Read the symmetry before calculating, integrals from easy to advanced lesson

The integral $\int_{-1}^{1} (x^3 + \cos x)\,dx$ can be found from an antiderivative, but its interval already does much of the work. The interval reaches exactly as far to the left of $0$ as to the right, so every point $x$ in it has a mirror point $-x$ . What happens to each term of the integrand when $x$ is replaced by $-x$ , and what does that do to the integral? This page answers both questions, shows that the cubic term contributes nothing and the cosine term doubles its right half, and arrives at the exact value $2\sin 1$ .

Key terms

Symmetric interval

Definition For a real number $a > 0$ , the interval $[-a, a]$ is symmetric about $0$ . Its midpoint is $0$ , and for every $x$ in the interval the point $-x$ lies in it too. More generally, the interval $[c - a, c + a]$ is symmetric about its midpoint $c$ . A set $D$ of real numbers is called symmetric about $0$ when $-x$ lies in $D$ for every $x$ in $D$ . Even and odd functions are defined on sets of this kind.

In plain words A symmetric interval reaches equally far on both sides of its center, so its points come in mirror pairs $x$ and $-x$ , and only the center is paired with itself. The symmetry rules on this page use intervals centered at $0$ , because $0$ is the point that the mirror $x \mapsto -x$ leaves in place.

Example The interval $[-1, 1]$ is symmetric about $0$ . The interval $[-1, 2]$ is not, because the point $2$ has no partner $-2$ in it. The interval $[0, 2]$ is symmetric about $1$ rather than $0$ . On it the substitution $t = x - 1$ moves the center to $0$ and changes the bounds to $-1$ and $1$ , so by the odd rule below

\begin{aligned} \int_0^2 (x - 1)^3\,dx &= \int_{-1}^{1} t^3\,dt \\[0.4em] &= 0. \end{aligned}

Odd function

Definition Let $f$ be defined on a set $D$ that is symmetric about $0$ . The function $f$ is odd if

f(-x) = -f(x)

for every $x$ in $D$ . The graph of an odd function is symmetric about the origin, which means that a half turn about the origin carries the graph onto itself. If $0$ lies in $D$ , then $f(0) = -f(0)$ , so $f(0) = 0$ .

In plain words Replacing $x$ by $-x$ keeps the size of the output and flips its sign. Heights at mirror points are equal in size and opposite in sign.

Example The function $x^3$ is odd, since $(-x)^3 = -x^3$ . For instance, $(\tfrac12)^3 = \tfrac18$ while $(-\tfrac12)^3 = -\tfrac18$ . The functions $\sin x$ , $x^5 - 4x$ , and $\tan x$ on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ are odd as well. The function $x^3 + 1$ is not odd, because its value at $0$ is $1$ instead of $0$ .

Even function

Definition Let $g$ be defined on a set $D$ that is symmetric about $0$ . The function $g$ is even if

g(-x) = g(x)

for every $x$ in $D$ . The graph of an even function is symmetric about the vertical axis, so reflecting the graph across that axis carries it onto itself.

In plain words Replacing $x$ by $-x$ leaves the output unchanged. Heights at mirror points are equal and have the same sign, whatever that sign is.

Example The function $\cos x$ is even, since $\cos(-x) = \cos x$ . So are $x^2$ , $x^4 - 3x^2$ , $\lvert x \rvert$ and $e^{-x^2}$ . An even function may be negative. The function $x^2 - 1$ is even and negative on $(-1, 1)$ , and its heights at $x$ and $-x$ still agree.

Parity

Definition The parity of a function defined on a set symmetric about $0$ records whether the function is even, odd or neither. Every such function $f$ is the sum of an even function and an odd function in exactly one way, with

\begin{aligned} f_{\text{even}}(x) &= \frac{f(x) + f(-x)}{2}, \\[0.4em] f_{\text{odd}}(x) &= \frac{f(x) - f(-x)}{2}. \end{aligned}

Sums and constant multiples of odd functions are odd, and sums and constant multiples of even functions are even. A product of two even functions or of two odd functions is even, and a product of an even function and an odd function is odd. The zero function is the only function that is both even and odd.

In plain words Parity describes how a function responds to the mirror $x \mapsto -x$ . A function that is neither even nor odd can still be split into an even part and an odd part, and each part can then use its own symmetry. In products, parity combines like signs, with even playing the role of plus and odd the role of minus.

Example The integrand $f(x) = x^3 + \cos x$ is neither even nor odd. Its values $f(1) = 1 + \cos 1 \approx 1.540$ and $f(-1) = -1 + \cos 1 \approx -0.460$ are neither equal nor opposite. Since $f(-x) = -x^3 + \cos x$ , the formulas give the even part $\cos x$ and the odd part $x^3$ . In the same way the exponential splits as $e^x = \cosh x + \sinh x$ , with even part $\cosh x$ and odd part $\sinh x$ . Among products, $x^2 \sin x$ is odd and $x \sin x$ is even.

Symmetry rule for definite integrals

Definition Let $a > 0$ and let $f$ be integrable on $[-a, a]$ , for example continuous there. If $f$ is odd, then

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

and if $f$ is even, then

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx.

