Challenging +1.3 This is a Further Maths question requiring integration by parts twice (standard technique for exponential-trig products) and solving a first-order linear ODE with integrating factor. Both parts follow well-established procedures taught in Further Maths, though part (b) requires recognizing the connection to part (a) and careful application of the integrating factor method. The 12 total marks reflect moderate length rather than exceptional difficulty.
Given that \(I = \int_a^b e^{2t} \sin t \, dt\), show that
$$I = \left[ qe^{2t} \sin t + re^{2t} \cos t \right]_a^b$$
where \(q\) and \(r\) are rational numbers to be found.
[6 marks]
A small object is initially at rest. The subsequent motion of the object is modelled by the differential equation
$$\frac{dv}{dt} + v = 5e^t \sin t$$
where \(v\) is the velocity at time \(t\).
Find the speed of the object when \(t = 2\pi\), giving your answer in exact form.
[6 marks]
πΌπΌ = οΏ½ e sinπ‘π‘β e cosπ‘π‘οΏ½ β πΌπΌ
2 4 ππ 4
ππ
2 2π‘π‘ 1 2π‘π‘
πΌπΌ = οΏ½ e sinπ‘π‘β e cosπ‘π‘οΏ½
5 5 ππ
Obtains correct result of
Answer
Marks
Guidance
integration by parts.
1.1b
A1
Uses integration by parts a
second time, consistent with
their choice of and in
their first integration by
Answer
Marks
Guidance
parts. π’π’ π£π£β²
1.1a
M1
Obtains correct result of
second integration by parts
FT their first integration by
Answer
Marks
Guidance
parts.
1.1b
A1
Deduces that the second
integration by parts gives an
equation in which can be
Answer
Marks
Guidance
solved
2.2a
M1
πΌπΌ
Completes rigorous
argument to show that the
result of the second
integration by parts gives
Answer
Marks
Guidance
ππ
2.1
R1
Answer
Marks
12(b)
Selects a method to
solve the differential
equation by finding an
Answer
Marks
Guidance
integrating factor.
3.1a
M1
+v=5etsint
dt
β«1dt
IF: e =et
d ( vet) =5e2tsint
dt
vet = 2e2tsint βe2t cost +c
When ,
π‘π‘ = 0 π£π£ = 0 β΄ ππ = 1
vet = 2e2tsintβe2tcost+1
ve2Ο =2e4Οsin2Οβe4Οcos2Ο+1
=βe4Ο +1
v = βe2Ο +eβ2Ο
Speed = e2Ο βeβ2Ο
Multiplies the differential
equation by their
Answer
Marks
Guidance
integrating factor
1.1a
M1
Integrates LHS correctly
Answer
Marks
Guidance
to obtain
1.1b
A1
Integrates thπ‘π‘eir RHS
correctly wπ£π£ππith or without
limits
Answer
Marks
Guidance
FT their and from (a)
1.1b
B1F
Uses the initial
conditionππs wheππn
evaluating a definite
integral or determining
the constant of
Answer
Marks
Guidance
integration.
3.4
M1
States the correct exact
value of the speed,
which must be positive
FT their and from (a)
provided v < 0
Answer
Marks
Guidance
when t =ππ 2 ππ
3.2a
A1
Total
12
Q
Marking Instructions
AO
Question 12:
--- 12(a) ---
12(a) | Selects a method to find the
required result by integrating
by parts. | 3.1a | M1 | ππ
1 2π‘π‘ 1 ππ 2π‘π‘
πΌπΌ = οΏ½2e sinπ‘π‘οΏ½ β2β«ππ ππ cosπ‘π‘πππ‘π‘
ππ
ππ
1 2π‘π‘
πΌπΌ = οΏ½ e sinπ‘π‘οΏ½
2 ππ
ππ
1 1 2π‘π‘
β οΏ½οΏ½ e cosπ‘π‘οΏ½
2 2 ππ
ππ
1 2π‘π‘
+ οΏ½ e sinπ‘π‘πππ‘π‘οΏ½
2 ππ
ππ
1 2π‘π‘ 1 2π‘π‘ 1
πΌπΌ = οΏ½ e sinπ‘π‘β e cosπ‘π‘οΏ½ β πΌπΌ
2 4 ππ 4
ππ
2 2π‘π‘ 1 2π‘π‘
πΌπΌ = οΏ½ e sinπ‘π‘β e cosπ‘π‘οΏ½
5 5 ππ
Obtains correct result of
integration by parts. | 1.1b | A1
Uses integration by parts a
second time, consistent with
their choice of and in
their first integration by
parts. π’π’ π£π£β² | 1.1a | M1
Obtains correct result of
second integration by parts
FT their first integration by
parts. | 1.1b | A1
Deduces that the second
integration by parts gives an
equation in which can be
solved | 2.2a | M1
πΌπΌ
Completes rigorous
argument to show that the
result of the second
integration by parts gives
ππ | 2.1 | R1
--- 12(b) ---
12(b) | Selects a method to
solve the differential
equation by finding an
integrating factor. | 3.1a | M1 | dv
+v=5etsint
dt
β«1dt
IF: e =et
d ( vet) =5e2tsint
dt
vet = 2e2tsint βe2t cost +c
When ,
π‘π‘ = 0 π£π£ = 0 β΄ ππ = 1
vet = 2e2tsintβe2tcost+1
ve2Ο =2e4Οsin2Οβe4Οcos2Ο+1
=βe4Ο +1
v = βe2Ο +eβ2Ο
Speed = e2Ο βeβ2Ο
Multiplies the differential
equation by their
integrating factor | 1.1a | M1
Integrates LHS correctly
to obtain | 1.1b | A1
Integrates thπ‘π‘eir RHS
correctly wπ£π£ππith or without
limits
FT their and from (a) | 1.1b | B1F
Uses the initial
conditionππs wheππn
evaluating a definite
integral or determining
the constant of
integration. | 3.4 | M1
States the correct exact
value of the speed,
which must be positive
FT their and from (a)
provided v < 0
when t =ππ 2 ππ | 3.2a | A1
Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Given that $I = \int_a^b e^{2t} \sin t \, dt$, show that
$$I = \left[ qe^{2t} \sin t + re^{2t} \cos t \right]_a^b$$
where $q$ and $r$ are rational numbers to be found.
[6 marks]
\item A small object is initially at rest. The subsequent motion of the object is modelled by the differential equation
$$\frac{dv}{dt} + v = 5e^t \sin t$$
where $v$ is the velocity at time $t$.
Find the speed of the object when $t = 2\pi$, giving your answer in exact form.
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2020 Q12 [12]}}