Standard +0.8 This question requires integrating a hyperbolic function (tanh), applying the mean value theorem for integrals, and careful numerical evaluation. While the setup is straightforward, students must recall the integral of tanh(ax), handle the constant correctly, and evaluate sinh at a non-standard argument—more demanding than routine Further Maths integration but not requiring novel insight.
6 In this question you must show detailed reasoning.
The power output, \(p\) watts, of a machine at time \(t\) hours after it is switched on can be modelled by the equation \(\mathrm { p } = 20 - 20 \tanh ( 1.44 \mathrm { t } )\) for \(t \geqslant 0\).
Determine, according to the model, the mean power output of the machine over the first half hour after it is switched on. Give your answer correct to \(\mathbf { 2 }\) decimal places.
M e a n v a l u e = 0 ( 2 0 − 2 0 t a n h (1 .4 4 t ) ) d t
0 .5
0 .5
4 0 ( )
= 4 0 t − l n e 1 .4 4 t + e − 1 .4 4 t
1 . 4 4
0
= 2 0 − 4 0 ( l n ( e 0 .7 2 + e ) − 0 .7 2 − l n 2 )
1 . 4 4
4 0 2 . 5 4 1 2
= 2 0 − l n = 1 3 . 3 5 ( W )
Answer
Marks
1 . 4 4 2
B1
M1
A1
Answer
Marks
A1
For using the definition of the mean value of p wrt t, correct limits and
1/0.5.
Answer
Marks
Guidance
For ∫tanh1.44𝑡d𝑡 = 𝑘ln
e1.44𝑡 +e−1.44𝑡
(+𝑐) k can = 1
Or ∫tanh1.44𝑡d𝑡 = 𝑘ln
cosh1.44𝑡
(+𝑐)
May see integration by substitution, eg.𝑢 = e1.44𝑡 +e−1.44𝑡or 𝑢 =
Answer
Marks
Guidance
cosh1.44𝑡. If so, award this mark for 𝑘ln
𝑢
seen
For fully correct integration of tanh1.44𝑡.
Answer
Marks
Guidance
Either for ∫tanh1.44𝑡d𝑡 = 1 ln
e1.44𝑡 +e−1.44𝑡
(+𝑐)
1.44
1
Answer
Marks
Guidance
or ∫tanh1.44𝑡d𝑡 = ln
cosh1.44𝑡
(+𝑐)
1.44
1
Answer
Marks
Guidance
or ∫tanh1.44𝑡d𝑡 = ln
𝑢
(+𝑐) with u as above.
1.44
Condone missing modulus.
cao, with clear working.
[4]
Question 6:
6 | DR
1 0 .5
M e a n v a l u e = 0 ( 2 0 − 2 0 t a n h (1 .4 4 t ) ) d t
0 .5
0 .5
4 0 ( )
= 4 0 t − l n e 1 .4 4 t + e − 1 .4 4 t
1 . 4 4
0
= 2 0 − 4 0 ( l n ( e 0 .7 2 + e ) − 0 .7 2 − l n 2 )
1 . 4 4
4 0 2 . 5 4 1 2
= 2 0 − l n = 1 3 . 3 5 ( W )
1 . 4 4 2 | B1
M1
A1
A1 | For using the definition of the mean value of p wrt t, correct limits and
1/0.5.
For ∫tanh1.44𝑡d𝑡 = 𝑘ln|e1.44𝑡 +e−1.44𝑡|(+𝑐) k can = 1
Or ∫tanh1.44𝑡d𝑡 = 𝑘ln|cosh1.44𝑡|(+𝑐)
May see integration by substitution, eg.𝑢 = e1.44𝑡 +e−1.44𝑡or 𝑢 =
cosh1.44𝑡. If so, award this mark for 𝑘ln|𝑢| seen
For fully correct integration of tanh1.44𝑡.
Either for ∫tanh1.44𝑡d𝑡 = 1 ln|e1.44𝑡 +e−1.44𝑡|(+𝑐)
1.44
1
or ∫tanh1.44𝑡d𝑡 = ln|cosh1.44𝑡|(+𝑐)
1.44
1
or ∫tanh1.44𝑡d𝑡 = ln|𝑢|(+𝑐) with u as above.
1.44
Condone missing modulus.
cao, with clear working.
[4]
6 In this question you must show detailed reasoning.
The power output, $p$ watts, of a machine at time $t$ hours after it is switched on can be modelled by the equation $\mathrm { p } = 20 - 20 \tanh ( 1.44 \mathrm { t } )$ for $t \geqslant 0$.
Determine, according to the model, the mean power output of the machine over the first half hour after it is switched on. Give your answer correct to $\mathbf { 2 }$ decimal places.
\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q6 [4]}}