AQA Further Paper 1 2023 June — Question 16 11 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric integration
TypeSurface area with arc length identity
DifficultyChallenging +1.8 Part (a) requires recognizing integration by parts for ln t / t, which is a standard Further Maths technique. Part (b) involves surface area of revolution with parametric equations, requiring the formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt, careful algebraic manipulation, and connecting back to part (a). The multi-step nature, parametric context, and need to simplify to the exact form elevate this above routine questions, though the techniques themselves are standard Further Maths content.
Spec4.08d Volumes of revolution: about x and y axes
  1. Show that $$\int_{0.5}^4 \frac{1}{t} \ln t \, \mathrm{d}t = a(\ln 2)^2$$ where \(a\) is a rational number to be found. [4 marks]
  2. A curve C is defined parametrically for \(t > 0\) by $$x = 2t \quad y = \frac{1}{2}t^2 - \ln t$$ The arc formed by the graph of C from \(t = 0.5\) to \(t = 4\) is rotated through \(2\pi\) radians about the \(x\)-axis to generate a surface with area \(S\) Find the exact value of \(S\), giving your answer in the form $$S = \pi\left(b + c \ln 2 + d(\ln 2)^2\right)$$ where \(b\), \(c\) and \(d\) are rational numbers to be found. [7 marks]
Question 16:
AnswerMarks
16(a)Selects a suitable method
to find the required result
by integration by parts
or
making an appropriate
AnswerMarks Guidance
substitution/inspection3.1a M1
u =lnt =
dt t
du 1
= v=lnt
dt t
∫ 4 1 lntdt =(lnt)2 4 −∫ 4 1 lntdt
 
0.5t 0.5 0.5t
∫ 4 1 lntdt = 1 (lnt)2 4
 
0.5t 2 0.5
1
= (ln4)2 −(ln0.5)2
 
2
1
=  4(ln2)2 −(−ln2)2
 
2
∫ 4 1 lntdt = 3 (ln2)2
0.5t 2
or
let u=lnt
1
...du=... dt
t
4 1 ln4
∫ lntdt = ∫ udu
t
0.5 ln0.5
ln4
u2
=
 
 2 
ln0.5
(ln4)2 (ln0.5)2
= −
2 2
(2ln2)2 (−ln2)2
= −
2 2
3
= (ln2)2
2
Applies their correct
integration method to
obtain
1
2∫ lntdt =( lnt )2
 
t
or
1 u2
∫ lntdt =  
t  2 
AnswerMarks Guidance
OE1.1b A1
Substitutes limits and
uses the laws of logs
correctly to simplify their
result of integration in
AnswerMarks Guidance
terms of ln21.1a M1
Completes a rigorous
argument to show the
required result.
AnswerMarks Guidance
NMS = 0/42.1 R1
Subtotal4
QMarking instructions AO
AnswerMarks
16(b)Correctly obtains
dx2 dy2
  + 
dt dt
AnswerMarks Guidance
in any form.1.1b B1
x=2t y= t2−lnt
2
dx dy 1
=2 =t−
dt dt t
dx2 dy2 1  12
  +  =t2+2+ =t+ 
dt dt t2  t
dx2 dy2
S=2π∫y   +  dt
dt dt
=2π∫ 4   1 t2−lnt    t+ 1  2 dt
0.52   t
S=2π∫ 4   1 t3+ 1 t−tlnt− 1 lnt  dt
0.52 2 t 
dv
u=lnt =t
dt
du 1 1
= v= t2
dt t 2
1 1 1 1
∫tlntdt= t2lnt−∫ tdt= t2lnt− t2
2 2 2 4
S=2π   1 t4+ 1 t2− 1 t2lnt+ 1 t2  4 − 3 (ln2)2  
8 4 2 4  2 
0.5
S=2π  32+4− 1 − 1 +4− 1 −8ln4+ 1 ln   1 − 3 (ln2)2  
 128 16 16 8 2 2 
S=π   5103 − 129 ln2−3(ln2)2  
 64 4 
Substitutes y and their
dx2 dy2
  + 
dt dt
into the integrand of the
formula for Surface Area
AnswerMarks Guidance
of Revolution.1.2 M1
Obtains correctly
AnswerMarks Guidance
expanded integrand1.1b A1
Selects integration by
parts to calculate an
integral of form
AnswerMarks Guidance
k∫tlntdt3.1a M1
Obtains correct result of
integration by parts for
their k∫tlntdt.
No limits needed at this
AnswerMarks Guidance
stage.1.1b A1
Substitutes limits into
their expression of the
form at4 +bt2 +ct2lnt
and use of their answer
AnswerMarks Guidance
to part (a).1.1a M1
Obtains
5103 129 
π − ln2−3(ln2)2 
AnswerMarks Guidance
 64 4 2.1 R1
Subtotal7
Question total11
Paper total100 
Question 16:
--- 16(a) ---
16(a) | Selects a suitable method
to find the required result
by integration by parts
or
making an appropriate
substitution/inspection | 3.1a | M1 | dv 1
u =lnt =
dt t
du 1
= v=lnt
dt t
∫ 4 1 lntdt =(lnt)2 4 −∫ 4 1 lntdt
 
