1. The problem involves finding the area between two curves using integral calculus. 2. The general formula for the area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Usually, if $f(x) \geq g(x)$ on $[a,b]$, then: $$\text{Area} = \int_a^b (f(x) - g(x)) \, dx$$ 3. For each problem, identify the curves, find the intersection points (limits of integration), and compute the integral. 4. Problem 1: Find the area between $y=x^2$ and $y=x+2$. - Find intersection points by solving $x^2 = x+2$: $$x^2 - x - 2 = 0$$ $$ (x-2)(x+1) = 0 \Rightarrow x=2, x=-1$$ - Set up the integral: $$\int_{-1}^2 ((x+2) - x^2) \, dx$$ - Compute the integral: $$\int_{-1}^2 (x+2 - x^2) \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2$$ - Evaluate at limits: $$\left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right) = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$$ $$= \left( 6 - \frac{8}{3} \right) - \left( -\frac{3}{2} + \frac{1}{3} \right) = \left( \frac{18}{3} - \frac{8}{3} \right) - \left( -\frac{9}{6} + \frac{2}{6} \right) = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$$ Final answer for problem 1: Area = 4.5 5. The graph shows the parabola $y=x^2$ and the line $y=x+2$ intersecting at $x=-1$ and $x=2$, with the shaded region between them.