The area is calculated by the integral $\int_{0}^{\pi} \sin x dx$. The antiderivative of $\sin x$ is $-\cos x$. Applying the limits: $[-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0)$. Since $\cos \pi = -1$ and $\cos 0 = 1$, the expression becomes $(-(-1)) - (-1) = 1 + 1 = 2$ sq units. Option 1 is incorrect because it represents the area from $0$ to $\pi/2$. Option 3 is incorrect because while the integral of $\sin x$ from $0$ to $2\pi$ is 0, the area from $0$ to $\pi$ is positive. Option 4 is a common miscalculation involving the limits.