Source: Chapter 5, Section 5.3 | Calculus (Texas A&M University)
Tags: even function, odd function, symmetric interval, integral shortcuts, parity, signed area cancellation
When integrating over a symmetric interval [−a, a], even and odd functions give you shortcuts. Odd functions integrate to zero (the negative and positive halves cancel). Even functions let you double the integral over [0, a] instead. Recognising parity saves serious time on exams.
Even function
A function satisfying f(−x) = f(x) for all x in its domain. Its graph is symmetric about the y-axis. Examples: x², x⁴, cos x, constants.
Odd function
A function satisfying f(−x) = −f(x) for all x in its domain. Its graph is symmetric about the origin. Examples: x³, x⁵, sin x, tan x, x.
Symmetric interval
An interval of the form [−a, a], centred at the origin. The even/odd shortcuts only apply on symmetric intervals.
To test: replace x with −x in the formula and simplify.
If f(−x) = f(x): even.
If f(−x) = −f(x): odd.
If neither: the function has no parity.
Common examples (Example 5.3.1):
x³: odd (since (−x)³ = −x³)
x⁴: even (since (−x)⁴ = x⁴)
3 (constant): even (3 = 3 regardless of sign)
x + 1: neither (−x + 1 is not x + 1, nor is it −(x + 1))
x² + 1: even
sin x: odd
cos x: even
tan x: odd (sin x / cos x = odd/even = odd)
A useful rule of thumb: products and quotients of functions with known parity follow sign-multiplication rules. Even × even = even, odd × odd = even, even × odd = odd.
On a symmetric interval [−a, a]:
If f is odd:
∫₋ₐᵃ f(x) dx = 0
The area under the curve on [0, a] cancels exactly with the (negative) area on [−a, 0]. The signed areas are equal and opposite.
If f is even:
∫₋ₐᵃ f(x) dx = 2 ∫₀ᵃ f(x) dx
The two halves contribute equally, so you can compute one half and double it.
∫₋₃³ sin(x³) dx
sin(x³) is a composition of two odd functions: sin(u) is odd, and u = x³ is odd. An odd function of an odd function is odd. (Check: sin((−x)³) = sin(−x³) = −sin(x³).)
On the symmetric interval [−3, 3], the integral is 0.
∫₋₁¹ x² dx
x² is even. So ∫₋₁¹ x² dx = 2 ∫₀¹ x² dx = 2 [x³/3]₀¹ = 2/3.
∫₋₅⁵ 1/(x⁵ + x³ + x) dx
The denominator is x⁵ + x³ + x. Check parity: (−x)⁵ + (−x)³ + (−x) = −x⁵ − x³ − x = −(x⁵ + x³ + x). The denominator is odd.
So 1/(odd function) is also odd (since 1/(−g(x)) = −1/g(x)).
On the symmetric interval [−5, 5], the integral is 0.
Even/odd integration rules on [−a, a]:
f odd: ∫₋ₐᵃ f(x) dx = 0
f even: ∫₋ₐᵃ f(x) dx = 2 ∫₀ᵃ f(x) dx
Parity multiplication table:
even × even = even
odd × odd = even
even × odd = odd
⚠️ These shortcuts only work on symmetric intervals [−a, a]. If the interval is not centred at the origin, you cannot apply them.
⚠️ On exams, look for integrands that are clearly odd over symmetric intervals. These are designed to be zero, saving you from a messy anti-derivative.
⚠️ Functions like x + 1 or x² + x are neither even nor odd. You cannot apply the shortcut to the whole function, but you can sometimes split it: x² + x = (even part) + (odd part) and handle each piece separately.
⚠️ Compositions: sin(x³) looks complex, but checking parity takes one line and gives an instant answer. Train yourself to check parity before reaching for anti-derivatives.
Q: Is x⁷ + x³ even, odd, or neither?
A: Odd. (−x)⁷ + (−x)³ = −x⁷ − x³ = −(x⁷ + x³).
Q: Evaluate ∫₋₂² (x⁵ − 3x) dx.
A: x⁵ − 3x is odd (check: −x⁵ + 3x = −(x⁵ − 3x)). Symmetric interval. The integral is 0.
Q: Evaluate ∫₋₁¹ (x⁴ + x³) dx.
A: Split into even and odd parts. ∫₋₁¹ x⁴ dx + ∫₋₁¹ x³ dx. The second integral is 0 (x³ is odd). The first: 2 ∫₀¹ x⁴ dx = 2 [x⁵/5]₀¹ = 2/5.
Q: Can you use the odd-function shortcut to evaluate ∫₀³ sin x dx?
A: No. The interval [0, 3] is not symmetric about the origin. The shortcut requires [−a, a].
even function, odd function, symmetric interval, parity of functions, integral of odd function, integral of even function, y-axis symmetry, origin symmetry, signed area cancellation, integration shortcut, function symmetry, composition parity