Does the integral$\displaystyle \int^3_0(x^2-1)dx$ represent the total area bounded by $\displaystyle f ( x ) = x^2 - 1$ and the x-axis , between x = 0 and x = 3? or is the area larger?
$\displaystyle x^2-1=0\Rightarrow{x=\pm{1}}$
1 being the only relevant result since -1 is outside of our domain
$\displaystyle \therefore\int_0^{3}|x^2-1|dx=\int_1^{3}x^2-1dx-\int_0^{1}x^2-1dx=\frac{22}{3}\approx{7.3333}$
Now
$\displaystyle \int_0^{3}x^2-1dx=\bigg[\frac{x^3}{3}-x\bigg]\bigg|_{x=0}^{3}=6$
So now what can you conclude?