Question 13:
--- 13(a) ---
13(a) | Explains that one of the
limits is infinity (or that
the interval of
integration is infinite) | AO2.4 | R1 | The upper limit is infinity, so it is an
improper integral.
(b) | Uses integration by
parts twice | AO3.1a | M1 | 2 ′ −2𝑎𝑎
𝑢𝑢 = 𝑎𝑎 𝑣𝑣 = e −2𝑎𝑎
−e
′
𝑢𝑢 = 2𝑎𝑎 𝑣𝑣 = 2
2 −2𝑥𝑥
2 −2𝑥𝑥 𝑎𝑎 e −2𝑥𝑥
� 𝑎𝑎 e d𝑎𝑎 = − + � 𝑎𝑎e d𝑎𝑎
2
′ −2𝑎𝑎
𝑢𝑢 = 𝑎𝑎 𝑣𝑣 = e −2𝑎𝑎
−e
′
𝑢𝑢 = 1 𝑣𝑣 = 2
−2𝑥𝑥 −2𝑥𝑥
−2𝑥𝑥 𝑎𝑎e e
� 𝑎𝑎e d𝑎𝑎 = �− �+ � d𝑎𝑎
2 2
−2𝑥𝑥 −2𝑥𝑥
𝑎𝑎e e
= − −
2 4
2 −2𝑥𝑥 −2𝑥𝑥 −2𝑥𝑥
2 −2𝑥𝑥 𝑎𝑎 e 𝑎𝑎e e
∴ �𝑎𝑎 e d𝑎𝑎=− − −
2 2 4
∞ 𝑛𝑛
2 −2𝑥𝑥 2 −2𝑥𝑥
�𝑎𝑎 e d𝑎𝑎 =𝑛𝑛li→m∞�𝑎𝑎 e d𝑎𝑎
3 3
2 −2𝑥𝑥 −2𝑥𝑥 −2𝑥𝑥 𝑛𝑛
𝑎𝑎 e 𝑎𝑎e e
=𝑛𝑛li→m∞��− − − � �
2 2 4 3
∞
−6 −6 −6
2 −2𝑎𝑎 9e 3e e
∴ �𝑎𝑎 e d𝑎𝑎 = + +
2 2 4
3
−6
25e
=
4
Obtains the correct
expressions for u’ and v
when integrating the
first time | AO1.1b | B1
Correctly applies
integration by parts
formula the first time | AO1.1b | B1
Correctly applies
integration by parts
formula to an integral of
the form
−2𝑥𝑥 | AO1.1a | M1
𝑘𝑘∫ 𝑎𝑎e d𝑎𝑎
Finds complete correct
expression for integral
with or without c. No
limits needed at this
stage.
PI by later work | AO1.1b | A1
Defines the improper
integral as a limit | AO2.4 | E1
Applies the limiting
process correctly, using
2 −𝑛𝑛 and
lim𝑛𝑛→∞(𝑛𝑛 e )= 0
−𝑛𝑛 | AO2.2a | M1
lSimub 𝑛𝑛 s → t ∞ itu(𝑛𝑛te− es𝑛𝑛 co)r=re0ct
lloimw 𝑛𝑛 e → r ∞ lim(eit )= 0
correctly into their
three-term expression | AO1.1a | M1
Obtains correct exact
value or awrt 0.0155 | AO1.1b | A1
Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution