Moderate -0.3 Part (i) is a straightforward improper integral requiring basic integration of an exponential and taking a limit—routine A-level technique. Part (ii)(a) involves integrating sine and cosine over a full period, which is standard calculus. Part (ii)(b) is a simple interpretation question. While this covers multiple techniques, each step is mechanical with no problem-solving insight required, making it slightly easier than average.
Evaluate the improper integral
$$\int_1^{\infty} 2e^{-\frac{1}{2}x} dx$$
[3]
The air temperature, \(\theta ^{\circ}C\), on a particular day in London is modelled by the equation
$$\theta = 8 - 5\sin\left(\frac{\pi}{12}t\right) - \cos\left(\frac{\pi}{6}t\right) \quad 0 \leq t \leq 24$$
where \(t\) is the number of hours after midnight.
Use calculus to show that the mean air temperature on this day is \(8^{\circ}C\), according to the model.
[3]
Given that the actual mean air temperature recorded on this day was higher than \(8^{\circ}C\),
2 e − 1 2 x d x = lt i→ m − 4 e − 1 2 t − − 4 e − 1 2
Answer
Marks
Guidance
1
M1
2.1
12
−
Answer
Marks
Guidance
= 4 e
A1
1.1b
(3)
Answer
Marks
(ii)(a)
2 4
1
M e a n t e m p e r a t u r e = 8 5 s i n t c o s t d t − −
2 4 1 2 6
Answer
Marks
Guidance
0
B1
1.2
2 4
1 6 0 6 1
8 t c o s t s i n t . . . = + − =
2 4 1 2 6 2 4
Answer
Marks
Guidance
0
M1
1.1b
1 6 0 6 6 0
8 ( 2 4 ) 0 = + − − = 8 * cso
Answer
Marks
Guidance
2 4
A1*
2.1
(3)
Answer
Marks
Guidance
(ii)(b)
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:
(i)
B1: Correct integration.
1
x −
M1: Attempt to integrate to a form e 2 where 2 , and applies correct limits with some
consideration of the infinite limit given (e.g. with the limit statement). Only allow with used as
the limit if subsequent work shows the term is zero.
A1: Correct value
(ii)(a)
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 t c o s t s i n t + + and uses the limits of 0 and 24 (the correct
1 2 6
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
60 60
some evidence of the limits used, minimum required for substitution is 8 ( 24 )+ − .
(ii)(b)
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”.
Answer
Marks
Guidance
Question
Scheme
Marks
Question 5:
--- 5(i) ---
5(i) | 12 12
− x − x
2 e d x = − 4 e | B1 | 1.1b
2 e − 1 2 x d x = lt i→ m − 4 e − 1 2 t − − 4 e − 1 2
1 | M1 | 2.1
12
−
= 4 e | A1 | 1.1b
(3)
(ii)(a) | 2 4
1
M e a n t e m p e r a t u r e = 8 5 s i n t c o s t d t − −
2 4 1 2 6
0 | B1 | 1.2
2 4
1 6 0 6 1
8 t c o s t s i n t . . . = + − =
2 4 1 2 6 2 4
0 | M1 | 1.1b
1 6 0 6 6 0
8 ( 2 4 ) 0 = + − − = 8 * cso
2 4 | A1* | 2.1
(3)
(ii)(b) | 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:
(i)
B1: Correct integration.
1
x −
M1: Attempt to integrate to a form e 2 where 2 , and applies correct limits with some
consideration of the infinite limit given (e.g. with the limit statement). Only allow with used as
the limit if subsequent work shows the term is zero.
A1: Correct value
(ii)(a)
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 t c o s t s i n t + + and uses the limits of 0 and 24 (the correct
1 2 6
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
60 60
some evidence of the limits used, minimum required for substitution is 8 ( 24 )+ − .
(ii)(b)
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”.
Question | Scheme | Marks | AOs
\begin{enumerate}[label=(\roman*)]
\item Evaluate the improper integral
$$\int_1^{\infty} 2e^{-\frac{1}{2}x} dx$$
[3]
\item The air temperature, $\theta ^{\circ}C$, on a particular day in London is modelled by the equation
$$\theta = 8 - 5\sin\left(\frac{\pi}{12}t\right) - \cos\left(\frac{\pi}{6}t\right) \quad 0 \leq t \leq 24$$
where $t$ is the number of hours after midnight.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that the mean air temperature on this day is $8^{\circ}C$, according to the model.
[3]
Given that the actual mean air temperature recorded on this day was higher than $8^{\circ}C$,
\item explain how the model could be refined.
[1]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2021 Q5 [7]}}