Using Property (5) of Definite Integrals

1. Find an upper estimate and a lower estimate for the area \(A\) under the graph of \(f(x) =\sin x\) from \(0\) to \(\pi \).
2. Find the actual area under the graph.

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:

\[ 0\leq \int_{0}^{\pi }\sin x~dx\leq \pi \]

(b) The actual area under the graph is \[ A= \int_{0}^{\pi}\sin x\,dx= \bigg[-\cos x\bigg] _{0}^{\pi }=-\cos \pi +\cos 0=1+1=2\hbox{ square units} \]