Find the total area between $f(x)=x^3-x$ and the x-axis on the interval $[0,3]$.
Here is my work: $$\int_{0}^{3}(x^3-x)dx$$ $$\left(\frac{(3)^4}{4}-\frac{(3)^2}{2}\right) - \left(\frac{(0)^4}{4}-\frac{(0)^2}{2}\right) = \frac{63}{4}$$
When I submitted this answer, it was marked as incorrect, why?
Your calculation is correct indeed we have
$$\int_{0}^{3}(x^3-x)dx=\left[\frac{x^4}4-\frac{x^2}2\right]_{0}^{3}=\frac{81}4-\frac 9 2=\frac{63}4$$
but maybe you are requested to find not the signed area but the effective area and since $x^3-x=x(x-1)(x+1)$ changes sign in the interval we have
$$A=\int_{0}^{3}|x^3-x|dx=\int_{0}^{1}(x-x^3) \,dx+\int_{1}^{3} (x^3-x)\, dx=\frac{65}4$$