Some definite integrals can be answered before any antiderivative is found at all, just by recognizing symmetry in the integrand. An even or odd function integrated over a symmetric interval collapses to something far simpler than the general computation would suggest. This shortcut resurfaces constantly once double and triple integrals are set up over regions that are themselves symmetric.
An odd function's graph looks the same after a 180° spin through the origin. Whatever area sits above the axis on the right shows up as an equal area below the axis on the left. Signed area cancels the two exactly. An even function mirrors left-right instead of rotating. Both sides land on the same side of the axis, so they add together instead of canceling. Recognizing which kind of symmetry a graph has tells you the shape of the answer before touching any algebra.
Symmetry, here, is a property of an integrand's graph that lets some definite integrals be answered before any antiderivative is found at all. The even odd shortcuts below are the two patterns worth recognizing on sight: an even or odd function integrated over a symmetric interval collapses to something far simpler than the general computation would suggest.
For an EVEN function (f(-x)=f(x)) on a symmetric interval [-a,a]: \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx. The two symmetric halves contribute equally, so computing one half and doubling it is just as good as computing the whole.
For an ODD function (f(-x)=-f(x)) on the same interval: \int_{-a}^a f(x)\,dx = 0 always. The two halves are exact negatives of each other and cancel completely, with no computation required.
Neither shortcut applies to an interval that isn't symmetric about 0. Neither applies to a function that's neither even nor odd. Most functions are one of the two only after some algebraic massaging, or not at all.
Break the symmetry and watch the shortcut disappear
Set c to 0 — the curve is purely odd. Slide c away from 0 and watch that symmetry (and the shortcut) vanish.
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An odd integrand's running total returns to zero
Drag x to the far right edge — the running total ends back at 0, the signature of an odd integrand over a symmetric interval.
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Worked example
Using odd symmetry to skip the computation
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1
Evaluate \displaystyle\int_{-2}^{2} (x^3-4x)\,dx by recognizing symmetry, without antidifferentiating.
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2
Check each term: (-x)^3=-x^3 and -4(-x)=4x=-(-4x). Both terms are odd, so their sum is odd.
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3
An odd function integrated over a symmetric interval [-a,a] is always zero.
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4
\boxed{\displaystyle\int_{-2}^{2}(x^3-4x)\,dx = 0}. No antiderivative is needed at all.
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Where this goes next
Later chapters: symmetric regions
A double integral over a disk centered at the origin, or a triple integral over a ball, is set up over a region that's symmetric in this sense. An odd integrand over such a region collapses to 0 by the identical argument. That shortcut spares a computation that could otherwise run to several pages of iterated integration.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Given \int_0^3 x^2\,dx = 9, use even symmetry to find \int_{-3}^{3} x^2\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
x^2 is even, so \int_{-3}^3 x^2\,dx = 2\int_0^3 x^2\,dx = 2(9)=18.
Correct answer
Also accepted:
Why
x^2 is an even function -- its integral over [-3,3] is twice its integral over [0,3].
From Calculus II — Integral Calculus & Infinite Series
For an EVEN function f on [-a,a], which shortcut is valid?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That shortcut is for ODD functions, not even ones.
This misses the doubling — both symmetric halves contribute equally, not just one.
A real shortcut does exist for even functions — doubling the half-interval's integral.
Why
An even function's two symmetric halves contribute equally, so doubling one half's integral gives the whole.
An even function's two halves over [-a,a] contribute equally -- how does that turn one half's integral into the whole?
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\int_{-2}^{2}(x^3-4x)\,dx using symmetry.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Both terms are odd, so the whole integrand is odd — an odd function over a symmetric interval always integrates to zero.
Correct answer
Also accepted:
Why
Check whether each term of x^3-4x is odd or even, then think about what an odd integrand does over a symmetric interval.