A2 October 2021 Paper 1 Q5

EdexcelCurrent spec7 marksIntegration

(i) Evaluate the improper integral\[\int_1^{\infty} 2\mathrm{e}^{-\frac{1}{2}x}\,\mathrm{d}x\] (3)

(ii) The air temperature, \(\theta\,{}^{\circ}\mathrm{C}\), on a particular day in London is modelled by the equation\[\theta = 8 - 5\sin\left(\frac{\pi}{12}t\right) - \cos\left(\frac{\pi}{6}t\right) \qquad\qquad 0 \leqslant t \leqslant 24\]where \(t\) is the number of hours after midnight.

(a) Use calculus to show that the mean air temperature on this day is \(8\,{}^{\circ}\mathrm{C}\), according to the model. (3)

Given that the actual mean air temperature recorded on this day was higher than \(8\,{}^{\circ}\mathrm{C}\),

(b) explain how the model could be refined. (1)

Mark scheme (i)

Scheme	Marks	AO
\(\displaystyle\int 2\mathrm{e}^{-\frac{1}{2}x}\,\mathrm{d}x = -4\mathrm{e}^{-\frac{1}{2}x}\)	B1	1.1b
\(\displaystyle\int_1^{\infty} 2\mathrm{e}^{-\frac{1}{2}x}\,\mathrm{d}x = \lim_{t \to \infty}\left[\left(-4\mathrm{e}^{-\frac{1}{2}t}\right) - \left(-4\mathrm{e}^{-\frac{1}{2}}\right)\right]\)	M1	2.1
\(= 4\mathrm{e}^{-\frac{1}{2}}\)	A1	1.1b
	(3)

Notes

B1: Correct integration.

M1: Attempt to integrate to a form \(\lambda\mathrm{e}^{-\frac{1}{2}x}\) where \(\lambda \neq 2\), and applies correct limits with some consideration of the infinite limit given (e.g. with the limit statement). Only allow with \(\infty\) used as the limit if subsequent work shows the term is zero.

A1: Correct value

Mark scheme (ii)(a)

Scheme	Marks	AO
Mean temperature \(= \dfrac{1}{24}\displaystyle\int_0^{24}\left(8 - 5\sin\left(\frac{\pi}{12}t\right) - \cos\left(\frac{\pi}{6}t\right)\right)\mathrm{d}t\)	B1	1.2
\(= \dfrac{1}{24}\left[\left(8t + \dfrac{60}{\pi}\cos\left(\dfrac{\pi}{12}t\right) - \dfrac{6}{\pi}\sin\left(\dfrac{\pi}{6}t\right)\right)\right]_0^{24} = \dfrac{1}{24}[\ldots]\)	M1	1.1b
\(= \dfrac{1}{24}\left[\left(8(24) + \dfrac{60}{\pi} - \dfrac{6}{\pi} \times 0\right) - \left(\dfrac{60}{\pi}\right)\right] = 8\) * cso	A1*	2.1
	(3)

Notes

B1: Recalls the correct formula for finding the mean value of a function. You may see the division by “24” only at the end. No integration is necessary, just a correct statement with an integral.

M1: Integrates to a form \(\alpha t + \beta\cos\left(\dfrac{\pi}{12}t\right) + \delta\sin\left(\dfrac{\pi}{6}t\right)\) and uses the limits of 0 and 24 (the correct way around). If no explicit substitution is seen, accept any value following the integral as an attempt. Answers from a calculator with no correct integral seen score M0 as the question requires calculus to be used.

A1*cso: Achieves 8 with no errors seen following a full attempt at the substitution. Must have seen some evidence of the limits used, minimum required for substitution is \(\left[\left(8(24) + \dfrac{60}{\pi}\right) - \left(\dfrac{60}{\pi}\right)\right]\).

Mark scheme (ii)(b)

Scheme	Marks	AO
E.g. increase the value of the constant 8 / adapt the constant 8 to a function which takes values greater than 8.	B1	3.5c
	(1)
	(7 marks)

Notes

B1: Accept any reasonable adaptation to the equation that will increase the mean value. E.g. as in scheme, or introduce another positive term, or decrease the constant 5 etc. It must be clear which constant they are referring to in their reason, not just “increase the constant”.