Mathematics · May / June 2024 · Paper 41 · Question 9
YMA01
As printed in the paper
The question as text
9. y C R O 3 x 1 3 Figure 3 The curve C, shown in Figure 3, has equation 1 4 x y 1 x arctan x The region R, shown shaded in Figure 3, is bounded by C, the line with equation x = 3, the x‑axis and the line with equation x = 1 3 The region R is rotated through 360° about the x‑axis to form a solid. Using the substitution tan u = x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[6]
1
3
x
−
2
Volume, V
=
π
y
2
d
x
=
π
d
x
(
)
2
(
)
1
1
+
x
arct
an
x
3
1
3
d
u
1
1
tan
u
x
sec
u
x
−
oe
=
2
=
2
d
x
2
1
1
2
x
−
×
2
2
2
V
=
(
π
)
d
x
=
( )
π
sec
2
u u
d
(
)
(
) (
2
)
u
2
1
+
ta
n
2
u
arc
t
a
n
x
1
+
x
2
1
=
π
sec u
2
d
u
=
2
π
d
u
u
2
sec u
2
u
2
π
✓
3
arctan
3
x
=
u
=
=
π
3
1
3
V
2
d
u
=
π
1
1
π
u
2
x
=
u
=
arctan
=
π
6
3
3
6
π
9(b)[3]
π
1
3
V
=
(2 )
−
u
π
6
1
1
(
)
=
2π
−
−−
π
/ 3
π
/ 6
3
1
1
1
Or
=
(2
π
)
−
=
(2
π
)
−
−−
arctan
arctan
3
1
x
arctan
1
3
3
3
6
=
2
π
−
+
=
6
π
π
B1: For a fully correct integral in x or u with π, dx or du , correct limits and bracketing.
May be implied by a fully correct (simplified or un simplified) integral in u
May be seen at any stage in the working following correct work.
The limits must be correct for the variable they are working in. So for an integral in u the limits must
be in radians
3
=
=
3
1
1
2
−
x
x
−
2
4
Look for
V
π
d
x
OR
V
π
d
x
(
)
(
)
2
(
)
1
+
x
arctan
x
1
+
x
arctan
x
1
1
3
3
π
2
3
If the correct integral in x is not seen score for
V
=
π
sec
2
u u
d
o.e.
(
)
u
2
1
+
tan
2
u
π
6
M1: Attempts the differentiation of the given substitution achieving the correct form with suitable
placement for du and dx.
Examples;
1
−
t
a u
n
=
x
se
c
2
u u
d
=
α
x
d
x
2
d
x
x
=
tan
2
u
=
...tan sec
u
2
u
d
u
d
u
1
1
−
u
=
arctan
x
=
×
...
x
2
d
x
1
+
x
A1: Correct differentiation, any form. Score when first seen, don’t penalise an incorrect rearrangement.
Examples
1
1
−
tan
u
=
x
s
ec
2
u
d
u
=
x
d
x
2
2
d
x
x
=
tan
2
u
u =
2tan sec
u
2
u
d
d
u
1
1
1
−
u
=
arctan
x
= +
×
x
2
d
x
1
x
2
2 d
dM1: Makes a complete substitution for x into
y
x
.
No need for the π or limits for this mark. It is dependent upon the previous M
c
Look for the form
s
ec
2
u
(d
u
)
(which may be unsimplified) where
k =
(
)
1
tan
u
k
+
2
u
Condone a missing du. It can be implied if the substitution for dx has been attempted.
Condone missing brackets if the intention is clear.
A1: Correct simplified integral in terms of u including the π from correct working.
Allow recovery from missing brackets. No need for limits for this mark.
Condone a missing du. It can be implied if the substitution for dx has been attempted
B1: Correct limits for the integral found. Allow where seen, even in part (b) and even if not attached to
the integral. Must be in radians
(b)
M1: Carries out the integration.
1
a
The π etc may be missing or incorrect for this mark. Accept
→±
u
2
u
dM1: Substitutes appropriate limits into their integral and subtracts either way round.
If the integral is in terms of u the limits must be in radians, not degrees.
1
If in terms of x they should be using arctan
3 and
arctan
3
A1: cso. Cannot be scored if k is guessed or found incorrectly
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.