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?

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

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

$$A=\int_{0}^{3}|x^3-x|dx=\int_{0}^{1}(x-x^3) \,dx+\int_{1}^{3} (x^3-x)\, dx=\frac{65}4$$

  • $\begingroup$ That answer was incorrect when I submitted it. $\endgroup$ – maxgonz Nov 30 '19 at 22:49
  • $\begingroup$ Ah ok maybe you were requested to find the effective area. I fix that. $\endgroup$ – user Nov 30 '19 at 22:50
  • $\begingroup$ @user505767 So the effective area is the area of each critical point to the to x-axis added together? And what would be the signed area? $\endgroup$ – maxgonz Nov 30 '19 at 23:02
  • $\begingroup$ @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 $$\int_{-1}^1 |x| dx =1$$ $\endgroup$ – user Nov 30 '19 at 23:04

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