1. **Problem:** Find the area of the region bounded by the curves $y = x^2 + 2x$, $y = x + 6$, and the vertical lines $x = -4$ and $x = 4$. Set up the area as the sum of three integrals and find the total area. 2. **Step 1: Find points of intersection between the curves $y = x^2 + 2x$ and $y = x + 6$** Set $x^2 + 2x = x + 6$: $$x^2 + 2x = x + 6$$ $$x^2 + 2x - x - 6 = 0$$ $$x^2 + x - 6 = 0$$ Factor: $$ (x + 3)(x - 2) = 0 $$ So, $x = -3$ or $x = 2$. 3. **Step 2: Determine which curve is on top in each interval** - For $x$ in $[-4, -3]$, test $x = -3.5$: - $y_1 = (-3.5)^2 + 2(-3.5) = 12.25 - 7 = 5.25$ - $y_2 = -3.5 + 6 = 2.5$ So $y = x^2 + 2x$ is above $y = x + 6$ here. - For $x$ in $[-3, 2]$, test $x = 0$: - $y_1 = 0 + 0 = 0$ - $y_2 = 0 + 6 = 6$ So $y = x + 6$ is above $y = x^2 + 2x$ here. - For $x$ in $[2, 4]$, test $x = 3$: - $y_1 = 9 + 6 = 15$ - $y_2 = 3 + 6 = 9$ So $y = x^2 + 2x$ is above $y = x + 6$ here. 4. **Step 3: Set up the area as sum of three integrals** $$\text{Area} = \int_{-4}^{-3} \left[(x^2 + 2x) - (x + 6)\right] dx + \int_{-3}^{2} \left[(x + 6) - (x^2 + 2x)\right] dx + \int_{2}^{4} \left[(x^2 + 2x) - (x + 6)\right] dx$$ Simplify the integrands: - For $[-4, -3]$ and $[2,4]$: $$ (x^2 + 2x) - (x + 6) = x^2 + 2x - x - 6 = x^2 + x - 6 $$ - For $[-3, 2]$: $$ (x + 6) - (x^2 + 2x) = x + 6 - x^2 - 2x = -x^2 - x + 6 $$ 5. **Step 4: Compute each integral** - First integral: $$\int_{-4}^{-3} (x^2 + x - 6) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 6x \right]_{-4}^{-3}$$ Calculate: At $x = -3$: $$\frac{(-3)^3}{3} + \frac{(-3)^2}{2} - 6(-3) = \frac{-27}{3} + \frac{9}{2} + 18 = -9 + 4.5 + 18 = 13.5$$ At $x = -4$: $$\frac{(-4)^3}{3} + \frac{(-4)^2}{2} - 6(-4) = \frac{-64}{3} + 8 + 24 = -21.3333 + 8 + 24 = 10.6667$$ Difference: $$13.5 - 10.6667 = 2.8333$$ - Second integral: $$\int_{-3}^{2} (-x^2 - x + 6) dx = \left[ -\frac{x^3}{3} - \frac{x^2}{2} + 6x \right]_{-3}^{2}$$ Calculate: At $x = 2$: $$ -\frac{8}{3} - 2 + 12 = -2.6667 - 2 + 12 = 7.3333$$ At $x = -3$: $$ -\frac{-27}{3} - \frac{9}{2} + (-18) = 9 - 4.5 - 18 = -13.5$$ Difference: $$7.3333 - (-13.5) = 20.8333$$ - Third integral: $$\int_{2}^{4} (x^2 + x - 6) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} - 6x \right]_{2}^{4}$$ Calculate: At $x = 4$: $$\frac{64}{3} + 8 - 24 = 21.3333 + 8 - 24 = 5.3333$$ At $x = 2$: $$\frac{8}{3} + 2 - 12 = 2.6667 + 2 - 12 = -7.3333$$ Difference: $$5.3333 - (-7.3333) = 12.6667$$ 6. **Step 5: Sum all areas** $$2.8333 + 20.8333 + 12.6667 = 36.3333$$ So, the total shaded area is approximately $36.33$ square units. --- **Final answer:** $$\boxed{\text{Area} \approx 36.33}$$