OCR MEI Further Pure Core Specimen — Question 15 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeIntegration by parts with inverse trig
DifficultyChallenging +1.8 This is a Further Maths integration by parts question requiring knowledge of the arcsinh function and its derivative. While the technique is standard (integration by parts with u = arcsinh 2x), students must correctly handle the derivative of arcsinh, perform algebraic manipulation with the resulting integral, and evaluate definite integral limits involving arcsinh(2π/3). The 8-mark allocation and 'show detailed reasoning' instruction indicate substantial working, but the path is relatively clear for Further Maths students who know the standard approach to integrating inverse hyperbolic functions.
Spec1.08i Integration by parts4.07f Inverse hyperbolic: logarithmic forms

In this question you must show detailed reasoning. Show that $$\int_0^{\frac{\pi}{3}} \operatorname{arcsinh} 2x \, dx = \frac{2}{3} \ln 3 - \frac{1}{3}.$$ [8]

Question 15:
AnswerMarks
15DR
dv
Let u(cid:32)arsinh2x, (cid:32)1
dx
(cid:159)v(cid:32)x
2
derivative of arsinh 2x is
1(cid:14)4x2
2 2 2x
(cid:62)x arsinh 2x(cid:64)3 (cid:16)(cid:179)3 dx
0 0 1(cid:14)4x2
2
(cid:170) 1 (cid:186)3
x arsinh 2x (cid:16) 1(cid:14)4x2
(cid:171) (cid:187)
(cid:172) 2 (cid:188)
0
2 4 1 5 1
arsinh (cid:16)0(cid:16) (cid:117) (cid:14)
3 3 2 3 2
2 (cid:167)4 16(cid:183) 1 5 1
ln(cid:168) (cid:14) 1(cid:14) (cid:184)(cid:16) (cid:117) (cid:14) e
(cid:168) (cid:184)
3 3 9 2 3 2
(cid:169) (cid:185)
2 1 p
= ln3(cid:16) AG
AnswerMarks
3 3M1
M1
A1A1
A1
A1
c
M1
A1
AnswerMarks
[8]3.1a
1.1
1.1 1.1
m2.1
1.1
i
1.1
AnswerMarks
2.1Use parts 1 and arsinh 2x
n
A1 for each term
e
must be seen
converting to logarithms
AG so must be convincing
AnswerMarks Guidance
155 2
QuestionAO1 AO2
16iA1 0
16iB1 0
16iiA2 1
16iiB0 0
16iii1 0
16iv2 1
16v0 0
Totals74 42
Question 15:
15 | DR
dv
Let u(cid:32)arsinh2x, (cid:32)1
dx
(cid:159)v(cid:32)x
2
derivative of arsinh 2x is
1(cid:14)4x2
2 2 2x
(cid:62)x arsinh 2x(cid:64)3 (cid:16)(cid:179)3 dx
0 0 1(cid:14)4x2
2
(cid:170) 1 (cid:186)3
x arsinh 2x (cid:16) 1(cid:14)4x2
(cid:171) (cid:187)
(cid:172) 2 (cid:188)
0
2 4 1 5 1
arsinh (cid:16)0(cid:16) (cid:117) (cid:14)
3 3 2 3 2
2 (cid:167)4 16(cid:183) 1 5 1
ln(cid:168) (cid:14) 1(cid:14) (cid:184)(cid:16) (cid:117) (cid:14) e
(cid:168) (cid:184)
3 3 9 2 3 2
(cid:169) (cid:185)
2 1 p
= ln3(cid:16) AG
3 3 | M1
M1
A1A1
A1
A1
c
M1
A1
[8] | 3.1a
1.1
1.1 1.1
m2.1
1.1
i
1.1
2.1 | Use parts 1 and arsinh 2x
n
A1 for each term
e
must be seen
converting to logarithms
AG so must be convincing
15 | 5 | 2 | 1 | 0 | 8
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Totals
16iA | 1 | 0 | 1 | 1 | 3
16iB | 1 | 0 | 0 | 0 | 1
16iiA | 2 | 1 | 0 | 0 | 3
16iiB | 0 | 0 | 0 | 2 | 2
16iii | 1 | 0 | 1 | 1 | 3
16iv | 2 | 1 | 0 | 1 | 4
16v | 0 | 0 | 0 | 2 | 2
Totals | 74 | 42 | 21 | 7 | 144
In this question you must show detailed reasoning.

Show that
$$\int_0^{\frac{\pi}{3}} \operatorname{arcsinh} 2x \, dx = \frac{2}{3} \ln 3 - \frac{1}{3}.$$ [8]

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q15 [8]}}