Question 11:
11 | ( )
Equates x2 −8x lnxto zero
PI by 1 or 8 or ±108.(2…)
May be seen on diagram or
integral | 3.1a | M1 | ( )
2 x − x 8 l n x = 0
x ( x − ) 8 l n x = 0
x = 8 , l n x = 0  x = 1
8 ( )
 2 x − x 8 l n x d x
1
1
u = l n x  u =
x
3 x
v  = x 2 − x 8 = v − 4 x 2
3
 3 x  3  x  1
− 4 2 x  l n x −   − 4 x 2 d x
3 3  x
 3 x  3  x 
= − 2 x 4 l n x −  − 2 x 2
3 9 
8( )
 x2 −8x lnx dx
1
 3   3 
8 8
= −48 2 ln8− −28 2 
 3   9 
 3   3 
1 1
− −41 2 ln1− −21 2 
 3   9 
256 640 17
=− ln8+ −
3 9 9
623
= −256ln2
9
6 2 3
 A r e a = − + 2 5 6 l n 2
9
Obtains at least one of
x = 1 or x = 8
PI by ±108.(2…)
May be seen on diagram or
integral | 1.1b | A1
Deduces the limits are 1 and 8
PI by ±108.(2…)
May be seen on integral or
substituted into their integrated
expression | 2.2a | R1
States u = l n x and v  = x 2 − 8 x
Condone v =lnx and
u  = x 2 − 8 x | 3.1a | M1
1 x 3
Finds u= and v = − 4 x 2
x 3 | 3.1a | A1
Applies integration by parts
formula correctly by substituting
their u, u’ and v
PI by
 x3   x3 
 −4x2 lnx− −2x2 or
 3   9 
 x 3   x 3 
− + 4 x 2 l n x + − 2 x 2
3 9
Condone missing brackets | 1.1a | M1
Obtains
x3  x3 
 −4x2 lnx− −2x2  OE
 3   9 
or
x3  x3 
− +4x2 lnx+ −2x2 OE
 3   9  | 1.1b | A1
Substitutes their non-zero limits
correctly into their integrated
expression (the subtraction does
not need to be seen)
or
obtains exact values for their
integrated expression using
their non-zero limits
256 640 17
eg− ln8+ and
3 9 9 | 1.1a | M1
6 2 3
Obtains − 2 5 6 l n 2 or
9
623 256
− ln8
9 3
ACF must be exact form with
two terms
6 2 3
PI − + 2 5 6 l n 2
9 | 1.1b | A1
Completes a reasoned
argument to obtain
623
− +256ln2
9
To be awarded R1, all marks
must be scored | 2.1 | R1
Question 11 Total | 10
Q | Marking instructions | AO | Marks | Typical solution