Let \(f(x) = \cos x\), for \(0\) \(\le \) \(x\) \( \le \) \(2\pi \). The following diagram shows the graph of \(f\).

There are \(x\)-intercepts at \(x = \frac{\pi }{2},{\text{ }}\frac{{3\pi }}{2}\).

The shaded region \(R\) is enclosed by the graph of \(f\), the line \(x = b\), where \(b > \frac{{3\pi }}{2}\), and the \(x\)-axis. The area of \(R\) is \(\left( {1 - \frac{{\sqrt 3 }}{2}} \right)\). Find the value of \(b\).

attempt to set up integral (accept missing or incorrect limits and missing \({\text{d}}x\))     M1

eg\(\;\;\;\int_{\frac{{3\pi }}{2}}^b {\cos x{\text{d}}x,{\text{ }}\int_a^b {\cos x{\text{d}}x,{\text{ }}\int_{\frac{{3\pi }}{2}}^b {f{\text{d}}x,{\text{ }}\int {\cos x} } } } \)

correct integration (accept missing or incorrect limits)     (A1)

eg\(\;\;\;[\sin x]_{\frac{{3\pi }}{2}}^b,{\text{ }}\sin x\)

substituting correct limits into their integrated function and subtracting (in any order)     (M1)

eg\(\;\;\;\sin b - \sin \left( {\frac{{3\pi }}{2}} \right),{\text{ }}\sin \left( {\frac{{3\pi }}{2}} \right) - \sin b\)

\(\sin \left( {\frac{{3\pi }}{2}} \right) =  - 1\;\;\;\)(seen anywhere)     (A1)

setting their result from an integrated function equal to \(\left( {1 - \frac{{\sqrt 3 }}{2}} \right)\)     M1

eg\(\;\;\;\sin b =  - \frac{{\sqrt 3 }}{2}\)

evaluating \({\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{3}\) or \({\sin ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) =  - \frac{\pi }{3}\)     (A1)

eg\(\;\;\;b = \frac{\pi }{3},{\text{ }} - 60^\circ \)

identifying correct value     (A1)

eg\(\;\;\;2\pi  - \frac{\pi }{3},{\text{ }}360 - 60\)

\(b = \frac{{5\pi }}{3}\)     A1     N3

[8 marks]

Most candidates recognised that a definite integral was required and many were able to set up a correct equation. Incorrect integration leading to \( - \sin x\) was quite common and poor notation was frequently seen. Some candidates appeared to guess their value from the graph, showing little supporting work.