Question Number
Scheme 1
t = (3x + 1) 2 ⇒
(c)
dt = dx
− 21
dt =3 dx
dt 3 3 = = 1 dx 2.(3x + 1) 2 2t
∫
∴I = e
∴I=
∫
2 3
(3 x +1)
A(3x + 1)
.3.(3x + 1)
1 2
… or t 2 = 3 x + 1 ⇒ 2t
so
Marks
∫
dx = et
⇒
3 2
dx 2t = dt 3
(3x + 1)
− 21
dt =A dx dt or 2t =3 dx
− 21
or t
M1 A1
Candidate obtains either dt or dx in terms of t … dx dt
∫
dx 2t . dt = e t . .dt dt 3
… and moves on to dM1 substitute this into I to convert an integral wrt x to an integral wrt t.
∫
t et dt
2 3
t et
A1
changes limits x → t so B1 that 0 → 1 and 5 → 4
change limits: when x = 0, t = 1 & when x = 5, t = 4 4
Hence I =
∫
2 3
tet dt ; where a = 1, b = 4, k =
2 3
1
[5] (d)
Let k be any constant for the first three marks of this part.
du ⎪⎧u = t ⇒ dt = 1 ⎪⎫ ⎨ dv t t⎬ ⎩⎪ dt = e ⇒ v = e ⎭⎪
(
∫
∫
k t et dt = k t e t − e t .1 dt
(
= k t et − et
4
∴
∫ 1
2 3
tet dt =
{(
)
Use of ‘integration by parts’ formula in the M1 correct direction. Correct expression with a A1 constant factor k.
)
Correct integration with/without A1 a constant factor k
+c
) (
2 4e4 − e4 − e1 − e1 3
Substitutes their changed limits into the integrand dM1 oe and subtracts oe.
)}
= 32 (3e4 ) = 2e 4 = 109.1963...
either 2e4 or awrt 109.2 A1 [5] 15 marks
• •
Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. 20