Sketch the region bounded by the graphs of the equations. Sketch the region bounded by the graphs of the equations. Find the area of the region. \( \left.\qquad \begin{array}{r}f(x)=x^{2}+6 x \\ g(x)\end{array}\right)=x+6 \)

To find the area of the region bounded by the graphs of the equations \( f(x) = x^2 + 6x \) and \( g(x) = x + 6 \), we need to determine the intersection points of the two graphs and then calculate the area of the region. First, let's find the intersection points by setting the two equations equal to each other: \[ x^2 + 6x = x + 6 \] Simplifying the equation: \[ x^2 + 5x - 6 = 0 \] Now, we can solve this quadratic equation to find the intersection points. Let's solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \( a = 1 \), \( b = 5 \), and \( c = -6 \). Substitute these values into the quadratic formula: \[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-6)}}{2(1)} \] \[ x = \frac{-5 \pm \sqrt{25 + 24}}{2} \] \[ x = \frac{-5 \pm \sqrt{49}}{2} \] \[ x = \frac{-5 \pm 7}{2} \] Therefore, the intersection points are \( x = 1 \) and \( x = -6 \). Now, we can sketch the region bounded by the graphs of the equations and calculate the area of the region. The area of the region can be calculated by integrating the difference between the two functions over the interval from the smaller intersection point to the larger intersection point. \[ \text{Area} = \int_{-6}^{1} (x + 6) - (x^2 + 6x) \, dx \] Let's calculate the area of the region. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-6}^{1} \left(\left(x+6\right)-\left(x^{2}+6x\right)\right) dx\) - step1: Remove the parentheses: \(\int_{-6}^{1} \left(x+6-\left(x^{2}+6x\right)\right) dx\) - step2: Subtract the terms: \(\int_{-6}^{1} \left(-5x+6-x^{2}\right) dx\) - step3: Evaluate the integral: \(\int \left(-5x+6-x^{2}\right) dx\) - step4: Use properties of integrals: \(\int -5x dx+\int 6 dx+\int -x^{2} dx\) - step5: Evaluate the integral: \(-\frac{5x^{2}}{2}+\int 6 dx+\int -x^{2} dx\) - step6: Evaluate the integral: \(-\frac{5x^{2}}{2}+6x+\int -x^{2} dx\) - step7: Evaluate the integral: \(-\frac{5x^{2}}{2}+6x-\frac{x^{3}}{3}\) - step8: Return the limits: \(\left(-\frac{5x^{2}}{2}+6x-\frac{x^{3}}{3}\right)\bigg |_{-6}^{1}\) - step9: Calculate the value: \(\frac{343}{6}\) The area of the region bounded by the graphs of the equations \( f(x) = x^2 + 6x \) and \( g(x) = x + 6 \) is \( \frac{343}{6} \) square units.