Question 9:
--- 9(a) ---
9(a) | 3

1
3 −
 x 2
Volume, V = y2dx= dx
1 3  ( 1+x ) arctan ( x )  2
1  
3 | B1
d u 1 − 12
t a n u = x  s e c 2 u = x oe
d x 2 | M1
A1
 1 − 1
2 x 2
V =() 2 dx=()  2 sec2udu
  ( ) 2 ( ) u2 ( 1+tan2u )
 arctan x  1+x
  | dM1
2 1  
 = s e c 2 u d u 2 d u  =
u 2 s e c 2 u u 2 | A1

x=3u =arctan 3 =  
3   3 1
V =2 du
1 1  u2
x= u =arctan =  
3 3 6 6
 | B1
(6)
(b) | 
1 3  
V ( 2 )  = −
u

6 | M1
 
( ) 1  1 
= 2 − −− 
 /3  /6
 
  
3     
1 1 1
Or ( 2 )  = − =(2)− −− 
a r c t a n x  arctan 3  arctan 1 
13  3 
   | dM1
3 6  
2 6  = − + =
  | A1
(3)
(9 marks)
B1: For a fully correct integral in x or u with , 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
 1 −
x 2
Look for V = dx OR
 ( 1+x ) arctan ( x )   2
1  
3
V =

3
13

1 + x a
x
r
−
c
1
4
t a n ( x )
 2
d x 
If the correct integral in x is not seen score for V =

6
3
u 2
(
1 +
2
t a n 2 u
)
s e c 2 u d u

 o.e.

M1: Attempts the differentiation of the given substitution achieving the correct form with suitable
placement for du and dx.
Examples;
• t a n u = x  s e c 2 u d u = x
−
12
d x 
• x = t a n 2 u 
d
d
x
u
= ... t a n u s e c 2 u
• u = a r c t a n x 
d
d
u
x
=
1
1
+ x
 ... x
− 12
A1: Correct differentiation, any form. Score when first seen, don’t penalise an incorrect rearrangement.
Examples
• t a n u = x  s e c 2 u d u =
1
2
x
− 12
d x
• x = t a n 2 u 
d
d
x
u
= 2 t a n u s e c 2 u
• u = a r c t a n x 
d
d
u
x
=
1
1
+ x

1
2
x
− 12
dM1: Makes a complete substitution for x into

y 2 d x .
No need for the  or limits for this mark. It is dependent upon the previous M
Look for the form

u k
(
1 +
c
t a n 2 u
)
s e c 2 u ( d u ) (which may be un-simplified) where k = 1 o r 2
PMT
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.
The π etc may be missing or incorrect for this mark. Accept
1
u 2
→ 
a
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.
If in terms of x they should be using a r c t a n 3 and a r c t a n
1
3
PMT
A1: cso. Cannot be scored if k is guessed or found incorrectly
PMT
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