# Taking symmetric integral not 0 for symmetric function?

Let there be a function

$$f(x) = x^3 - x$$

Ffter calculating the integral, I get this:

$$1/4x^4 - 1/2x^2 = \text{Integral from x to x}$$

If I now put in $$1/3$$ and $$-1/3$$ I will not get $$0$$ as a result even though it is a symmetric function. The integral is the same as an online calculator put out. And I checked the calculation 5 times now. Whats wrong?

• What are the bounds? Can you use MathJax to write out the integral? – N. Bar Jul 19 at 19:28
• Welcome to MSE. Are you plugging in $1/3$ and then subtracting the value when you plug in $-1/3$? That is what you should be doing and, if you do, you should get $0$. – John Omielan Jul 19 at 19:28

$$\int_x^{x'} t^3-t\,dt$$ is not $$\frac14x^4-\frac12x^2$$. It's $$\frac14(x')^4-\frac12(x')^2-\frac14x^4+\frac12x^2$$
$$f(x)=x^3-x$$ has odd symmetry: $$f(x)=-f(-x)$$.
Its antiderivative $$g(x)=\dfrac{x^4}4-\dfrac{x^2}2$$ has even symmetry: $$g(x)=g(-x)$$.
The integral $$\int_{-a}^a f(x)dx$$ is therefore $$g(a)-g(-a)=g(a)-g(a)=0$$.