# Finding the total area between curve and x-axis over a closed interval

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?

• Do you mean total signed area or total area? – user658409 Nov 30 '19 at 22:48
• @user658409 I'm not sure, that was all the information I was given in the problem. – maxgonz Nov 30 '19 at 22:50

$$\int_{0}^{3}(x^3-x)dx=\left[\frac{x^4}4-\frac{x^2}2\right]_{0}^{3}=\frac{81}4-\frac 9 2=\frac{63}4$$
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
$$A=\int_{0}^{3}|x^3-x|dx=\int_{0}^{1}(x-x^3) \,dx+\int_{1}^{3} (x^3-x)\, dx=\frac{65}4$$
• @maxgonz By the defined integral we compute with + sign the area above the $x$ axis and with - sign the area below. For e.g. think to $$\int_{-1}^1 x dx =0$$ but $$\int_{-1}^1 |x| dx =1$$ – user Nov 30 '19 at 23:04