1. **State the problem:** We want to find the area of the region $R$ enclosed by the curves $y = -x + 1$, $y = e^x$, $y = 0$, and the vertical line $x = 2$. 2. **Understand the region:** The region is bounded below by $y=0$, on the left and right by the curves $y = -x + 1$ and $y = e^x$, and the vertical line $x=2$. 3. **Find intersection points:** To set up the integral, find where $y = -x + 1$ and $y = e^x$ intersect. Solve $-x + 1 = e^x$: This cannot be solved algebraically, but by inspection: At $x=0$, $-0 + 1 = 1$ and $e^0 = 1$, so they intersect at $x=0$. 4. **Determine which curve is on top between $x=0$ and $x=2$:** At $x=1$, $-1 + 1 = 0$ and $e^1 = e \\approx 2.718$, so $e^x$ is above $-x + 1$ for $x > 0$. 5. **Set up the double integral:** The area can be found by integrating the vertical distance between the curves over the interval $x=0$ to $x=2$. Since $y=0$ is the lower boundary and the upper boundary changes: - From $x=0$ to $x=1$, the upper curve is $-x + 1$ (since $-x + 1 > e^x$ for $x<0$ but we confirmed $e^x$ is above for $x>0$, so check carefully). Check at $x=0.5$: $-0.5 + 1 = 0.5$, $e^{0.5} \\approx 1.65$, so $e^x$ is above $-x + 1$ for $x>0$. Therefore, the region bounded by $y=0$, $y=-x+1$, and $y=e^x$ between $x=0$ and $x=2$ is split: - From $x=0$ to $x=1$, the top curve is $e^x$ and bottom is $0$. - From $x=1$ to $x=2$, the top curve is $e^x$ and bottom is $-x + 1$. 6. **Find intersection of $e^x$ and $-x + 1$ between 0 and 2:** We already found they intersect at $x=0$. Check if they intersect again between 0 and 2: At $x=1$, $e^1 = 2.718$, $-1 + 1 = 0$, no intersection. At $x=2$, $e^2 = 7.389$, $-2 + 1 = -1$, no intersection. So only intersection at $x=0$. 7. **Set up the integral for the area:** Area $= \int_0^2 \left( e^x - \max(0, -x + 1) \right) dx$ Since $-x + 1$ is positive between $x=0$ and $x=1$ and negative after $x=1$, split the integral: - From $0$ to $1$, bottom curve is $0$ (since $-x + 1$ is above 0 but $y=0$ is also a boundary, the region is between $y=0$ and $y=e^x$). - From $1$ to $2$, bottom curve is $-x + 1$ (since $-x + 1$ is below 0, but $y=0$ is boundary, so the region is between $y=-x + 1$ and $y=e^x$). 8. **Calculate the area:** $$\text{Area} = \int_0^1 (e^x - 0) dx + \int_1^2 (e^x - (-x + 1)) dx$$ $$= \int_0^1 e^x dx + \int_1^2 (e^x + x - 1) dx$$ 9. **Evaluate the integrals:** $$\int e^x dx = e^x + C$$ $$\int x dx = \frac{x^2}{2} + C$$ $$\int 1 dx = x + C$$ So, $$\int_0^1 e^x dx = e^1 - e^0 = e - 1$$ $$\int_1^2 (e^x + x - 1) dx = \left[e^x + \frac{x^2}{2} - x \right]_1^2$$ Calculate at bounds: At $x=2$: $$e^2 + \frac{2^2}{2} - 2 = e^2 + 2 - 2 = e^2$$ At $x=1$: $$e^1 + \frac{1^2}{2} - 1 = e + \frac{1}{2} - 1 = e - \frac{1}{2}$$ Subtract: $$e^2 - (e - \frac{1}{2}) = e^2 - e + \frac{1}{2}$$ 10. **Sum the two parts:** $$\text{Area} = (e - 1) + (e^2 - e + \frac{1}{2}) = e^2 - 1 + \frac{1}{2} = e^2 - \frac{1}{2}$$ 11. **Final answer:** $$\boxed{\text{Area} = e^2 - \frac{1}{2}}$$