Definite Integrals Of Symmetric Functions

If the graph of a function is symmetric, its definite integral over a symmetric interval $[- a, a]$ can be simplified, and in the odd case it can be written down with no computation at all.

Formula	Name	Test
$\int_{- a}^{a} f (x) d x = 0$	Odd Function Rule	$f (- x) = - f (x)$
$\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$	Even Function Rule	$f (- x) = f (x)$
$x, x^{3}, x^{5}, \sin x, \tan x$	Standard Odd Functions	Symmetric about the origin
$1, x^{2}, x^{4}, \cos x, \| x \|$	Standard Even Functions	Symmetric about the $y$ -axis
(odd) $\times$ (even) is odd; (odd) $\times$ (odd) is even	Products	Same sign rules as multiplication

The Symmetry Theorem

Suppose that $f$ is continuous on the symmetric interval $[- a, a]$ .

If $f$ is an odd function, then $\int_{- a}^{a} f (x) d x = 0$ .
If $f$ is an even function, then $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ .

Both statements are visually obvious. For an odd function, the piece of the region to the left of the $y$ -axis is the mirror image of the piece on the right, rotated 180 degrees about the origin; whatever net area one contributes, the other contributes with the opposite sign, so they cancel. For an even function, the two halves are reflections in the $y$ -axis and contribute equally, so the whole is twice either half.

Graph of an odd function on a symmetric interval, with equal areas above and below the x-axis cancelling — For an odd function, the integral from $- a$ to $a$ is zero.

Graph of an even function on a symmetric interval, with two mirror-image regions of equal area — For an even function, the integral from $- a$ to $a$ is twice the integral from $0$ to $a$ .

Proof of the Symmetry Theorem.

Proof

We can write

\int_{-a}^{a}f(x)\,dx=\int_{-a}^{0}f(x)\,dx+\int_{0}^{a}f(x)\,dx. \tag{i}

For the first integral on the right-hand side, let $u = - x$ . Then $d u = - d x$ ; we have $u = a$ when $x = - a$ , and $u = 0$ when $x = 0$ . Therefore

\int_{-a}^{0}f(x)\,dx=-\int_{a}^{0}f(-u)\,du. \tag{ii}

(a) The odd case. If $f$ is an odd function, then $f (- u) = - f (u)$ and the right-hand side of Equation (ii) can be written as

\begin{aligned} -\int_{a}^{0}f(-u)\,du &=\int_{a}^{0}f(u)\,du\\ &=-\int_{0}^{a}f(u)\,du. \end{aligned}

For the last equation, we used the fact that $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ . Because the variable of integration in a definite integral is a dummy variable, we can write $x$ instead of $u$ :

- \int_{0}^{a} f (u) d u = - \int_{0}^{a} f (x) d x .

So we have shown

\int_{-a}^{0}f(x)\,dx=-\int_{0}^{a}f(x)\,dx. \tag{iii}

Substituting (iii) into (i) gives

\begin{aligned} \int_{-a}^{a}f(x)\,dx &=-\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(x)\,dx\\ &=0. \end{aligned}

(b) The even case. If $f$ is even, then

\begin{aligned} \int_{-a}^{a}f(x)\,dx &=\int_{-a}^{0}f(x)\,dx+\int_{0}^{a}f(x)\,dx &&{\small(\text{Eq. (i)})}\\ &=-\int_{a}^{0}f(-u)\,du+\int_{0}^{a}f(x)\,dx &&{\small(\text{let }u=-x)}\\ &=-\int_{a}^{0}f(u)\,du+\int_{0}^{a}f(x)\,dx &&{\small(f\text{ is even})}\\ &=\int_{0}^{a}f(u)\,du+\int_{0}^{a}f(x)\,dx &&{\small\left(\int_{0}^{a}f=-\int_{a}^{0}f\right)}\\ &=\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(x)\,dx &&{\small(u\text{ is a dummy variable})}\\ &=2\int_{0}^{a}f(x)\,dx. \end{aligned}

Applying the Rules

For example, because $f (x) = \sin (x^{3})$ is an odd function,

f (- x) = \sin ((- x)^{3}) = \sin (- x^{3}) = - \sin (x^{3}) = - f (x),

we know immediately that

\int_{- π}^{π} \sin (x^{3}) d x = 0.

This is worth appreciating: the function $\sin (x^{3})$ has no elementary antiderivative, so no amount of algebra would evaluate this integral. Symmetry answers it in one line.

Evaluate $\int_{- 2}^{2} (x^{5} - 3 x^{3} + x) d x$ .

