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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?

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    $\begingroup$ What are the bounds? Can you use MathJax to write out the integral? $\endgroup$ – N. Bar Jul 19 at 19:28
  • $\begingroup$ 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 $0$. $\endgroup$ – John Omielan Jul 19 at 19:28
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$\int_x^{x'} t^3-t\,dt$ is not $\frac14x^4-\frac12x^2$. It's $$\frac14(x')^4-\frac12(x')^2-\frac14x^4+\frac12x^2$$

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$f(x)=x^3-x$ 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 $g(a)-g(-a)=g(a)-g(a)=0$.

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