# How do you find the area under the graph of f(x)=x^2 on the interval [-3,3] ?

This is an integration problem. We will find the area under the curve ${x}^{2}$ over the interval [-3,3]. This included both -3 and 3 because of the square brackets.
The integration of ${x}^{2}$ is found by incrementing the power to $3$ and using $3$ as the denominator.
${\int}_{a}^{b} {x}^{n} \mathrm{dx} = {\left[{x}^{n + 1} / \left(n + 1\right)\right]}_{a}^{b} = {b}^{n + 1} / \left(n + 1\right) - {a}^{n + 1} / \left(n + 1\right)$
${\int}_{-} {3}^{3} {x}^{2} \mathrm{dx} = {\left[{x}^{3} / 3\right]}_{-} {3}^{3} = \left[{\left(3\right)}^{3} / 3 - {\left(- 3\right)}^{3} / 3\right] = \frac{27}{3} - \frac{- 27}{3} = \frac{27}{3} + \frac{27}{3} = 9 + 9 = 18$