Area between curves $y=x^3$ and $y=x$ I've tried to done one of my homework problems for several times, but the answer doesn't make sense to me. The question asks to find the area between $y=x^3$ and $y=x$. Those are odd functions, and I'm pretty sure that the area between their graphs should be 0. However, I keep getting 1/2 as my answer. Can anyone please check my work to see if I did something wrong? 
$$
\begin{align}
A &= \int_0^1 \left(x-x^3\right) \,\operatorname{d}\!x + \int_{-1}^0 \left(x^3-x\right) \,\operatorname{d}\!x \\
  &= \left. \left( \frac{x^2}{2} - \frac{x^4}{4} \right) \right\vert_0^1 + \left. \left( \frac{x^4}{4} - \frac{x^2}{2} \right) \right\vert_{-1}^0 \\
  &= \left( \frac{1}{2} - \frac{1}{4} - 0 \right) + \left[ 0 - \left( \frac{1}{4} - \frac{1}{2} \right) \right] \\
  &= \left(\frac{1}{4}\right)+\left[-\left(\frac{-1}{4}\right)\right] \\
  &= \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\end{align}
$$
Thanks ahead!

The area between two curves is always positive. See the below graph.

The area in green and orange is the area you are finding. It is always going to be positive because it exists. When you subtract the two curves, you are finding the area between the curves, regardless of their position relative to the x axis. 
Also, you could have just multiplied one of your integrals by 2 to get the answer because they are odd functions :P
