# What is the area under the curve in the interval [-3, 3] for the function f(x)=x^3-9x^2?

I found $162$ units of area.
$A r e a = {\int}_{-} {3}^{3} \left({x}^{3} - 9 {x}^{2}\right) \mathrm{dx} = {\left[{x}^{4} / 4 - {\cancel{9}}^{3} {x}^{3} / \cancel{3}\right]}_{-} {3}^{3} = - \frac{243}{4} - \frac{405}{4} = - 162$
Negative as below the $x$ axis.