Find the area of the region under the graph of the function f on the interval [0, 9]. f(x) = 9x - x²

Solution:

Given, the function is f(x) = 9x - x²

We have to find the area of the region under the graph of the function f on the interval [0, 9].

Area = \(\int_{a}^{b}f(x)\, dx\)

So, area = \(\int_{0}^{9}(9x-x^{2})\, dx\)

= \([\frac{9}{2}x^{2}-\frac{x^{3}}{3}]_{0}^{9}\)

=\([\frac{9}{2}(9)^{2}-\frac{(9)^{3}}{3}]-[\frac{9}{2}(0)^{2}-\frac{(0)^{3}}{3}]\\=[\frac{9}{2}(81)-\frac{729}{3}-0]\\=[\frac{729}{2}-\frac{729}{3}]\\=[\frac{729(3-2)}{6}]\\=[\frac{729(1)}{6}]\\=\frac{729}{6}\)

=\(\frac{243}{2}\) square units.

Therefore, the area of the region is \(\frac{243}{2}\) square units.