Even and odd functions

even: f(−t) = f(t)   odd: f(−t) = −f(t)

t

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Choose a function from the dropdown: the even functions are listed first, then the odd functions.

Points marks \(f(t)\) and \(f(-t)\) for the value of \(t\) on the slider. For an even function the two points are at the same height; for an odd function they are at equal and opposite heights.

Areas shades the area between the graph and the \(t\)-axis over \([-t, 0]\) and \([0, t]\): blue above the axis, red below. It is not available for functions that are undefined somewhere in the interval, such as \(1/t\).

Animate: the part of the graph with \(t \lt 0\) fades away, and a copy of the part with \(t \gt 0\) moves onto it, leaving a faint shadow behind. Shaded areas move with the graph.

Overview

A function \(f\) is even if

\[ f(-t) = f(t) \quad \text{for all } t, \]

and odd if

\[ f(-t) = -f(t) \quad \text{for all } t. \]

Taking \(t = 0\) shows that an odd function defined at \(0\) has \(f(0) = 0\).

Symmetry of the graph

Integrals over \([-a, a]\)

\[ \begin{aligned} f \text{ even:} \quad & \int_{-a}^{a} f(t) \, dt = 2 \int_0^a f(t) \, dt \\ f \text{ odd:} \quad & \int_{-a}^{a} f(t) \, dt = 0 \end{aligned} \]

For an even function the areas over \([-a, 0]\) and \([0, a]\) are equal; for an odd function they are equal in size and opposite in sign, so they cancel. The integral must exist: \(\int_{-1}^{1} \frac{1}{t^2} \, dt\) diverges, and \(\int_{-1}^{1} \frac{1}{t} \, dt\) is not \(0\), because it does not exist.

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