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How can this integral be equal to $0$?

$$\int_{-1}^{1} \left(x^2 - \frac{1}{3}\right) \; dx$$

The integrand is even, and the bounds are symmetric, how could this be $0$?

Michael Hardy
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gem
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    Why not? It is even, not positive. The fact that it is even simply means it is twice $\int_0^1 x^2- \frac{1}{3}$ which is also zero.... – N. S. Mar 19 '12 at 03:55
  • I know you're asking related to symmetry. But here's the plot. Visually, it seems about right that the areas on approximately $x \in [-1, -0.5],$ $[-0.5, 0],$ $[0, 0.5],$ $[0.5, 1]$ almost cancel each others. –  Mar 19 '12 at 04:06
  • Haha. I've just seen Emile Okada's answer. –  Mar 19 '12 at 04:08

4 Answers4

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There is nothing that prevents an even function from having a $0$ integral over a symmetric interval, so long as the integral from $0$ to $a$ is also $0$.

Note that $\int_0^1\left(x^2-\frac{1}{3}\right)\,dx = 0$ as well (draw the function to see why this is reasonable), so there is no problem here: $$\int_{-1}^1\left(x^2 - \frac{1}{3}\right)\,dx = 2\int_0^1\left(x^2-\frac{1}{3}\right)\,dx = 2(0) = 0.$$

For another function that is even and has integral $0$ on $[-1,1]$, take $y=|x|-\frac{1}{2}$.

Arturo Magidin
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enter image description here

This may help.

Edit: If it still isn't clear try $\displaystyle\int_0^{\frac{1}{2}}x^2-\frac{1}{3}dx$ and then $\displaystyle\int_{\frac{1}{2}}^{1}x^2-\frac{1}{3}dx$ and you will see that one is the negative of the other and thus will cancel out. They may not look like they have the same area, but they do. Just like a circle with radius $5$ may not look like it has the same area as an rectangle with width $5$ and length $5\pi$, but they still do.

E.O.
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Even integrals can easily be zero. For example, pick you favorite even integrable function $f(x)$.

Let $$g(x)= f(x) -\frac{1}{2a}\int_{-a}^a f(t) dt \,.$$

Then $g$ is even and

$$\int_{-a}^a g(x) dx =0$$

N. S.
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Why not? It is true that this is twice the integral from $0$ to $1$, by symmetry. But the integral from $0$ to $1$ is $0$.

André Nicolas
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