1. **State the problem:** Find the area of the shaded region bounded by the curve, the line $y=2$, and the vertical line $x=\pi$. 2. **Identify the boundaries:** The region is bounded above by $y=2$, below by the curve (which we will call $f(x)$), and vertically between $x=0$ and $x=\pi$. 3. **Set up the integral for the area:** The area between two curves $y = g(x)$ (upper) and $y = f(x)$ (lower) from $x=a$ to $x=b$ is given by $$\text{Area} = \int_a^b [g(x) - f(x)] \, dx$$ Here, $g(x) = 2$ and $f(x)$ is the curve given (not explicitly stated, but from the description, the curve starts at $y=2$ at $x=0$ and curves downward). Since the exact function $f(x)$ is not provided, we assume the curve is $y = \cos x$ (a common curve starting at 2 is unlikely, so we must rely on the description; however, since the problem is about the shaded area bounded by $y=2$, $x=\pi$, and the curve, the shaded area is the area between $y=2$ and the curve from $0$ to $\pi$). 4. **Assuming the curve is $y = 2 - \sin x$ (to start at 2 and curve downward), the area is:** $$\text{Area} = \int_0^{\pi} [2 - (2 - \sin x)] \, dx = \int_0^{\pi} \sin x \, dx$$ 5. **Evaluate the integral:** $$\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$$ 6. **Final answer:** The area of the shaded region is $2$. **Note:** Since the exact curve function is not given, this solution assumes the curve is $y = 2 - \sin x$ based on the description. If the curve is different, please provide the explicit function for precise calculation.