(i)     Sketch the graphs of \(y = \sin x\) and \(y = \sin 2x\) , on the same set of axes, for \(0 \leqslant x \leqslant \frac{\pi }{2}\) .

(ii)     Find the x-coordinates of the points of intersection of the graphs in the domain \(0 \leqslant x \leqslant \frac{\pi }{2}\) .

(iii)     Find the area enclosed by the graphs.

Find the value of \(\int_0^1 {\sqrt {\frac{x}{{4 - x}}} }{{\text{d}}x} \) using the substitution \(x = 4{\sin ^2}\theta \) .

The increasing function f satisfies \(f(0) = 0\) and \(f(a) = b\) , where \(a > 0\) and \(b > 0\) .

(i)     By reference to a sketch, show that \(\int_0^a {f(x){\text{d}}x = ab - \int_0^b {{f^{ - 1}}(x){\text{d}}x} } \) .

(ii)     Hence find the value of \(\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x} \) .

(i)

    A2

Note: Award A1 for correct \(\sin x\) , A1 for correct \(\sin 2x\) .

Note: Award A1A0 for two correct shapes with \(\frac{\pi }{2}\) and/or 1 missing.

Note: Condone graph outside the domain.

(ii)     \(\sin 2x = \sin x\) , \(0 \leqslant x \leqslant \frac{\pi }{2}\)

\(2\sin x\cos x - \sin x = 0\)     M1

\(\sin x(2\cos x - 1) = 0\)

\(x = 0,\frac{\pi }{3}\)     A1A1     N1N1

(iii)     area \( = \int_0^{\frac{\pi }{3}} {(\sin 2x - \sin x){\text{d}}x} \)     M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with \(\sin x\) and \(\sin 2x\) subtracted in either order.

\( = \left[ { - \frac{1}{2}\cos 2x + \cos x} \right]_0^{\frac{\pi }{3}}\)     A1

\( = \left( { - \frac{1}{2}\cos \frac{{2\pi }}{3} + \cos \frac{\pi }{3}} \right) - \left( { - \frac{1}{2}\cos 0 + \cos 0} \right)\)     (M1)

\( = \frac{3}{4} - \frac{1}{2}\)

\( = \frac{1}{4}\)     A1

[9 marks]

\(\int_0^1 {\sqrt {\frac{x}{{4 - x}}} } {\text{d}}x = \int_0^{\frac{\pi }{6}} {\sqrt {\frac{{4{{\sin }^2}\theta }}{{4 - 4{{\sin }^2}\theta }}}  \times 8\sin \theta \cos \theta {\text{d}}\theta } \)     M1A1A1 

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of \({\text{d}}\theta \) , first A1 for correct limits, second A1 for correct substitution for dx .

\(\int_0^{\frac{\pi }{6}} {8{{\sin }^2}\theta {\text{d}}\theta } \)     A1

\(\int_0^{\frac{\pi }{6}} {4 - 4\cos 2\theta {\text{d}}\theta } \)     M1

\( = [4\theta - 2\sin 2\theta ]_0^{\frac{\pi }{6}}\)     A1

\( = \left( {\frac{{2\pi }}{3} - 2\sin \frac{\pi }{3}} \right) - 0\)     (M1)

\( = \frac{{2\pi }}{3} - \sqrt 3 \)     A1

[8 marks]

(i)     

     M1

from the diagram above

the shaded area \( = \int_0^a {f(x){\text{d}}x = ab - \int_0^b {{f^{ - 1}}(y){\text{d}}y} } \)     R1

\({ = ab - \int_0^b {{f^{ - 1}}(x){\text{d}}x} }\)     AG

(ii)     \(f(x) = \arcsin \frac{x}{4} \Rightarrow {f^{ - 1}}(x) = 4\sin x\)     A1

\(\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x = \frac{\pi }{3} - \int_0^{\frac{\pi }{6}} {4\sin x{\text{d}}x} } \)     M1A1A1

Note: Award A1 for the limit \(\frac{\pi }{6}\) seen anywhere, A1 for all else correct.

\( = \frac{\pi }{3} - [ - 4\cos x]_0^{\frac{\pi }{6}}\)     A1

\( = \frac{\pi }{3} - 4 + 2\sqrt 3 \)     A1

Note: Award no marks for methods using integration by parts.

[8 marks]

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by \(\sin x\) and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by \(\sin x\) and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by \(\sin x\) and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.