a sufficient condition for integrability. The theorem needs integrability; continuity is a convenient way to ensure that condition in common examples. The case \(a=0\) is trivial, since both integrals have equal endpoints and are zero.
Here is the geometric reason for the rule. For an even function, the graph over the negative half is a reflection of the graph over the positive half across the vertical axis. Matching vertical slices have the same height, and the reflected pieces have the same signed area. Thus the contribution from \([-a,0]\) equals the contribution from \([0,a]\). Adding those two equal contributions gives the factor of \(2\).
This reasoning concerns signed area: portions of the graph below the horizontal axis contribute negatively. The equality still holds if the function takes negative values, because reflected parts have the same heights and therefore the same signed contributions.
Worked Example: A Polynomial on a Symmetric Interval
Worked Example: A Polynomial on a Symmetric Interval
Evaluate \(\int_{-3}^{3}(x^2+1)\,dx\).
First check the symmetry. If \(f(x)=x^2+1\), then $$ f(-x)=(-x)^2+1=x^2+1=f(x). $$ So \(f\) is even. It is a polynomial, hence continuous and integrable on \([-3,3]\). The symmetry rule applies:
To check the antiderivative, differentiate \(F(x)=x^3/3+x\): $$ F'(x)=x^2+1. $$ A direct endpoint check gives \(F(3)-F(-3)=12-(-12)=24\), agreeing with the result.
The reduction did not change the function being integrated or the interval’s total signed contribution. It used the equality of the two halves to avoid computing both separately.
Worked Example: The Absolute-Value Function
Worked Example: The Absolute-Value Function
Evaluate \(\int_{-2}^{2}|x|\,dx\).
The absolute-value identity \(|-x|=|x|\) shows that \(f(x)=|x|\) is even. It is continuous, including at \(x=0\), so it is integrable on the interval. Apply the symmetry rule, and note that \(|x|=x\) on \([0,2]\):
On the positive half, the antiderivative check is \(\frac{d}{dx}(x^2/2)=x\). As an independent check of the symmetry, split the original integral at zero. On \([-2,0]\), \(|x|=-x\), so that part is \(\int_{-2}^{0}-x\,dx=2\). On \([0,2]\), it is \(\int_{0}^{2}x\,dx=2\). The two parts add to \(2+2=4\).
This example also shows that an even function need not be given by one simple formula that behaves identically on both sides. Absolute value has different expressions on the two halves, but its values still satisfy the defining relation \(f(-x)=f(x)\).
Worked Example: A Rational Function
Worked Example: A Rational Function
Evaluate \(\int_{-1}^{1}\frac{1}{1+x^2}\,dx\).
Let \(f(x)=1/(1+x^2)\). Then $$ f(-x)=\frac{1}{1+(-x)^2}=\frac{1}{1+x^2}=f(x). $$ The denominator is positive for every real \(x\), so the function is continuous on \([-1,1]\) and therefore integrable. Reduce the integral to the positive half:
For the antiderivative check, \(\frac{d}{dx}\arctan x=1/(1+x^2)\). Checking the original endpoints directly gives \(\arctan(1)-\arctan(-1)=\pi/4-(-\pi/4)=\pi/2\), as required.
A Practical Checklist
Before using the symmetry rule, check each of its ingredients. A symmetric-looking formula or a pair of symmetric limits is not enough by itself.
The limits must be \(-a\) and \(a\) for some \(a>0\), so the interval is symmetric about zero.
Verify \(f(-x)=f(x)\) throughout the interval. Do not infer evenness just from a few values.
The function must be integrable on the full interval. Continuity on the closed interval is a common sufficient condition.
Evaluate the integral from \(0\) to \(a\), then multiply that value by \(2\).
The factor \(2\) applies to the value of the half-interval integral, including its sign. It does not mean that the integrand itself should be doubled before integration, and it does not remove the need to evaluate the integral over the correct half.
Common Mistake: Applying the Rule to a Non-Even Function
Consider the incorrect calculation $$ \int_{-1}^{1}(x+2)\,dx \overset{\text{wrong}}{=}2\int_{0}^{1}(x+2)\,dx =2\left(\frac{1}{2}+2\right)=5. $$ The interval is symmetric, but the function is not even: \(f(-x)=-x+2\), which is generally not equal to \(f(x)=x+2\). So the rule’s hypotheses fail.
Compute the integral correctly using the antiderivative \(F(x)=x^2/2+2x\):
The mistake is not an arithmetic error in evaluating the positive-half integral. It is using the symmetry rule without checking that the function is even. Checking the defining condition first prevents this error.
Why the Reduction Is Useful
When the hypotheses hold, the rule can reduce the work in two ways. First, it replaces an integral over an interval of width \(2a\) with one over an interval of width \(a\). Second, it can make the integrand easier to handle. For example, on the positive half, \(|x|\) becomes \(x\), as in the second worked example.
It is also a useful check on a calculation. If an even function is integrated over \([-a,a]\), the full integral must equal twice the positive-half integral. If a separate calculation disagrees, check the limits, the parity test, the antiderivative, and the factor of \(2\).
Check Your Understanding
For each question, identify the relevant symmetry condition before calculating.
- Is \(f(x)=x^4-3x^2+1\) even? State the check using \(f(-x)\).
- Use the symmetry rule to evaluate \(\int_{-2}^{2}(x^2+3)\,dx\).
- Explain why \(|x|\) is even, and evaluate \(\int_{-3}^{3}|x|\,dx\).
- Can the rule \(\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx\) be used for \(f(x)=x+1\)? Why or why not?
- State the interval, parity, and integrability conditions needed to use the even-function rule.
