1. **State the problem:** Find the area of the region $R$ enclosed by the curves $y = -x + 1$, $y = e^x$, $y = 0$, and the vertical line $x = 2$ using a double integral. 2. **Understand the region:** The region is bounded below by $y=0$, above by $y = -x + 1$ and $y = e^x$, and vertically between $x=0$ (where the curves intersect) and $x=2$. 3. **Find intersection points:** Solve $-x + 1 = e^x$ to find the intersection in $[0,2]$. 4. **Check intersection at $x=0$:** $-0 + 1 = 1$ and $e^0 = 1$, so they intersect at $(0,1)$. 5. **Check if $-x + 1$ and $e^x$ intersect again between 0 and 2:** Since $-x + 1$ decreases and $e^x$ increases, they only intersect at $x=0$. 6. **Determine which curve is on top between 0 and 2:** For $x>0$, $e^x > -x + 1$, so $y = e^x$ is above $y = -x + 1$ only at $x=0$, but for $x>0$, $e^x$ is above $-x + 1$. 7. **Set up the integral:** The region is bounded by $y=0$ below and $y = -x + 1$ for $x$ in $[0,1]$ (since $-x+1$ hits zero at $x=1$), and by $y=0$ below and $y = e^x$ for $x$ in $[1,2]$. 8. **Calculate area:** $$\text{Area} = \int_0^1 (-x + 1) \, dx + \int_1^2 e^x \, dx$$ 9. **Evaluate the integrals:** $$\int_0^1 (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_0^1 = \left(-\frac{1}{2} + 1\right) - 0 = \frac{1}{2}$$ $$\int_1^2 e^x \, dx = \left[ e^x \right]_1^2 = e^2 - e$$ 10. **Sum the areas:** $$\text{Area} = \frac{1}{2} + e^2 - e$$ **Final answer:** $$\boxed{\frac{1}{2} + e^2 - e}$$