Solution The graph of \(f\) is shown in Figure 27. Since \(f(x) \geq 0\) for all \(x\) in the closed interval \([ 0,\pi ] \), the area \(A\) under its graph is given by the definite integral, \(\int_{0}^{\pi }\sin x~dx\).
(a) From the Extreme Value Theorem, \(f\) has an absolute minimum value and an absolute maximum value on the interval \([0,\pi] .\) The absolute maximum of \(f\) occurs at \(x=\dfrac{\pi}{2},\) and its value is \(f \left( \dfrac{\pi }{2}\right) =\sin \dfrac{\pi }{2}=1.\) The absolute minimum occurs at \(x=0\) and at \(x=\pi \); the absolute minimum value is \(f (0) =\sin 0=0=f (\pi)\). Using the inequalities in (5), the area under the graph of \(f\) is bounded as follows:
The Extreme Value Theorem is discussed in Section 4.2, pp. xx-xx.
\[ 0\leq \int_{0}^{\pi }\sin x~dx\leq \pi \]
(b) The actual area under the graph is \[ A= \int_{0}^{\pi}\sin x\,dx= \bigg[-\cos x\bigg] _{0}^{\pi }=-\cos \pi +\cos 0=1+1=2\hbox{ square units} \]