In this question after integration pi is the upper boundary and 0 is the lower solve using the intervals the way I've enlisted

thanks for clarifying the limits! Let's integrate $\sin x$ with $0$ as the lower boundary and $\pi$ as the upper boundary.

The integral is $\int_{0}^{\pi} \sin x \, dx$.

Step 1: Find the antiderivative of $\sin x$. The antiderivative of $\sin x$ is $-\cos x$.

Step 2: Evaluate the antiderivative at the limits of integration. $\left[ -\cos x \right]_{0}^{\pi}$ Substitute the upper limit ($\pi$): $-\cos(\pi) = -(-1) = 1$ Substitute the lower limit ($0$): $-\cos(0) = -1$

Step 3: Subtract the value at the lower limit from the value at the upper limit. $(1) - (-1) = 1 + 1 = 2$

The final answer is $\boxed{2}$.