Answer
The area is $A = 2\ln 2 - 1$.
Work Step by Step
At $t=1$, $c\left( 1 \right) = \left( {0,1} \right)$; whereas at $t=2$, $c\left( 2 \right) = \left( {\ln 2,0} \right)$. Plotting several points in $t$-interval $1 \le t \le 2$ and connecting them, we obtain the graph.
We have
$x\left( t \right) = \ln t$, $x'\left( t \right) = \frac{1}{t}$,
$y\left( t \right) = 2 - t$.
Using Eq. (11) the area is
$A = \mathop \smallint \limits_{t = 1}^2 \left( {2 - t} \right)\frac{1}{t}{\rm{d}}t = \mathop \smallint \limits_{t = 1}^2 \left( {\frac{2}{t} - 1} \right){\rm{d}}t$.
Since $\mathop \smallint \limits_{}^{} \frac{1}{t}{\rm{d}}t = \ln t$, we get
$A = 2\mathop \smallint \limits_{t = 1}^2 \frac{1}{t}{\rm{d}}t - \mathop \smallint \limits_{t = 1}^2 {\rm{d}}t = 2\ln t|_1^2 - t|_1^2 = 2\ln 2 - 1$.
