Evaluate the following integral using the Fundamental Theorem of Calculus. Sketch the graph of the intogrand and shade the region whose net area you have found. \( \int_{-2}^{3}\left(x^{2}-x-6\right) d x=\square \) (Simplify your answer.)

Calculate the integral \( \int_{-2}^{3}(x^{2}-x-6)dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-2}^{3} \left(x^{2}-x-6\right) dx\) - step1: Evaluate the integral: \(\int \left(x^{2}-x-6\right) dx\) - step2: Use properties of integrals: \(\int x^{2} dx+\int -x dx+\int -6 dx\) - step3: Evaluate the integral: \(\frac{x^{3}}{3}+\int -x dx+\int -6 dx\) - step4: Evaluate the integral: \(\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int -6 dx\) - step5: Evaluate the integral: \(\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x\) - step6: Return the limits: \(\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x\right)\bigg |_{-2}^{3}\) - step7: Calculate the value: \(-\frac{125}{6}\) The integral \( \int_{-2}^{3}(x^{2}-x-6)dx \) evaluates to \( -\frac{125}{6} \) or \( -20\frac{5}{6} \) or \( -20.8\dot{3} \). To sketch the graph of the integrand and shade the region whose net area you have found, you would plot the function \( f(x) = x^{2} - x - 6 \) over the interval \( [-2, 3] \) and shade the region below the curve. The net area under the curve from \( -2 \) to \( 3 \) is the value of the integral, which is \( -\frac{125}{6} \).