Mathematics · May / June 2023 · Paper 41 · Question 5
YMA01
As printed in the paper
Show the answer and mark scheme
5(i)[4]
(
)
x
2
e d
x
x
=
x
2
e
x
−
2 e d
x
x
x
+
c
e d
x
e
x
e d
x
(
)
x
x
=
x
−
x
+
c
(
)
x
2
e d
x
x
=
x
2
e
x
−
2 e
x
x
+
2e
x
+
c
(
)(
)
Also allow
x
2
e d
x
x
=
x
2
e
x
−
2 e
x
x
−
2e
x
+
c
5(ii)[6]
1
d
u
(
)
u
=
1 3
−
x
2
u
2
= −
1 3
x
2
u x
= −
3
d
or
1
1
d
u
3
−
(
)
(
)
u
=
1 3
−
x
2
= −
1 3
−
x
2
d
x
2
27
x
1
−
u
2
2
u
d
x
=
27
3
−
d
u
1 3
−
x
u
3
or
27
x
1
−
u
2
2
d
x
=
27
3
−
1 3
−
x
d
u
3
1 3
−
x
1 3
−
x
(
)
(
)
6
u
2
−
1 d
u
or
−
6
1
−
u
2
d
u
u
3
(
)
6
−
u
+
k
3
3
(
)
1 3
−
x
2
1
1
(
)
(
)
(
) (
)(
)
6
−
1 3
−
x
+
k
=
2 1 3
−
x
1 3
−
x
−
3
+
k
2
2
3
1
1
(
) (
)(
)
(
) (
)(
)
= −
2 1 3
−
x
2
+
3
x
+
k
or
= −
2 1 3
−
x
3
x
+
2
+
k
2
2
(i) Condone missing d 'x s here
(
)
M1: For applying parts to obtain
x
2e
x
−
k
x
e d
x
x
+
c
where k > 0
A1: Correct expression. No requirement for + c
e d
x
e d
x
e
x
e d
x
(
)
M1: Applies parts again to
x
x
and obtains
x
x
=
x
−
α
x
+
c
where
α
>
0
Note that if the whole expression is written out then this mark is implied by
x
2
e d
x
x
=
x
2
e
x
−
α
x
e
x
±
β
e d
x
x
x
2
e d
x
x
=
x
2
e
x
−
α
x
e
x
±
β
e
x
or
where
α
>
0
A1: Correct answer (+ c not required)
(
)(
2
e d
x
2
e
x
2 e
x
2e
x
)
Watch for
x
x
=
x
−
x
−
+
c
which scores all 4 marks. ISW after sight of this
Watch for D & I method which may be seen. You will just see the answer here.
D
I
2x
ex
+
=
x
2e
x
−
2 e
x
x
+
2e
x
2x
−
ex
2
Score M2 for
x
2e
x
±
2 e
x
x
±
2e
x
ex
+
(
)
and then A2 for
x
2e
x
−
2 e
x
x
+
2e
x
+
c
0
ex
(ii)
B1: Any correct expression involving d
u
3
1
−
(
)
x which may be ''split'' E.g.
d
u
= −
1 3
−
x
2
d
x
d
2
(
)
f
2
27
x
u
(
)
M1: Substitutes
x
=
f
u
2
to fully change the integrand from
d
x
to
k
u u
d
o.e.
1 3
x
u
−
Note that the u's may have been cancelled.
Condone the du not being present but it cannot appear as dx
A1: Correct simplified integral. Look for the u's being cancelled and some simplification/collection of
the constant terms
(
)
(
)
6
u
2
1 d ,
u
6
u
2
6d ,
u
6
1
u
2
d
u
Look for, for example. integrals such as
−
−
−
−
or even
2
(
)
−
9
−
9
u
2
d
u
The brackets may be implied by subsequent work
3
Condone the du not being present as long as it is implied by subsequent integration
A1ft: Correct follow through integration on their simplified integral.
It is dependent upon having achieved an integral of the form above
(
)
(
)
E.g.
α
u
2
−
1 d
u
or
β
1
−
u
2
d
u
αβ
,
>
0
. No requirement for + k
M1: Achieves an integral of
Pu
3
Qu
and
+
1
''correctly'' takes out a common factor of u or (
1 3x
−
)
2
for their
Pu
3
+
Qu
or
3
1
1
(
)
(
)
(
)
AND proceeds to the form (
)
(
)
α
1 3
−
x
+
β
1 3
−
x
1 3
−
x
1 3
−
x
+
or better. Condone
2
2
2
slips on coefficients
1
(
)
substitutes
u
=
1 3
−
x
2
A1: CAO. No requirement for + k
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