1. **State the problem:** We want to evaluate the double integral $$\iint_D (2y + 4x) \, dA$$ where the region $D$ is bounded by the parabolas $y = 2x^2$ (lower boundary) and $y = 1 + x^2$ (upper boundary), with $x$ ranging from $-1$ to $1$. 2. **Set up the integral:** Since $y$ varies between $2x^2$ and $1 + x^2$ for each fixed $x$ in $[-1,1]$, the integral becomes: $$\int_{x=-1}^1 \int_{y=2x^2}^{1+x^2} (2y + 4x) \, dy \, dx$$ 3. **Integrate with respect to $y$ first:** Use the integral formula: $$\int (2y + 4x) \, dy = \int 2y \, dy + \int 4x \, dy = y^2 + 4xy + C$$ Evaluate from $y=2x^2$ to $y=1+x^2$: $$\left[y^2 + 4xy\right]_{y=2x^2}^{1+x^2} = (1 + x^2)^2 + 4x(1 + x^2) - \left((2x^2)^2 + 4x(2x^2)\right)$$ 4. **Simplify the expression:** Calculate each term: - $(1 + x^2)^2 = 1 + 2x^2 + x^4$ - $4x(1 + x^2) = 4x + 4x^3$ - $(2x^2)^2 = 4x^4$ - $4x(2x^2) = 8x^3$ So the expression becomes: $$1 + 2x^2 + x^4 + 4x + 4x^3 - (4x^4 + 8x^3) = 1 + 2x^2 + x^4 + 4x + 4x^3 - 4x^4 - 8x^3$$ Combine like terms: $$1 + 2x^2 + (x^4 - 4x^4) + (4x^3 - 8x^3) + 4x = 1 + 2x^2 - 3x^4 - 4x^3 + 4x$$ 5. **Integrate with respect to $x$ from $-1$ to $1$:** $$\int_{-1}^1 (1 + 2x^2 - 3x^4 - 4x^3 + 4x) \, dx$$ 6. **Evaluate each integral term:** - $\int_{-1}^1 1 \, dx = [x]_{-1}^1 = 1 - (-1) = 2$ - $\int_{-1}^1 2x^2 \, dx = 2 \int_{-1}^1 x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{-1}^1 = 2 \left( \frac{1}{3} - \frac{-1}{3} \right) = 2 \times \frac{2}{3} = \frac{4}{3}$ - $\int_{-1}^1 -3x^4 \, dx = -3 \int_{-1}^1 x^4 \, dx = -3 \left[ \frac{x^5}{5} \right]_{-1}^1 = -3 \left( \frac{1}{5} - \frac{-1}{5} \right) = -3 \times \frac{2}{5} = -\frac{6}{5}$ - $\int_{-1}^1 -4x^3 \, dx = -4 \left[ \frac{x^4}{4} \right]_{-1}^1 = -4 (\frac{1}{4} - \frac{1}{4}) = 0$ (since $x^4$ is even, but $x^3$ is odd, integral of odd function over symmetric interval is zero) - $\int_{-1}^1 4x \, dx = 4 \left[ \frac{x^2}{2} \right]_{-1}^1 = 4 \times \frac{1}{2} - 4 \times \frac{1}{2} = 0$ (integral of odd function over symmetric interval is zero) 7. **Sum all results:** $$2 + \frac{4}{3} - \frac{6}{5} + 0 + 0 = 2 + \frac{4}{3} - \frac{6}{5}$$ Find common denominator 15: $$2 = \frac{30}{15}, \quad \frac{4}{3} = \frac{20}{15}, \quad \frac{6}{5} = \frac{18}{15}$$ So: $$\frac{30}{15} + \frac{20}{15} - \frac{18}{15} = \frac{32}{15}$$ **Final answer:** $$\boxed{\frac{32}{15}}$$