1. Problem: Find the area of region R1 under the curve $y = x^2$ from $x = -2$ to $x = 0$. 2. Formula: The area under a curve $y = f(x)$ from $a$ to $b$ is given by the definite integral $$\text{Area} = \int_a^b f(x)\,dx.$$ Important: Since $y = x^2$ is always positive, the integral directly gives the area. 3. Calculate the area of R1: $$\int_{-2}^0 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^0 = \frac{0^3}{3} - \frac{(-2)^3}{3} = 0 - \frac{-8}{3} = \frac{8}{3}.$$ 4. Explanation: We integrated $x^2$ from $-2$ to $0$. The negative lower limit cubed is negative, but subtracting a negative gives a positive area. 5. Final answer for R1 area: $\boxed{\frac{8}{3}}$. Note: The user asked multiple questions but per instructions, only the first problem is solved here.