Area Between Functions B0C41C

1. **State the problem:** Find the area of the region enclosed by the line $$y = x + 4$$ and the parabola $$y = x^2 + x - 5$$. 2. **Find the points of intersection:** Set the two functions equal to find limits $$a$$ and $$b$$: $$x + 4 = x^2 + x - 5$$ Simplify: $$x + 4 - x^2 - x + 5 = 0$$ $$-x^2 + 9 = 0$$ Multiply both sides by $$-1$$: $$x^2 - 9 = 0$$ Factor: $$(x - 3)(x + 3) = 0$$ So, $$x = 3$$ or $$x = -3$$. 3. **Set up the integral for the area:** The area between two curves $$y = f(x)$$ and $$y = g(x)$$ from $$a$$ to $$b$$ is: $$\text{Area} = \int_a^b (f(x) - g(x)) \, dx$$ Here, the line $$y = x + 4$$ is above the parabola $$y = x^2 + x - 5$$ between $$x = -3$$ and $$x = 3$$. 4. **Write the integral:** $$\text{Area} = \int_{-3}^3 \big((x + 4) - (x^2 + x - 5)\big) \, dx$$ Simplify the integrand: $$ (x + 4) - (x^2 + x - 5) = x + 4 - x^2 - x + 5 = -x^2 + 9$$ 5. **Evaluate the integral:** $$\int_{-3}^3 (-x^2 + 9) \, dx = \int_{-3}^3 -x^2 \, dx + \int_{-3}^3 9 \, dx$$ Calculate each: $$\int_{-3}^3 -x^2 \, dx = - \int_{-3}^3 x^2 \, dx = - \left[ \frac{x^3}{3} \right]_{-3}^3 = - \left( \frac{3^3}{3} - \frac{(-3)^3}{3} \right) = - \left( 9 - (-9) \right) = -18$$ $$\int_{-3}^3 9 \, dx = 9[x]_{-3}^3 = 9(3 - (-3)) = 9 \times 6 = 54$$ 6. **Sum the results:** $$-18 + 54 = 36$$ 7. **Final answer:** The area enclosed between the line and the parabola is $$\boxed{36}$$.