1. **State the problem:** Find the area of the shaded region bounded by the curves $$x = y^2 - 4$$ and $$x = e^y$$, and the horizontal lines $$y = 1$$ and $$y = -1$$. 2. **Set up the integral:** The area between two curves with respect to $$y$$ is given by $$\text{Area} = \int_{y=a}^{y=b} \left( x_{\text{right}} - x_{\text{left}} \right) dy$$ where $$x_{\text{right}}$$ and $$x_{\text{left}}$$ are the rightmost and leftmost functions in terms of $$y$$. 3. **Identify the bounds and functions:** - Lower bound: $$y = -1$$ - Upper bound: $$y = 1$$ - Left curve: $$x = y^2 - 4$$ (parabola) - Right curve: $$x = e^y$$ (exponential) 4. **Write the integral:** $$\text{Area} = \int_{-1}^{1} \left( e^y - (y^2 - 4) \right) dy = \int_{-1}^{1} \left( e^y - y^2 + 4 \right) dy$$ 5. **Evaluate the integral:** Break the integral into parts: $$\int_{-1}^{1} e^y dy - \int_{-1}^{1} y^2 dy + \int_{-1}^{1} 4 dy$$ Calculate each: - $$\int_{-1}^{1} e^y dy = e^y \Big|_{-1}^{1} = e^1 - e^{-1} = e - \frac{1}{e}$$ - $$\int_{-1}^{1} y^2 dy = \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$$ - $$\int_{-1}^{1} 4 dy = 4y \Big|_{-1}^{1} = 4(1) - 4(-1) = 8$$ 6. **Combine results:** $$\text{Area} = \left(e - \frac{1}{e}\right) - \frac{2}{3} + 8 = e - \frac{1}{e} + \frac{22}{3}$$ 7. **Final answer:** $$\boxed{\text{Area} = e - \frac{1}{e} + \frac{22}{3}}$$