Solution

Every term has an odd power of $x$ , so the whole integrand is odd:

f (- x) = (- x)^{5} - 3 (- x)^{3} + (- x) = - x^{5} + 3 x^{3} - x = - f (x) .

The interval $[- 2, 2]$ is symmetric about the origin, so

\int_{- 2}^{2} (x^{5} - 3 x^{3} + x) d x = 0.

Evaluate $\int_{- 1}^{1} (x^{4} + 3 x^{2} + 2) d x$ .

Solution

All powers are even and the constant is even, so the integrand is even. By the Even Function Rule,

\begin{aligned} \int_{-1}^{1}\left(x^{4}+3x^{2}+2\right)dx &=2\int_{0}^{1}\left(x^{4}+3x^{2}+2\right)dx\\ &=2\left[\frac{x^{5}}{5}+x^{3}+2x\right]_{0}^{1}\\ &=2\left(\frac{1}{5}+1+2\right)=\frac{32}{5}. \end{aligned}

The saving here is modest, but the rule also removes a common source of sign errors: substituting a negative lower limit into odd powers.

A warning. Both rules require the interval to be symmetric about the origin. Nothing at all follows about $\int_{1}^{3}$ of an odd function, and nothing follows for $\int_{- 2}^{3}$ either.

Exercises

Classify each function as even, odd, or neither:

(a) $f (x) = x^{3} \cos x$

(b) $f (x) = x^{2} \sin x$

(c) $f (x) = \frac{x}{1 + x^{2}}$

(d) $f (x) = e^{x}$

Answer

(a) odd (b) odd (c) odd (d) neither

Solution

(a) $x^{3}$ is odd and $\cos x$ is even; odd times even is odd.

(b) $x^{2}$ is even and $\sin x$ is odd; even times odd is odd.

(c) The numerator is odd and the denominator is even, so the quotient is odd: $f (- x) = \frac{- x}{1 + x^{2}} = - f (x)$ .

(d) $e^{- x}$ equals neither $e^{x}$ nor $- e^{x}$ , so $e^{x}$ is neither. (It does split into an even part $\cosh x$ and an odd part $\sinh x$ .)

Evaluate without antidifferentiating:

(a) $\int_{- 3}^{3} \frac{x}{1 + x^{4}} d x$

(b) $\int_{- π / 2}^{π / 2} x^{2} \sin x d x$

(c) $\int_{- 5}^{5} x^{7} \cos (x^{2}) d x$

Answer

All three equal $0$ .

Solution

Each integrand is odd and each interval is symmetric about the origin.

(a) Odd numerator over even denominator.

(b) Even times odd.

(c) $x^{7}$ is odd and $\cos (x^{2})$ is even, since $\cos ((- x)^{2}) = \cos (x^{2})$ .

None of these has an antiderivative that is pleasant to find, which is exactly why the symmetry rule is worth checking first.

Evaluate $\int_{- 2}^{2} (3 x^{2} + 1) d x$ using the Even Function Rule.

Answer

$20$

Solution

The integrand is even, so

\int_{- 2}^{2} (3 x^{2} + 1) d x = 2 \int_{0}^{2} (3 x^{2} + 1) d x = 2 {[x^{3} + x]}_{0}^{2} = 2 (8 + 2) = 20.

Evaluate $\int_{- a}^{a} (x^{3} + 2 x^{2} - 5 x + 4) d x$ in terms of $a$ .

Answer

$\frac{4 a^{3}}{3} + 8 a$

Solution

Split the integrand into its odd part $x^{3} - 5 x$ and its even part $2 x^{2} + 4$ . The odd part contributes 0, so

\begin{aligned} \int_{-a}^{a}\left(x^{3}+2x^{2}-5x+4\right)dx &= 2\int_{0}^{a}\left(2x^{2}+4\right)dx\\ &= 2\left[\frac{2x^{3}}{3}+4x\right]_{0}^{a} = \frac{4a^{3}}{3}+8a. \end{aligned}

This split-into-parts technique works for any polynomial: keep the even-degree terms, discard the odd-degree ones.

Evaluate $\int_{- 1}^{1} | x |^{3} d x$ .

Answer

$\frac{1}{2}$

Solution

Since $| - x | = | x |$ , the function $| x |^{3}$ is even (in contrast to $x^{3}$ , which is odd). Therefore

\int_{- 1}^{1} | x |^{3} d x = 2 \int_{0}^{1} x^{3} d x = 2 \cdot \frac{1}{4} = \frac{1}{2} .

On $[0, 1]$ we could replace $| x |$ by $x$ because $x \geq 0$ there. Note how the absolute value converts an odd function into an even one, changing the answer from 0 to $\frac{1}{2}$ .

