Mathematics · Oct / Nov 2021 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. y y = x ln x R O e x Figure 3 ∫ 2 ln d
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[3]
3
3
x
1
1
3
ln
3
(
)
=
x
x
−
x
+
c
3
9
∫
1
∫
1
1
2
8(b)[4]
x
2
ln
2
x
d
x
=
x
3
ln
x
−
x
3
×
2ln
x
×
d
x
3
3
x
1
2 1
1
3
2
3
3
=
x
ln
x
−
x
ln
x
−
x
3
3 3
9
∫π
e
e
1
2 1
1
2
x
2
ln
2
x
d
x
=
π
×
x
3
ln
x
−
x
3
ln
x
−
x
3
Volume =
3
3 3
9
1
1
5
2
3
=
π
e
−
π
27
27
(a) Note that no marks are available for integrating incorrect functions
∫
x
2
ln
x x
d
=
...
x
3
M1: Attempts to integrate by parts the correct way around. Look for
3
3
dM1: And then integrates again to a form
px
ln
x
−
qx
where p and q are positive constants.
1
1
A1:
x
3
ln
x
−
x
3
+
c
Condone a missing + c.
3
9
x
3
1
x
4
1
ln
3
ln
3
Candidates who go on to write
x
×
−
x
=
−
x
o.e. withhold this final A1
3
9
3
9
(b)
∫
2
2
3
2
M1: Integrates
x
ln
x
d
x
by parts, the correct way around to reach a form
...
x
ln
x
You may see a π or 2π in front of this expression which is fine.
∫
1
2
3
2
A1ft: Achieves
x
2
ln
2
x
d
x
=
x
ln
x
−
×
their answer to ( )
a
.
3
3
∫π
1
2
2
2
2
=
π
x
3
ln
x
−
×
π
×
their answer to ( )
a
Accept
x
ln
x
d
x
or other incorrect factors such as
3
3
∫
2
4
3
2
2
π
x
2
ln
2
x
d
x
=
π
x
ln
x
−
×
π
×
their answer to ( )
a
3
3
1
2 1
1
3
2
3
3
FYI The correct answer is
x
ln
x
−
x
ln
x
−
x
3
3 3
9
2
dM1: Applies the limits 1 and e to a function of the form
π
×
α
x
3
ln
x
−
β
x
3
ln
x
±
χ
x
and subtracts where
α β χ are positive constants. The π must now be seen.
,
,
(
)
5
2
π
3
π
e
−
π or exact simplified equivalent such as
3
A1:
5e
−
2
27
27
27
Alt (b)
∫
∫
2
(b)
x
2
ln
2
x
d
x
=
ln
x
×
x
ln
x
d
x
1
1
∫
1
1
1
3
3
3
3
=
''
x
ln
x
−
x
''ln
x
−
''
x
ln
x
−
x
''
×
d
x
3
9
3
9
x
∫
1
1
1
1
=
''
x
3
ln
x
−
x
3
''ln
x
−
x
2
ln
x
−
x
2
d
x
3
9
3
9
1
1
1
1
1
1
=
''
x
3
ln
x
−
x
3
''ln
x
−
''
x
3
ln
x
−
x
3
''
+
x
3
3
9
3
3
9
27
Volume =
e
∫π
e
1
1
1 1
1
1
2
x
ln
2
x
d
x
=
π
×
x
3
ln
x
−
x
3
ln
x
−
x
3
ln
x
−
x
3
+
x
3
3
9
3 3
9
27
1
1
5
2
3
=
π
e
−
π
27
27
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