Area Enclosed Region 90E744

1. **Problem Statement:** Find the area of the region enclosed by the curves $$y = x + 4$$ and $$y = x^2 + x - 3$$. 2. **Formula and Important Rules:** The area between two curves $$y = f(x)$$ and $$y = g(x)$$ from $$x = a$$ to $$x = b$$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ If one curve is always above the other in the interval, the absolute value can be removed by subtracting the lower function from the upper function. 3. **Find the points of intersection:** Set the two functions equal to find limits of integration: $$x + 4 = x^2 + x - 3$$ Simplify: $$x + 4 - x^2 - x + 3 = 0$$ $$-x^2 + 7 = 0$$ Multiply both sides by $$-1$$: $$x^2 - 7 = 0$$ $$x^2 = 7$$ $$x = \pm \sqrt{7}$$ So the curves intersect at $$x = -\sqrt{7}$$ and $$x = \sqrt{7}$$. 4. **Determine which function is on top between the intersection points:** Test at $$x=0$$: $$y_1 = 0 + 4 = 4$$ $$y_2 = 0^2 + 0 - 3 = -3$$ Since $$y = x + 4$$ is above $$y = x^2 + x - 3$$ in the interval $$[-\sqrt{7}, \sqrt{7}]$$, the area is: $$\text{Area} = \int_{-\sqrt{7}}^{\sqrt{7}} \left[(x + 4) - (x^2 + x - 3)\right] dx$$ Simplify the integrand: $$ (x + 4) - (x^2 + x - 3) = x + 4 - x^2 - x + 3 = -x^2 + 7$$ 5. **Calculate the integral:** $$\int_{-\sqrt{7}}^{\sqrt{7}} (-x^2 + 7) \, dx = \int_{-\sqrt{7}}^{\sqrt{7}} -x^2 \, dx + \int_{-\sqrt{7}}^{\sqrt{7}} 7 \, dx$$ Calculate each integral separately: $$\int -x^2 \, dx = -\frac{x^3}{3}$$ $$\int 7 \, dx = 7x$$ Evaluate from $$-\sqrt{7}$$ to $$\sqrt{7}$$: $$\left[-\frac{x^3}{3} + 7x\right]_{-\sqrt{7}}^{\sqrt{7}} = \left(-\frac{(\sqrt{7})^3}{3} + 7\sqrt{7}\right) - \left(-\frac{(-\sqrt{7})^3}{3} + 7(-\sqrt{7})\right)$$ Calculate powers: $$(\sqrt{7})^3 = (7)^{3/2} = 7 \times \sqrt{7}$$ So: $$= \left(-\frac{7\sqrt{7}}{3} + 7\sqrt{7}\right) - \left(-\frac{-7\sqrt{7}}{3} - 7\sqrt{7}\right)$$ Simplify inside parentheses: $$= \left(-\frac{7\sqrt{7}}{3} + 7\sqrt{7}\right) - \left(\frac{7\sqrt{7}}{3} - 7\sqrt{7}\right)$$ Rewrite: $$= \left(7\sqrt{7} - \frac{7\sqrt{7}}{3}\right) - \left(\frac{7\sqrt{7}}{3} - 7\sqrt{7}\right)$$ Calculate each bracket: $$7\sqrt{7} - \frac{7\sqrt{7}}{3} = \frac{21\sqrt{7}}{3} - \frac{7\sqrt{7}}{3} = \frac{14\sqrt{7}}{3}$$ $$\frac{7\sqrt{7}}{3} - 7\sqrt{7} = \frac{7\sqrt{7}}{3} - \frac{21\sqrt{7}}{3} = -\frac{14\sqrt{7}}{3}$$ So the entire expression is: $$\frac{14\sqrt{7}}{3} - (-\frac{14\sqrt{7}}{3}) = \frac{14\sqrt{7}}{3} + \frac{14\sqrt{7}}{3} = \frac{28\sqrt{7}}{3}$$ 6. **Final answer:** $$\boxed{\frac{28\sqrt{7}}{3}}$$ This is the area of the region enclosed by the curves $$y = x + 4$$ and $$y = x^2 + x - 3$$.