Mathematics · May / June 2025 · Paper 41 · Question 9

YMA01

As printed in the paper

The question as text

9. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. y C 8 cm 4 x O Figure 1 Figure 2 Figure 1 shows a sketch of part of the curve C with equation 1 x y x cos 5 e The finite region, shown shaded in Figure 1, is bounded by C, the y‑axis, the x‑axis and the line with equation x = 4 The region is rotated through 2π radians about the x‑axis to form a solid S.

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9(a)[4]
4
2
1
( )
=
π
+
V
cos
x
e
x
d
x
5
0
2
1
2
1
cos
x
+
e
x
=
cos
2
x
+
e cos
x
x
+
e
2
x
5
5
25
+
cos2
x
1
2
1
=
+
e cos
x
x
+
e
2
x
2
5
25
4
1
cos2
x
2
1
=
π
+
+
e cos
x
x
+
e
2
x
d
x
2
2
5
25
0
9(b)[4]
Way I
Way Two
e cos
x
x
d
x
= ±
e sin
x
x
±
e sin
x
x
d
x
e cos
x
x
d
x
=
e cos
x
x
±
e sin
x
x
d
x
+
−
=
e sin
x
x
e sin
x
x
d
x
=
e cos
x
x
e sin
x
x
d
x
e cos
x
x
d
x
=
e cos
x
x
±
e sin
x
x
−
e cos
x
x
d
x
2 e cos
x
x
d
x
=
e cos
x
x
±
e sin
x
x
1
(
)
( )
e cos
x
x
d
x
=
e
x
cos
x
+
sin
x
+
c
2
9(c)[4]
4
1
cos2
x
2
1
π
+
+
e cos
x
x
+
e
2
x
d
x
2
2
5
25
0
4
1
sin 2
x
2 1
1
=
( )
π
+
+
(
+
)
+
x
e
x
sin
x
cos
x
e
2
x
2
4
5 2
50
0
sin8
2 1
1
11
(
( )
( )
)
π
+
+
4
+
+
8
−
2
2
e
sin 4
cos 4
e
4
5 2
50
50
=
awrt 290
(cm3)
(a)
4
1
2
π
+
B1:
A correct expression for the volume including limits and dx E.g.
cos
x
e
x
5
0
The correct limits may be implied by subsequent work. It can be awarded from a correct final answer (as it
is not a given answer). Don’t be concerned by the non appearance of a V. Condone it being termed ‘Area’
2
1
+
x
2
±
x
±
2
x
M1:
Attempts to multiply out
cos
x
e
leading to
cos
x
...e cos
x
...e
5
= ±
2
±. Dependent upon previous M
dM1: Uses the identity
cos2
x
2cos
x
1
4
1
cos2
x
2
1
π
+
+
e cos
x
x
+
e
2
x
d
x
A1:
Fully correct
2
2
5
25
0
For this final mark
condone either a missing dx or limits (but not both) if the candidate has already lost the B mark for
the same reason
condone missing limits if they have already appeared within an expression for the volume
anywhere in part (a)
(b) Condone lack of dx’s throughout part (b).
If you see other methods that you feel deserve credit, and you cannot apply the scheme, then please send
to review.
M1: Attempts integration by parts. Be generous unless they explicitly state the rule is
v u
d
Score for Way One e cos
x
x
d
x
= ±
e sin
x
x
±
e sin
x
x
d
x
x
=
x
±
x
Or Way Two e cos
x
d
x
e cos
x
e sin
x
d
x
A1:
Correct intermediate form
Score for Way One e cos
x
x
d
x
=
e sin
x
x
−
e sin
x
x
d
x
Or Way Two e cos
x
=
x
+
x
x
d
x
e cos
x
e sin
x
d
x
dM1: Attempts integration by parts a second time. It is dependent upon having scored the previous M.
Score for e cos
x
x
d
x
=
e cos
x
x
±
e sin
x
x
−
e cos
x
x
d
x
AND rearranges to 2 e cos
x
=
x
±
x
x
d
x
e cos
x
e sin
x
1
(
)
A1:
CSO
e cos
x
x
d
x
=
e
x
cos
x
+
sin
x
+
c
o.e. condone lack of +c and the dx
2
Withold this mark if there is incorrect or unexplained working within the proof
(c)
M1:
Attempts to integrate their (a) condoning transcription errors on coefficients achieving
1
(
)
α
+
β
+
x
+
+
2
x
where , ,
αβand are constants following through on their
x
sin 2
x
"
e
cos
x
sin
x
"
e
2
attempted integration of e cos
x
x
They must have an attempted integration for part (b), which may be incorrect, which is then used in (c).
1
sin 2
x
2 1
1
A1ft: ( )
(
)
π
x
+
+
e
x
cos
x
+
sin
x
+
e
2
x
2
4
5 2
50
Condone the absence of π and follow through their coefficients A, B, C and D from (a). The integration
from part (c) must be correct.
(
)
4
B
1
So, look for
A
+
B
cos2
x
+
C
e cos
x
x
+
D
e
2
x
d
x
→
(
Ax
+
sin 2
x
+
C
e
x
cos
x
+
sin
0
2
2
dM1: This requires
the limits 4 and 0 must be substituted into an expression of the form
(
)
ax
+
b
sin 2
x
+
c
e
x
cos
x
±
sin
x
+ e
d
2
x
either way around and subtracted. This may be implied by a
decimal answer which you may need to check.
the answer to the above multiplied by 2π (An appreciation that the shape consists of two solids)
awrt 290 cm3 (condone lack of units).
A1:
If they have correct integration, correct limits followed by awrt 290, the last two marks can be awarded.
Note that an answer of 460 is achieved when candidates are working in degrees and scores the method
mark but not the accuracy mark.

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