By the definition of an expectation trough the density, $ \mathsf E g(X) = \int\limits_{-\infty}^\infty g(t)f_X(t)\,dt $ where $f_X(t)$ is a density function of r.v. $X$. In your case $X\sim\mathcal E(\mu)$ so $f_X(t) = 0$ for $t<0$ and $f_X(t) = \mu\mathrm e^{-\mu t}$ for $t\geq 0$. As a result $ \mathsf E\max\{\mu/2,X\}= \int\limits_{0}^\infty \max\{\mu/2,t\}\mu\mathrm e^{-\mu t}\,dt = \int\limits_{0}^{\mu/2} \frac12\mu^2\mathrm e^{-\mu t}\,dt+\int\limits_{\mu/2}^\infty t\mu\mathrm e^{-\mu t}\,dt. $
Do you know how to find value of these integrals?