91Ó°ÊÓ

In order to apply the Fundamental Theorem of Calculus, we need to determine an antiderivative for the integrand function, \(x^2-9\). This is a simple polynomial function, and its antiderivative can be found by applying the power rule of integration: $$F(x) = \int(x^2 - 9) dx = \frac{x^3}{3} - 9x + C$$

Now that we have an antiderivative, \(F(x)\), we can apply the Fundamental Theorem of Calculus to compute the definite integral: $$\int_{0}^{5}\left(x^{2}-9\right) d x = F(5) - F(0)$$ Now we will plug in the values 5 and 0 into \(F(x)=\frac{x^3}{3} - 9x+C\): $$F(5) = \frac{5^3}{3} - 9(5) = \frac{125}{3} - 45$$ $$F(0) = \frac{0^3}{3} - 9(0) = 0$$ Therefore, the definite integral evaluates to: $$\int_{0}^{5}\left(x^{2}-9\right) d x = F(5) - F(0) = \left(\frac{125}{3} - 45\right) - 0= \frac{25}{3}$$

To sketch the graph of the integrand function, \(x^2 - 9\), first, identify the x-intercepts (when the function equals zero): $$x^2 - 9 = 0 \Rightarrow x = \pm 3 $$ Noting that this is a quadratic function with a positive leading coefficient, the graph is a parabola opening upwards, crossing the x-axis at x=-3 and x=3. By evaluation, we find that the function is negative for x between 0 and 3, and positive for x between 3 and 5. Therefore, the net area is the area below the x-axis, subtracted from the area above the x-axis. To visualize this, draw the graph of the quadratic function, and shade the region under the curve between x=0 and x=3 as well as the region above the curve between x=3 and x=5. The total area, as computed in the previous steps, is \(\frac{25}{3}\).