Answer
$\text{Area = $4\ln 2$}$
Work Step by Step
$\text{It is given that}$
\begin{align}
g(x) = \frac{4}{2-x}; \ y = 4; \ x = 0
\end{align}
$\text{The intersection of these functions is}$
\begin{align}
\frac{4}{2-x} = 4 \Rrightarrow x = 1
\end{align}
$\text{Thus, the area of the region is}$
\begin{align}
Area =\int_0^1 \frac{4}{2-x} dx
\end{align}
$\text{Apply u-substitution $u = 2-x \Rrightarrow du = -dx$:}$
\begin{align}
Area =\int_2^1 - \frac{4}{u} du = 4 \left[\ln x\right]_1^2 = 4\ln 2
\end{align}
