Write down a definite integral to represent the area of $A$.

Calculate the area of $A$.

METHOD 1

$2\arcsin (x - 1) - \frac{\pi }{4} = \frac{\pi }{4}$     (M1)

$x = 1 + \frac{1}{{\sqrt 2 }}\,\,\,( = 1.707 \ldots )$     (A1)

$\int\limits_0^{1 + \frac{1}{{\sqrt 2 }}} {\frac{\pi }{4} - \left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx}$   M1A1

Note:     Award M1 for an attempt to find the difference between two functions, A1 for all correct.

METHOD 2

when $x = 0,{\text{ }}y = \frac{{ - 5\pi }}{4}\,\,\,( = - 3.93)$     A1

$x = 1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)$    M1A1

Note:     Award M1 for an attempt to find the inverse function.

$\int_{\frac{{ - 5\pi }}{4}}^{\frac{\pi }{4}} {\left( {1 + \sin \left( {\frac{{4y + \pi }}{8}} \right)} \right){\text{d}}y}$     A1

METHOD 3

$\int_0^{1.38...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right){\text{d}}x} \left|  +  \right.\int\limits_0^{1.71...} {\frac{\pi }{4}{\text{d}}x - \int\limits_{1.38...}^{1.71...} {\left( {2\arcsin \left( {x - 1} \right) - \frac{\pi }{4}} \right)dx} }$    M1A1A1A1

Note:     Award M1 for considering the area below the $x$-axis and above the $x$-axis and A1 for each correct integral.

[4 marks]

${\text{area}} = 3.30{\text{ (square units)}}$     A2

[2 marks]