1. **Problem statement:** Evaluate the double integral $$\iint_R (x+1) \, dA$$ where $R$ is the region bounded by $y=1$ and $y=x^2$ for $x$ in $[-2,2]$. 2. **Set up the integral:** The region $R$ lies between the curves $y=x^2$ (lower curve) and $y=1$ (upper curve). For each fixed $x$ in $[-2,2]$, $y$ ranges from $x^2$ to $1$. So the integral becomes: $$\int_{-2}^2 \int_{x^2}^1 (x+1) \, dy \, dx$$ 3. **Integrate with respect to $y$ first:** Since $x+1$ does not depend on $y$, the inner integral is: $$\int_{x^2}^1 (x+1) \, dy = (x+1)(1 - x^2)$$ 4. **Rewrite the integral:** Now the integral reduces to: $$\int_{-2}^2 (x+1)(1 - x^2) \, dx$$ 5. **Expand the integrand:** $$(x+1)(1 - x^2) = (x+1) - (x+1)x^2 = x + 1 - x^3 - x^2$$ 6. **Simplify the integrand:** $$x + 1 - x^3 - x^2 = -x^3 - x^2 + x + 1$$ 7. **Integrate term-by-term:** $$\int_{-2}^2 (-x^3 - x^2 + x + 1) \, dx = \int_{-2}^2 -x^3 \, dx - \int_{-2}^2 x^2 \, dx + \int_{-2}^2 x \, dx + \int_{-2}^2 1 \, dx$$ 8. **Evaluate each integral:** - $$\int_{-2}^2 -x^3 \, dx = 0$$ (odd function over symmetric interval) - $$\int_{-2}^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^2 = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3}$$ - $$\int_{-2}^2 x \, dx = 0$$ (odd function over symmetric interval) - $$\int_{-2}^2 1 \, dx = 2 - (-2) = 4$$ 9. **Combine results:** $$0 - \frac{16}{3} + 0 + 4 = 4 - \frac{16}{3} = \frac{12}{3} - \frac{16}{3} = -\frac{4}{3}$$ **Final answer:** $$\boxed{-\frac{4}{3}}$$