CAIE P3 2012 June — Question 7 9 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeIterative or numerical methods after integration
DifficultyStandard +0.8 This question combines integration by substitution (standard P3 technique) with iterative numerical methods. Part (i) requires careful handling of the substitution and algebraic manipulation to reach the given result. Part (ii) involves multiple iterations with inverse sine, requiring precision and understanding of convergence. The integration itself is routine, but the combination with numerical methods and the need for accurate iteration elevates this above average difficulty.
Spec1.08h Integration by substitution1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7 \includegraphics[max width=\textwidth, alt={}, center]{e2cc23d2-f3ac-488b-97e1-79e2a98a87ba-3_421_885_251_628} The diagram shows part of the curve \(y = \cos ( \sqrt { } x )\) for \(x \geqslant 0\), where \(x\) is in radians. The shaded region between the curve, the axes and the line \(x = p ^ { 2 }\), where \(p > 0\), is denoted by \(R\). The area of \(R\) is equal to 1 .
  1. Use the substitution \(x = u ^ { 2 }\) to find \(\int _ { 0 } ^ { p ^ { 2 } } \cos ( \sqrt { } x ) \mathrm { d } x\). Hence show that \(\sin p = \frac { 3 - 2 \cos p } { 2 p }\).
  2. Use the iterative formula \(p _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 3 - 2 \cos p _ { n } } { 2 p _ { n } } \right)\), with initial value \(p _ { 1 } = 1\), to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

AnswerMarks Guidance
(i) Substitute for \(x\) and \(dx\) throughout the integralM1
Obtain \(\int 2u\cos u\,du\)A1
Integrate by parts and obtain answer of the form \(au\sin u + b\cos u\), where \(ab \neq 0\)M1
Obtain \(2u\sin u + 2\cos u\)A1
Use limits \(u = 0, u = p\) correctly and equate result to 1M1
Obtain the given answerA1 [6]
(ii) Use the iterative formula correctly at least onceM1
Obtain final answer \(p = 1.25\)A1
Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval \((1.245, 1.255)\)A1 [3]
**(i)** Substitute for $x$ and $dx$ throughout the integral | M1 |
Obtain $\int 2u\cos u\,du$ | A1 |
Integrate by parts and obtain answer of the form $au\sin u + b\cos u$, where $ab \neq 0$ | M1 |
Obtain $2u\sin u + 2\cos u$ | A1 |
Use limits $u = 0, u = p$ correctly and equate result to 1 | M1 |
Obtain the given answer | A1 | [6]

**(ii)** Use the iterative formula correctly at least once | M1 |
Obtain final answer $p = 1.25$ | A1 |
Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval $(1.245, 1.255)$ | A1 | [3]

---
7\\
\includegraphics[max width=\textwidth, alt={}, center]{e2cc23d2-f3ac-488b-97e1-79e2a98a87ba-3_421_885_251_628}

The diagram shows part of the curve $y = \cos ( \sqrt { } x )$ for $x \geqslant 0$, where $x$ is in radians. The shaded region between the curve, the axes and the line $x = p ^ { 2 }$, where $p > 0$, is denoted by $R$. The area of $R$ is equal to 1 .\\
(i) Use the substitution $x = u ^ { 2 }$ to find $\int _ { 0 } ^ { p ^ { 2 } } \cos ( \sqrt { } x ) \mathrm { d } x$. Hence show that $\sin p = \frac { 3 - 2 \cos p } { 2 p }$.\\
(ii) Use the iterative formula $p _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 3 - 2 \cos p _ { n } } { 2 p _ { n } } \right)$, with initial value $p _ { 1 } = 1$, to find the value of $p$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

\hfill \mbox{\textit{CAIE P3 2012 Q7 [9]}}