June 2024 Paper 1 Q8

OCR MEICurrent spec6 marksIntegration Radians

8 The equation of a curve is \(y = \sqrt{\sin 4x} + 2\cos 2x\), where \(x\) is in radians.

(a) Show that, for small values of \(x\), \(y \approx 2\sqrt{x} + 2 - 4x^2\). [2]

The diagram shows the region bounded by the curve \(y = \sqrt{\sin 4x} + 2\cos 2x\), the axes and the line \(x = 0.1\).

Graph of the curve for x from 0 to 0.2, starting at y = 2 and rising slowly; the region under the curve between x = 0 and x = 0.1 is shaded

(b) In this question you must show detailed reasoning.
Use the approximation in part (a) to estimate the area of this region. [4]

Mark scheme (a)

Scheme	Marks	AO
Using \(\sin x \approx x\) and \(\cos x \approx 1 - \tfrac{1}{2}x^2\) \(\sqrt{\sin 4x} + 2\cos 2x \approx \sqrt{4x} + 2\left(1 - \tfrac{1}{2}(2x)^2\right)\)	M1	1.1
So \(y \approx 2\sqrt{x} + 2 - 4x^2\)	A1	1.1
	[2]

Notes

M1: Uses both given small angle approximations in the expression. Must see \(\sqrt{4x}\) and \(\tfrac{1}{2}(2x)^2\) or \(\tfrac{1}{2} \times 4x^2\) condone missing brackets. Also allow for clear use of \(\cos 2x = 1 - 2\sin^2 x\) and \(\sin x \approx x\) used.

A1: AG Convincing argument to reach given answer

Mark scheme (b)

Scheme	Marks	AO
DR \(\displaystyle\int_0^{0.1}\left(\sqrt{\sin 4x} + 2\cos 2x\right)\mathrm{d}x \approx \int_0^{0.1}\left(2x^{\frac{1}{2}} + 2 - 4x^2\right)\mathrm{d}x\)	M1	3.1a
\(= \left[\dfrac{2x^{\frac{3}{2}}}{\frac{3}{2}} + 2x - \dfrac{4}{3}x^3\right]_0^{0.1}\)	A1 A1	1.1 1.1
\(= \left(\dfrac{4}{3}0.1^{\frac{3}{2}} + 0.2 - \dfrac{0.004}{3}\right) - 0 = 0.24083\ldots\)	A1	1.1
	[4]

Notes

M1: Attempts to integrate the expression in powers of \(x\)
Must be seen

A1: At least 2 correct terms (no FT from (a) as answer given)

A1: Fully correct indefinite integral. Need not be simplified

A1: Correct value from correct indefinite integral. Allow for 0.24 or better if the method is clear.
(Do not allow for 0.24059… which is obtained by integrating the original function by calculator.)

Alternative solution

Scheme	Marks	AO
Use trapezium rule to find an area	M1
Correct \(\frac{h}{2}\) for the number of strips used	A1
At least 3 correct ordinates used	A1
Area is approximately 0.24	A1

M1: Also allow for a single trapezium

A1: Soi For example \(\frac{0.1}{2}\ \frac{0.05}{2}, \frac{0.025}{2}, \frac{0.01}{2}\) used for 1, 2, 4 10 strips

A1: Must be ordinates from the approximating function
eg 2, 2.4372… and 2.5924… seen

A1: Accept awrt 0.24

Table of values

\(x\)	actual \(y\)	\(y\) from approx
0	2	2
0.01	2.199573	2.1996
0.02	2.281092	2.281243
0.03	2.342396	2.34281
0.04	2.39275	2.3936
0.05	2.435732	2.437214
0.06	2.473165	2.475498
0.07	2.506127	2.50955
0.08	2.535317	2.540085
0.09	2.561214	2.5676
0.1	2.584167083	2.592455532
0.025	2.313465	2.313728
0.075	2.52116	2.525223

Answers

	using approx	using original function
1 strip	0.229623	0.229208
2 strips	0.236672	0.235817
4 strips	0.23931	0.239061
10 strips	0.240434	0.240194