CAIE P3 2015 June — Question 6 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow integral transforms via substitution then evaluate (algebraic/exponential)
DifficultyStandard +0.3 This is a standard integration by substitution question with clear guidance. Part (i) requires routine manipulation of the given substitution to transform the integral, while part (ii) involves expanding and integrating a polynomial divided by u, then evaluating limits. The substitution is provided, and both parts are 'show that' questions with target answers given, making it slightly easier than average but requiring careful algebraic manipulation.
Spec1.08h Integration by substitution

6 Let \(I = \int _ { 0 } ^ { 1 } \frac { \sqrt { } x } { 2 - \sqrt { } x } \mathrm {~d} x\).
  1. Using the substitution \(u = 2 - \sqrt { } x\), show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 ( 2 - u ) ^ { 2 } } { u } \mathrm {~d} u\).
  2. Hence show that \(I = 8 \ln 2 - 5\).

AnswerMarks Guidance
(i) State or imply \(du = -\frac{1}{2\sqrt{x}}dx\), or equivalentB1
Substitute for \(x\) and \(dx\) throughoutM1
Obtain integrand \(\frac{\pm 2(2-u)^2}{u}\), or equivalentA1
Show correct working to justify the change in limits and obtain the given answer with no errors seenA1 [4]
(ii) Integrate and obtain at least two terms of the form \(a\ln u\), \(bu\), and \(cu^2\)M1✱
Obtain indefinite integral \(8\ln u - 8u + u^2\), or equivalentA1
Substitute limits correctlyM1(dep✱)
Obtain the given answer correctly having shown sufficient workingA1 [4]
**(i)** State or imply $du = -\frac{1}{2\sqrt{x}}dx$, or equivalent | B1 |
Substitute for $x$ and $dx$ throughout | M1 |
Obtain integrand $\frac{\pm 2(2-u)^2}{u}$, or equivalent | A1 |
Show correct working to justify the change in limits and obtain the given answer with no errors seen | A1 | [4]

**(ii)** Integrate and obtain at least two terms of the form $a\ln u$, $bu$, and $cu^2$ | M1✱ |
Obtain indefinite integral $8\ln u - 8u + u^2$, or equivalent | A1 |
Substitute limits correctly | M1(dep✱) |
Obtain the given answer correctly having shown sufficient working | A1 | [4]
6 Let $I = \int _ { 0 } ^ { 1 } \frac { \sqrt { } x } { 2 - \sqrt { } x } \mathrm {~d} x$.\\
(i) Using the substitution $u = 2 - \sqrt { } x$, show that $I = \int _ { 1 } ^ { 2 } \frac { 2 ( 2 - u ) ^ { 2 } } { u } \mathrm {~d} u$.\\
(ii) Hence show that $I = 8 \ln 2 - 5$.

\hfill \mbox{\textit{CAIE P3 2015 Q6 [8]}}