Integrate sinx on the intervals of pi and 0

let's integrate $\sin x$ on the interval from $\pi$ to $0$.

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

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. $\left[ -\cos x \right]_{\pi}^{0}$ Substitute the upper limit ($0$): $-\cos(0) = -1$ Substitute the lower limit ($\pi$): $-\cos(\pi) = -(-1) = 1$

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

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