91Ó°ÊÓ

Firstly, we need to find the antiderivative of the integrand, \(x^2 - x - 6\), to evaluate the definite integral: $$F(x) = \int (x^{2}-x-6) dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + C$$ Now, we can use the FTC to evaluate the definite integral: $$\int_{-2}^{3} (x^{2}-x-6) dx = F(3) - F(-2)$$

Now, we will go ahead and find the values of our antiderivative at the given limits: $$F(3) = \frac{1}{3}(3)^3 - \frac{1}{2}(3)^2 - 6(3)$$ $$F(3) = \frac{27}{3} - \frac{9}{2} - 18 = 9 - \frac{9}{2} - 18 = 9 - 4.5 - 18 = -13.5$$ $$F(-2) = \frac{1}{3}(-2)^3 -\frac{1}{2}(-2)^2 - 6(-2)$$ $$F(-2) = \frac{-8}{3} - \frac{4}{2} + 12 = \frac{-8}{3} - 2 + 12 = -\frac{2}{3} + 10 = \frac{28}{3}$$ Then, we subtract F(-2) from F(3): $$\int_{-2}^{3} (x^{2}-x-6) dx = F(3) - F(-2) = -13.5 - \frac{28}{3} = -13.5 - 9.333\dots = -22.833\dots$$

To find the region that corresponds to the definite integral, first sketch \(f(x) = x^2 - x - 6\) using the zeros of the quadratic equation to find the x-intercepts: \(x^2 - x - 6 = 0\) can be factored as \((x+2)(x-3) = 0\), so the x-intercepts are \(-2\) and \(3\). We see that the function is a parabola with the vertex located at the average of the x-intercepts and \(y\)-value found by substituting the value of the x-coordinate of the vertex into the function. Now, knowing the integral has a negative value, it's important to note that negative area corresponds to the part of the function that is below the x-axis. In our case, this region will be bounded by the x-axis, the two x-intercepts: -2 and 3, and the curve \(f(x) = x^2 - x - 6\) from \(x=-2\) to \(x=3\). Shade this region in your graph. By following these steps, we have successfully evaluated the integral and found the net area under the curve. The value of the integral is approximately -22.833, and the shaded region in the graph represents this net area.