The area between the curve $x^2+y^2=4$ and the lines $x=0,x=1$ is 

The given curve is  $x^2+y^2=4$ and the lines are $x=0,x=1$

Draw the rough sketch the required region

The required area is area of the shaded region

Observing the graph, the curve $x^2+y^2=4$ is symmetric about the $X-$ axis

The curve is 

          $\begin{array}{rcl}{y}^2& =& 4-x^2\\ y& =& \sqrt{4-x^2}\end{array}$

We know that the area of the  region bounded by the curve $y=f\left(x\right)$, $x=a,x=b$ and $X-$axis is $A={\int }_a^{b}f\left(x\right)dx$

Hence, the area of the shaded region is 

   $\begin{array}{rcl}A& =& 2{\int }_0^1\sqrt{4-x^2}dx\\ & =& 2{\left[\frac{x^2}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_0^1\\ & =& 2\left(\frac{1}{2}\left(0\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)\right)\\ & =& 4\frac{\pi }{6}\\ & =& \frac{2\pi }{3}\end{array}$

The area of the shaded region is $\boxed{\frac{2\pi }{3}}$