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 the curves $y = -x + 1$ 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:** At $x=1$, $-1 + 1=0$, $e^1 = e > 0$, so $e^x > -x + 1$ at $x=1$. At $x=2$, $-2 + 1 = -1$, $e^2 > 0$, so $e^x > -x + 1$ for $x>0$. Thus, $y = -x + 1$ is above $y = e^x$ only near $x=0$. 6. **Determine the region shape:** From $x=0$ to some $a$, $y = -x + 1$ is above $y = e^x$, and from $a$ to $2$, $y = e^x$ is above $y=0$. 7. **Find $a$ where $-x + 1 = e^x$:** Numerically, $a hickapprox 0.567$. 8. **Set up the integral for area:** $$\text{Area} = \int_0^a ((-x + 1) - 0) \, dx + \int_a^2 (e^x - 0) \, dx$$ 9. **Calculate the first integral:** $$\int_0^a (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_0^a = -\frac{a^2}{2} + a$$ 10. **Calculate the second integral:** $$\int_a^2 e^x \, dx = \left[ e^x \right]_a^2 = e^2 - e^a$$ 11. **Sum the integrals for total area:** $$\text{Area} = -\frac{a^2}{2} + a + e^2 - e^a$$ 12. **Substitute $a \approx 0.567$:** $$\text{Area} \approx -\frac{(0.567)^2}{2} + 0.567 + e^2 - e^{0.567}$$ 13. **Calculate numerical values:** $$-\frac{0.321}{2} + 0.567 + 7.389 - 1.763 \approx -0.1605 + 0.567 + 7.389 - 1.763 = 6.0325$$ **Final answer:** The area of the region $R$ is approximately $6.03$ square units.