The area of the upper half of the circle whose equation is ${\left(x-1\right)}^2+y^2=1$ is given by

The given equation of the curve is ${\left(x-1\right)}^2+y^2=1$

The upper half of the curve is $y=\sqrt{1-{\left(x-1\right)}^2}$

This cuts the $x-$ axis at $\left(0.0\right),\left(2,0\right)$

The rough sketch of required area is as below

The area of the shaded region is ${\int }_0^2\sqrt{1-{\left(x-1\right)}^2}dx$

Hence, 

$\begin{array}{rcl} & & {\int }_0^2\sqrt{1={\left(x-1\right)}^2}dx\end{array}$$\begin{array}{rcl} & =& {\left[\frac{{\left(x-1\right)}^2}{2}\sqrt{1={\left(x-1\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(\frac{x-1}{1}\right)\right]}_0^2\\ & =& \frac{\mathrm{\pi }}{2}\end{array}$

Therefore, the area of the upper half of the given circle is $\frac{\pi }{2}$ square units.