IBDP Maths analysis and approaches Topic: SL 5.5 Area of the region enclosed by a curve and x axis HL Paper 2

(i)     METHOD 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – \sin x + \cos x\)     A1

\(y\frac{{{\text{d}}y}}{{{\text{d}}x}} = (\cos x + \sin x)( – \sin x + \cos x)\)     M1

\( = {\cos ^2}x – {\sin ^2}x\)     A1

\( = \cos 2x\)     AG

METHOD 2

\({y^2} = {(\sin x + \cos x)^2}\)     A1

\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2(\cos x + \sin x)(\cos x – \sin x)\)     M1

\(y\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\cos ^2}x – {\sin ^2}x\)     A1

\( = \cos 2x\)     AG

(ii)     attempting to separate variables \(\int {y{\text{ d}}y = \int {\cos 2x{\text{ d}}x} } \)     M1

\(\frac{1}{2}{y^2} = \frac{1}{2}\sin 2x + C\)     A1A1

Note: Award A1 for a correct LHS and A1 for a correct RHS.

\(y =  \pm {(\sin 2x + A)^{\frac{1}{2}}}\)     A1

(iii)     \(\sin 2x + A \equiv {(\cos x + \sin x)^2}\)     (M1)

\({(\cos x + \sin x)^2} = {\cos ^2}x + 2\sin x\cos x + {\sin ^2}x\)

use of \(\sin 2x \equiv 2\sin x\cos x\).     (M1)

A = 1     A1

[10 marks]

(i)     substituting \(x = \frac{\pi }{4}\) and y = 2 into \(y = {(\sin 2x + A)^{\frac{1}{2}}}\)     M1

so \(g(x) = {(\sin 2x + 3)^{\frac{1}{2}}}\).     A1

range g is \(\left[ {\sqrt 2 ,{\text{ }}2} \right]\)     A1A1A1

Note: Accept [1.41, 2]. Award A1 for each correct endpoint and A1 for the correct closed interval.

(ii)     \(\int_0^{\frac{\pi }{2}} {{{(\sin 2x + 3)}^{\frac{1}{2}}}{\text{d}}x} \)     (M1)(A1)

= 2.99     A1

(iii)     \(\pi \int_0^{\frac{\pi }{2}} {(\sin 2x + 3){\text{d}}x – \pi (1)\left( {\frac{\pi }{2}} \right)} \) (or equivalent)     (M1)(A1)(A1)

Note: Award (M1)(A1)(A1) for \(\pi \int_0^{\frac{\pi }{2}} {(\sin 2x + 2){\text{d}}x} \)

\( = 17.946 – 4.935{\text{ }}( = \frac{\pi }{2}(3\pi  + 2) – \pi \left( {\frac{\pi }{2}} \right))\)     A1

Note: Award A1 for \(\pi (\pi  + 1)\).

[12 marks]

Part (a) was not well done and was often difficult to mark. In part (a) (i), a large number of candidates did not know how to verify a solution, \(y(x)\), to the given differential equation. Instead, many candidates attempted to solve the differential equation. In part (a) (ii), a large number of candidates began solving the differential equation by correctly separating the variables but then either neglected to add a constant of integration or added one as an afterthought. Many simple algebraic and basic integral calculus errors were seen. In part (a) (iii), many candidates did not realize that the solution given in part (a) (i) and the general solution found in part (a) (ii) were to be equated. Those that did know to equate these two solutions, were able to square both solution forms and correctly use the trigonometric identity \(\sin 2x = 2\sin x\cos x\). Many of these candidates however started with incorrect solution(s).

In part (b), a large number of candidates knew how to find a required area and a required volume of solid of revolution using integral calculus. Many candidates, however, used incorrect expressions obtained in part (a). In part (b) (ii), a number of candidates either neglected to state ‘π’ or attempted to calculate the volume of a solid of revolution of ‘radius’ \(f(x) – g(x)\).