Questions: Find the area of the banded region for the given graphs

Solution Steps

To find the intersection points, we set the two equations equal to each other: $x^2 - 1 = x + 1$ $x^2 - x - 2 = 0$ $(x-2)(x+1) = 0$ $x = 2$ or $x = -1$

The area of the bounded region is given by the integral of the top function minus the bottom function, from the left intersection point to the right intersection point. In this case, the top function is $y = x+1$ and the bottom function is $y = x^2 - 1$. The left intersection point is $x = -1$ and the right intersection point is $x=2$. Area $= \int_{-1}^2 [(x+1) - (x^2 - 1)] dx$

Area $= \int_{-1}^2 (-x^2 + x + 2) dx$ Area $= [-\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x]_{-1}^2$ Area $= (-\frac{8}{3} + 2 + 4) - (\frac{1}{3} + \frac{1}{2} - 2)$ Area $= -\frac{8}{3} + 6 - \frac{1}{3} - \frac{1}{2} + 2$ Area $= -3 + 8 - \frac{1}{2} = 5 - \frac{1}{2} = \frac{9}{2}$

Final Answer: The area of the bounded region is 9/2 or 4.5.