1. **State the problem:** Find the exact area of the region R bounded by the curve $C$ with parametric equations $x = 1 - \frac{1}{2}t$, $y = 2^t - 1$, the line $x = -1$, and the x-axis. 2. **Recall the parametric area formula:** The area under a parametric curve from $t=a$ to $t=b$ is given by $$\text{Area} = \int_a^b y \frac{dx}{dt} dt$$ where $\frac{dx}{dt}$ is the derivative of $x$ with respect to $t$. 3. **Find the limits for $t$:** - At $x=0$ (point A), solve $0 = 1 - \frac{1}{2}t$ gives $t=2$. - At $x=-1$, solve $-1 = 1 - \frac{1}{2}t$ gives $t=4$. So the region $R$ corresponds to $t$ from 4 to 2 (note the order). 4. **Calculate $\frac{dx}{dt}$:** $$\frac{dx}{dt} = -\frac{1}{2}$$ 5. **Set up the integral for the area:** $$\text{Area} = \int_4^2 (2^t - 1) \left(-\frac{1}{2}\right) dt = \int_2^4 \frac{1}{2}(2^t - 1) dt$$ 6. **Evaluate the integral:** $$\int_2^4 \frac{1}{2}(2^t - 1) dt = \frac{1}{2} \int_2^4 (2^t - 1) dt = \frac{1}{2} \left( \int_2^4 2^t dt - \int_2^4 1 dt \right)$$ 7. **Integrate each term:** - $$\int 2^t dt = \frac{2^t}{\ln 2}$$ - $$\int 1 dt = t$$ So, $$\frac{1}{2} \left[ \frac{2^t}{\ln 2} - t \right]_2^4 = \frac{1}{2} \left( \frac{2^4}{\ln 2} - 4 - \frac{2^2}{\ln 2} + 2 \right)$$ 8. **Simplify the expression:** $$= \frac{1}{2} \left( \frac{16}{\ln 2} - 4 - \frac{4}{\ln 2} + 2 \right) = \frac{1}{2} \left( \frac{12}{\ln 2} - 2 \right) = \frac{6}{\ln 2} - 1$$ **Final answer:** $$\boxed{\frac{6}{\ln 2} - 1}$$ This is the exact area of the shaded region $R$.