Answer
$$\frac{9\pi -16}{8}$$
Work Step by Step
Given
$$r=2+ \cos{\theta},\ \ \ \ \ $$
From the given graph, the region bounded by $ \pi/2\leq \theta \leq \pi $. Then area given by
\begin{aligned}
A&= \frac{1}{2}\int_{a}^{b} r^2d\theta\\
&= \frac{1}{2}\int_{\pi/2}^{\pi}(2+ \cos{\theta})^2d\theta
\\
&= \frac{1}{2}\int_{\pi/2}^{\pi}(4+2\cos \theta +\cos^2{\theta}) d\theta
\\
&= \frac{1}{2}\int_{\pi/2}^{\pi}(\frac{9}{2}+2\cos \theta +\frac{1}{2}\cos{2\theta}) d\theta
\\
&=\frac{1}{2} \left(\frac{9}{2}\theta+2\sin \theta +\frac{1}{4}\sin{2\theta}\right)\bigg|_{\pi/2}^{\pi} \\
&= \frac{1}{2}\left(2\pi -4+\frac{\pi }{4}\right)\\
&= \frac{9\pi -16}{8}
\end{aligned}
