Answer
\[\pi - 2\]
Work Step by Step
\[\begin{gathered}
\int_0^\pi {\left( {1 - \sin x} \right)} dx \hfill \\
\hfill \\
{\text{Integrate}} \hfill \\
\hfill \\
= \left[ {x + \cos x} \right]_0^\pi \hfill \\
\hfill \\
{\text{Use the fundamental theorem of calculus}} \hfill \\
\hfill \\
= \left[ {\pi + \cos \pi } \right] - \left[ {0 + \cos 0} \right] \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
\left( {\pi - 1} \right) - \left( {0 + 1} \right) \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\pi - 1 - 1 \hfill \\
\hfill \\
\pi - 2 \hfill \\
\end{gathered} \]
