1. **State the problem:** We need to find the area between the curves given by the integral $$\int_{-1}^{1} [(-2x^2 + 8x + 11) - (8x + 9)] \, dx$$. 2. **Simplify the integrand:** Subtract the functions inside the integral: $$(-2x^2 + 8x + 11) - (8x + 9) = -2x^2 + 8x + 11 - 8x - 9 = -2x^2 + 2$$ 3. **Rewrite the integral:** $$\int_{-1}^{1} (-2x^2 + 2) \, dx$$ 4. **Integrate term-by-term:** $$\int (-2x^2) \, dx = -2 \cdot \frac{x^3}{3} = -\frac{2}{3}x^3$$ $$\int 2 \, dx = 2x$$ So the antiderivative is: $$F(x) = -\frac{2}{3}x^3 + 2x$$ 5. **Evaluate the definite integral:** $$F(1) - F(-1) = \left(-\frac{2}{3}(1)^3 + 2(1)\right) - \left(-\frac{2}{3}(-1)^3 + 2(-1)\right)$$ $$= \left(-\frac{2}{3} + 2\right) - \left(\frac{2}{3} - 2\right)$$ 6. **Simplify the expression:** $$= \left(\frac{6}{3} - \frac{2}{3}\right) - \left(\frac{2}{3} - 2\right) = \frac{4}{3} - \left(\frac{2}{3} - 2\right)$$ $$= \frac{4}{3} - \frac{2}{3} + 2 = \frac{2}{3} + 2 = \frac{2}{3} + \frac{6}{3} = \frac{8}{3}$$ **Final answer:** The area is $$\frac{8}{3}$$.