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]