Suppose $f$ is even and $\int_{0}^{4} f (x) d x = 9$ . Find $\int_{- 4}^{4} f (x) d x$ and $\int_{- 4}^{0} f (x) d x$ .

Answer

$18$ and $9$ respectively.

Solution

By the Even Function Rule, $\int_{- 4}^{4} f = 2 \int_{0}^{4} f = 18$ . Then additivity gives

\int_{- 4}^{0} f = \int_{- 4}^{4} f - \int_{0}^{4} f = 18 - 9 = 9,

which is just the statement that the two halves contribute equally.

If instead $f$ were odd with $\int_{0}^{4} f = 9$ , the answers would be 0 and $- 9$ .

Prove that if $f$ is odd and continuous, then $F (x) = \int_{0}^{x} f (t) d t$ is an even function.

Answer

Substituting $t = - u$ in $F (- x)$ produces $F (x)$ .

Solution

Compute $F (- x)$ and substitute $t = - u$ , so $d t = - d u$ ; when $t = 0$ , $u = 0$ , and when $t = - x$ , $u = x$ :

\begin{aligned} F(-x) &= \int_{0}^{-x}f(t)\,dt = \int_{0}^{x}f(-u)(-du)\\ &= \int_{0}^{x}\left[-f(u)\right](-du) = \int_{0}^{x}f(u)\,du = F(x), \end{aligned}

where the third equality used oddness, $f (- u) = - f (u)$ , and the two minus signs cancelled.

An alternative one-line argument: F'(x)=f(x) is odd, and the antiderivative of an odd function is even provided it vanishes at 0, which $F$ does.

Evaluate $\int_{- π}^{π} (\sin^{3} x + \cos^{2} x) d x$ .

Answer

$π$

Solution

The term $\sin^{3} x$ is odd, since an odd power of an odd function is odd, so it contributes 0. The term $\cos^{2} x$ is even, so

\begin{aligned} \int_{-\pi}^{\pi}\left(\sin^{3}x+\cos^{2}x\right)dx &= 0+2\int_{0}^{\pi}\cos^{2}x\,dx\\ &= 2\int_{0}^{\pi}\frac{1+\cos2x}{2}\,dx\\ &= \left[x+\frac{\sin2x}{2}\right]_{0}^{\pi}=\pi. \end{aligned}

A student claims that $\int_{- 2}^{2} \frac{d x}{x^{3}} = 0$ because the integrand is odd. Is the claim correct?

Answer

No. The integrand is odd, but it is not continuous on $[- 2, 2]$ , so the theorem does not apply and the integral does not exist.

Solution

The function $1 / x^{3}$ is indeed odd, but it has a vertical asymptote at $x = 0$ , which lies inside the interval of integration. The symmetry theorem assumes $f$ is continuous on $[- a, a]$ , and that hypothesis fails.

Splitting at the origin gives $\int_{- 2}^{0} \frac{d x}{x^{3}}$ and $\int_{0}^{2} \frac{d x}{x^{3}}$ , neither of which is a finite number, so the proposed "cancellation" is a cancellation of infinities and carries no meaning.

This is the same trap as the invalid evaluation of $\int_{- 1}^{1} \frac{d x}{x}$ in Section 7.3: symmetry never rescues an integrand with an interior asymptote.

Show that every function $f$ defined on $[- a, a]$ can be written as the sum of an even function and an odd function, and use the decomposition to explain why only the even part matters for $\int_{- a}^{a} f$ .

Answer

$f (x) = \underset{even}{\underset{⏟}{\frac{f (x) + f (- x)}{2}}} + \underset{odd}{\underset{⏟}{\frac{f (x) - f (- x)}{2}}}$ ; the odd part integrates to zero.

Solution

Define

E (x) = \frac{f (x) + f (- x)}{2}, O (x) = \frac{f (x) - f (- x)}{2} .

Then $E (- x) = E (x)$ , so $E$ is even, and $O (- x) = - O (x)$ , so $O$ is odd, and clearly $E (x) + O (x) = f (x)$ .

Now integrate over the symmetric interval. By the Odd Function Rule the second piece vanishes, so

\int_{- a}^{a} f (x) d x = \int_{- a}^{a} E (x) d x = 2 \int_{0}^{a} E (x) d x = \int_{0}^{a} [f (x) + f (- x)] d x .

In words: on a symmetric interval, an integral can only "see" the even part of the integrand. For $f (x) = e^{x}$ , for example, the even part is $\cosh x$ , and indeed $\int_{- a}^{a} e^{x} d x = 2 \int_{0}^{a} \cosh x d x = 2 \sinh a$ .

The Symmetry Theorem

Applying the Rules

Exercises

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