0.5t 0.5 0.5t
∫ 4 1 lntdt = 1 (lnt)2 4
 
0.5t 2 0.5
1
= (ln4)2 −(ln0.5)2
 
2
1
=  4(ln2)2 −(−ln2)2
 
2
∫ 4 1 lntdt = 3 (ln2)2
0.5t 2
or
let u=lnt
1
...du=... dt
t
4 1 ln4
∫ lntdt = ∫ udu
t
0.5 ln0.5
ln4
u2
=
 
 2 
ln0.5
(ln4)2 (ln0.5)2
= −
2 2
(2ln2)2 (−ln2)2
= −
2 2
3
= (ln2)2
2
Applies their correct
integration method to
obtain
1
2∫ lntdt =( lnt )2
 
t
or
1 u2
∫ lntdt =  
t  2 
OE | 1.1b | A1
Substitutes limits and
uses the laws of logs
correctly to simplify their
result of integration in
terms of ln2 | 1.1a | M1
Completes a rigorous
argument to show the
required result.
NMS = 0/4 | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b) ---
16(b) | Correctly obtains
dx2 dy2
  + 
dt dt
in any form. | 1.1b | B1 | 1
x=2t y= t2−lnt
2
dx dy 1
=2 =t−
dt dt t
dx2 dy2 1  12
  +  =t2+2+ =t+ 
dt dt t2  t
dx2 dy2
S=2π∫y   +  dt
dt dt
=2π∫ 4   1 t2−lnt    t+ 1  2 dt
0.52   t
S=2π∫ 4   1 t3+ 1 t−tlnt− 1 lnt  dt
0.52 2 t 
dv
u=lnt =t
dt
du 1 1
= v= t2
dt t 2
1 1 1 1
∫tlntdt= t2lnt−∫ tdt= t2lnt− t2
2 2 2 4
S=2π   1 t4+ 1 t2− 1 t2lnt+ 1 t2  4 − 3 (ln2)2  
8 4 2 4  2 
0.5
S=2π  32+4− 1 − 1 +4− 1 −8ln4+ 1 ln   1 − 3 (ln2)2  
 128 16 16 8 2 2 
S=π   5103 − 129 ln2−3(ln2)2  
 64 4 
Substitutes y and their
dx2 dy2
  + 
dt dt
into the integrand of the
formula for Surface Area
of Revolution. | 1.2 | M1
Obtains correctly
expanded integrand | 1.1b | A1
Selects integration by
parts to calculate an
integral of form
k∫tlntdt | 3.1a | M1
Obtains correct result of
integration by parts for
their k∫tlntdt.
No limits needed at this
stage. | 1.1b | A1
Substitutes limits into
their expression of the
form at4 +bt2 +ct2lnt
and use of their answer
to part (a). | 1.1a | M1
Obtains
5103 129 
π − ln2−3(ln2)2 
 64 4  | 2.1 | R1
Subtotal | 7
Question total | 11
Paper total | 100
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\int_{0.5}^4 \frac{1}{t} \ln t \, \mathrm{d}t = a(\ln 2)^2$$

where $a$ is a rational number to be found.
[4 marks]

\item A curve C is defined parametrically for $t > 0$ by
$$x = 2t \quad y = \frac{1}{2}t^2 - \ln t$$

The arc formed by the graph of C from $t = 0.5$ to $t = 4$ is rotated through $2\pi$ radians about the $x$-axis to generate a surface with area $S$

Find the exact value of $S$, giving your answer in the form
$$S = \pi\left(b + c \ln 2 + d(\ln 2)^2\right)$$

where $b$, $c$ and $d$ are rational numbers to be found.
[7 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 1 2023 Q16 [11]}}
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