Mathematics · May / June 2024 · Paper 41 · Question 1
YMA01
As printed in the paper
The question as text
1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Find π 6 x cos3 d x x 0 giving your answer in simplest form. (5)
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Show the answer and mark scheme
x
cos3 d
x x
=
x
sin3
x
sin3
x
d
x
3
3
1
1
=
x
sin3
x
+
cos3
x
3
9
π
π
6
1
1
6
1
π
π
1
π
1
x
cos3
x
d
x
=
x
sin3
x
+
cos3
x
=
sin
+
cos
−
0
+
3
9
3 6
2
9
2
9
0
0
π
1
=
−
18
9
±
M1: Applies integration by parts in the correct direction to achieve
α
x
sin3
x
β
s n3 d
i
x
So do not penalise the appearance of a ‘+’ sign unless they state
vu
'd
x uv
uv
'd
=
+
dM1: Completes the process by integrating the sin3x term, achieving
x
sin3
x
cos3
x
±
α
±
β
A1: Correct integration which may be left un simplified. E.g. 1
1
1
x
sin3
x
−
−
cos3
x
3
3
3
May have been done in stages (allow if 1 sin3
1 cos3
x
x and
+
x
are seen on separate lines.)
3
9
Ignore spurious notation as long as the intention is clear.
M1: Applies both the limits to their integral which must be of the form
sin3
cos3
±
α
x
x
±
β
substitution must be seen, or clearly implied by the correct evaluation of their
±
α
x
sin3
π
If embedded limits are not seen first, then [
sin3
cos
3
]
( )
...
0
±α
x
x
±β
x
=
−
is M0
6
0
Condone the subtraction being done either way around.
π
−
2
A1: Correct simplified answer. ISW after a correct answer. Allow 18
Alt methods may be seen. E.g. D & I method
Differentiation
Integration
x
cos3x
+
1
1 sin3
x
3
−
1 cos3
0
−
x
9
Score M1 dM1 for sight of
±
α
x
sin3
x
±
β
cos3
x
7
−
2
a
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