Help calculus

$\displaystyle \int_{0}^{3} x^2 dx = 3^2 - 0, \mathrm{which\space is} \ \textcolor{red}{9}$, so the area under the curve from 0 to 3 is 9.

Hi Hipie,

$\displaystyle \int_{0}^{3} x^2 dx\\~\\ =\left[ \frac{x^3}{3} \right]_0^3=\frac{3^3}{3}-0=9u^2$

Yes it is 9.