OCR MEI Paper 2 Specimen — Question 13 6 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeSquare root substitution: definite integral
DifficultyChallenging +1.2 This requires the non-standard substitution u = √x (or u = 1 + √x) to transform the integral into a rational function, then partial fractions or algebraic manipulation, followed by integration and back-substitution. While the technique is A-level appropriate, recognizing the correct substitution and executing the multi-step process elevates this above routine integration questions.
Spec1.08h Integration by substitution

Evaluate \(\int_0^1 \frac{1}{1 + \sqrt{x}} \, dx\), giving your answer in the form \(a + b \ln c\), where \(a\), \(b\) and \(c\) are integers. [6]

Question 13:
AnswerMarks
131
(cid:16)
let u(cid:32)1(cid:14) x du(cid:32) 1x 2 dx
2
(cid:159) dx = 2(u – 1)du
1 1 22(u(cid:16)1)
(cid:159)(cid:179) dx(cid:32)(cid:179) du
01(cid:14) x 1 u
2(cid:167) 2(cid:183)
(cid:32)(cid:179) 2(cid:16) du
(cid:168) (cid:184)
1 (cid:169) u(cid:185)
(cid:62)2u(cid:16)2lnu(cid:64)2
=
1
AnswerMarks
= 4 – 2ln 2 – 2 = 2 – 2ln 2 or 2 – ln 4M1
A1
A1
M1
A1
A1
AnswerMarks
[6]3.1a
1.1
1.1
3.1a
m
1.1
i
AnswerMarks
1.1substituting u(cid:32)1(cid:14) x or w(cid:32) x
dx = 2(u – 1) du or dx = 2w dw
2(cid:11)u– 1(cid:12) 2w
(cid:11)du(cid:12)or (cid:11)dw(cid:12)
u (cid:11)w(cid:14)1(cid:12)
n
splitting fraction or dividing to get
2
2 –
e(cid:11)w(cid:14) 1(cid:12)
(or substituting u = w + 1 (cid:159)
2(cid:11)u– 1(cid:12)
and then splitting fraction)
u
(cid:62)2w(cid:16)2ln(w(cid:14)1)(cid:64)1
or if still in terms
0
of w
AnswerMarks
caoEvidence of method must be seen
Evidence of method must be seen
AnswerMarks Guidance
134 0
60
14 a5 2
14 b i2 0
02 0
14 b ii0 1
15 a1 0
02 0
15 b3 0
15 c0 0
02 2
15 d1 0
02 3
15 e2 2
16 a0 e
20 0
16 b i0 1
16 b ii0 p
00 1
16 b iii0 1
16 b ivS
01 0
16 c3 0
16 d i1 0
16 d ii1 0
16 d iii0 0
16 e i1 0
16 e ii0 0
16 f1 1
Totals51 27
4
Question 13:
13 | 1
(cid:16)
let u(cid:32)1(cid:14) x du(cid:32) 1x 2 dx
2
(cid:159) dx = 2(u – 1)du
1 1 22(u(cid:16)1)
(cid:159)(cid:179) dx(cid:32)(cid:179) du
01(cid:14) x 1 u
2(cid:167) 2(cid:183)
(cid:32)(cid:179) 2(cid:16) du
(cid:168) (cid:184)
1 (cid:169) u(cid:185)
(cid:62)2u(cid:16)2lnu(cid:64)2
=
1
= 4 – 2ln 2 – 2 = 2 – 2ln 2 or 2 – ln 4 | M1
A1
A1
M1
A1
A1
[6] | 3.1a
1.1
1.1
3.1a
m
1.1
i
1.1 | substituting u(cid:32)1(cid:14) x or w(cid:32) x
dx = 2(u – 1) du or dx = 2w dw
2(cid:11)u– 1(cid:12) 2w
(cid:11)du(cid:12)or (cid:11)dw(cid:12)
u (cid:11)w(cid:14)1(cid:12)
n
splitting fraction or dividing to get
2
2 –
e(cid:11)w(cid:14) 1(cid:12)
(or substituting u = w + 1 (cid:159)
2(cid:11)u– 1(cid:12)
and then splitting fraction)
u
(cid:62)2w(cid:16)2ln(w(cid:14)1)(cid:64)1
or if still in terms
0
of w
cao | Evidence of method must be seen
Evidence of method must be seen
13 | 4 | 0 | 2 | 0 | n
6 | 0
14 a | 5 | 2 | 1 | 0 | 8 | 0
14 b i | 2 | 0 | 0 | e
0 | 2 | 0
14 b ii | 0 | 1 | 1 | 0 | 2 | 0
15 a | 1 | 0 | 1 | m
0 | 2 | 0
15 b | 3 | 0 | 0 | 0 | 3 | 0
15 c | 0 | 0 | i
0 | 2 | 2 | 0
15 d | 1 | 0 | c
0 | 2 | 3 | 0
15 e | 2 | 2 | 0 | 1 | 5 | 0
16 a | 0 | e
2 | 0 | 0 | 2 | 0
16 b i | 0 | 1 | 0 | 0 | 1 | 0
16 b ii | 0 | p
0 | 0 | 1 | 1 | 0
16 b iii | 0 | 1 | 0 | 0 | 1 | 1
16 b iv | S
0 | 1 | 0 | 0 | 1 | 1
16 c | 3 | 0 | 0 | 0 | 3 | 0
16 d i | 1 | 0 | 1 | 0 | 2 | 0
16 d ii | 1 | 0 | 0 | 0 | 1 | 0
16 d iii | 0 | 0 | 1 | 0 | 1 | 1
16 e i | 1 | 0 | 0 | 1 | 2 | 0
16 e ii | 0 | 0 | 0 | 2 | 2 | 1
16 f | 1 | 1 | 1 | 0 | 3 | 0
Totals | 51 | 27 | 11 | 11 | 100 | LDS
4
Evaluate $\int_0^1 \frac{1}{1 + \sqrt{x}} \, dx$, giving your answer in the form $a + b \ln c$, where $a$, $b$ and $c$ are integers. [6]

\hfill \mbox{\textit{OCR MEI Paper 2  Q13 [6]}}