The reason is the substitution $x = -t$ on the left half, which turns the bounds $-a$ and $0$ into $a$ and $0$ and gives

\int_{-a}^{0} f(x)\,dx = \int_0^a f(-t)\,dt.

For odd $f$ the left half is therefore $-\int_0^a f(t)\,dt$ , and for even $f$ it is $+\int_0^a f(t)\,dt$ . For any $f$ integrable on $[-a, a]$ , adding the two halves gives

\begin{aligned} &\int_{-a}^{a} f(x)\,dx \\[0.4em] &\quad = \int_0^a \bigl(f(t) + f(-t)\bigr)\,dt \\[0.4em] &\quad = 2\int_0^a f_{\text{even}}(t)\,dt, \end{aligned}

so over a symmetric interval only the even part of an integrand contributes. Both conditions matter. The rule says nothing about an interval such as $[-1, 2]$ , and it does not apply to $1/x$ on $[-1, 1]$ , which is not integrable there.

In plain words Cut the region over a symmetric interval into thin strips that come in mirror pairs of equal width. For an odd function the two strips of a pair carry opposite signed areas and cancel. For an even function they carry equal signed areas, so the left half repeats the right half. Neither rule needs the function to be positive, and a zero integral of an odd function is a balance of signed area, not an absence of area.

Example Call the opening integral $I$ . Its integrand is continuous on $[-1, 1]$ , and angles are measured in radians. By linearity, and because $x^3$ is odd and $\cos x$ is even,

\begin{aligned} I &= \int_{-1}^{1} x^3\,dx \\[0.4em] &\quad + \int_{-1}^{1} \cos x\,dx \\[0.4em] &= 0 + 2\int_0^1 \cos x\,dx \\[0.4em] &= 2\bigl[\sin x\bigr]_0^1 = 2\sin 1. \end{aligned}

So the integral equals $2\sin 1 \approx 1.68294$ . The cubic term still encloses two regions of area $\tfrac14$ each, and the $0$ records only that their signed areas cancel. As a check on size, $\cos x$ decreases from $1$ to $\cos 1 \approx 0.540$ on $[0, 1]$ , so the integral lies between $2\cos 1 \approx 1.081$ and $2$ . Dropping the factor $2$ or reading the $1$ in $\sin 1$ as one degree would give $0.841$ or $0.035$ , both outside that range.

Check yourself

Evaluate $\int_{-2}^{2} (x^3 + x^2)\,dx$ .

Show the answer

Both terms are continuous on $[-2, 2]$ . The term $x^3$ is odd and $x^2$ is even, so the symmetry rule gives

\begin{aligned} &\int_{-2}^{2} (x^3 + x^2)\,dx \\[0.4em] &\quad = 0 + 2\int_0^2 x^2\,dx \\[0.4em] &\quad = 2 \cdot \frac{8}{3} = \frac{16}{3}. \end{aligned}

A direct antiderivative gives the same value, $(4 + \tfrac83) - (4 - \tfrac83) = \tfrac{16}{3}$ .

Evaluate the integral

\int_{-\pi/4}^{\pi/4} \bigl(x^4 \sin x + \sec^2 x\bigr)\,dx.

Show the answer

Both terms are continuous on $[-\tfrac{\pi}{4}, \tfrac{\pi}{4}]$ , where $\cos x > 0$ . The product $x^4 \sin x$ is an even function times an odd function, so it is odd and its integral over the symmetric interval is $0$ . The function $\sec^2 x = 1/\cos^2 x$ is even because $\cos x$ is even. Hence

\begin{aligned} &\int_{-\pi/4}^{\pi/4} \bigl(x^4 \sin x + \sec^2 x\bigr)\,dx \\[0.4em] &\quad = 2\int_0^{\pi/4} \sec^2 x\,dx \\[0.4em] &\quad = 2\bigl[\tan x\bigr]_0^{\pi/4} = 2. \end{aligned}

The term $x^4 \sin x$ would need integration by parts four times, and symmetry removes it without any of that work.

A classmate writes $\int_{-1}^{2} x^3\,dx = 0$ and $\int_{-1}^{1} \frac{1}{x}\,dx = 0$ , because $x^3$ and $1/x$ are odd. What goes wrong in each case? Find the first integral.

Show the answer

The interval $[-1, 2]$ is not symmetric about $0$ . Its symmetric part $[-1, 1]$ does contribute $0$ , but the unmatched piece $[1, 2]$ remains, so

\begin{aligned} \int_{-1}^{2} x^3\,dx &= \int_1^2 x^3\,dx \\[0.4em] &= \frac{16 - 1}{4} = \frac{15}{4}. \end{aligned}

In the second case the interval is symmetric and $1/x$ is odd, but $1/x$ is undefined at $0$ and unbounded near it, so it is not integrable on $[-1, 1]$ and the rule does not apply. Treated as an improper integral, the right half already fails,

\begin{aligned} \int_0^1 \frac{dx}{x} &= \lim_{t \to 0^+} \int_t^1 \frac{dx}{x} \\[0.4em] &= \lim_{t \to 0^+} (-\ln t) = \infty, \end{aligned}

so the integral diverges. Cutting out a gap $(-t, t)$ around $0$ and letting $t \to 0^+$ does give $0$ , a number called the Cauchy principal value, but that is a different quantity from the integral.