OCR H240/01 2020 November — Question 10 11 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2020
SessionNovember
Marks11
PaperDownload PDF ↗
TopicIntegration by Substitution
TypeShow substitution transforms integral, then apply integration by parts or further substitution
DifficultyStandard +0.8 This question requires numerical approximation with rectangles, then a non-trivial substitution followed by integration by parts (the 2t sin(t/2) integral). While substitution itself is standard A-level, the combination of techniques and the need to handle the composite function through multiple transformations makes this moderately challenging, though still within typical Further Maths scope.
Spec1.08h Integration by substitution1.08i Integration by parts1.09f Trapezium rule: numerical integration

10 \includegraphics[max width=\textwidth, alt={}, center]{febe231d-200a-4957-b41b-de5b9be98b0a-7_352_545_258_239} The diagram shows the curve \(y = \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right)\), for \(1 \leqslant x \leqslant 2\).
  1. Use rectangles of width 0.25 to find upper and lower bounds for \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x\). Give your answers correct to 3 significant figures.
    1. Use the substitution \(t = \sqrt { x - 1 }\) to show that \(\int \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = \int 2 t \sin \left( \frac { 1 } { 2 } t \right) \mathrm { d } t\).
    2. Hence show that \(\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = 8 \sin \frac { 1 } { 2 } - 4 \cos \frac { 1 } { 2 }\).

Question 10 - Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
\(0.25\{\sin 0 + \sin(\frac{1}{2}\sqrt{0.25}) + \sin(\frac{1}{2}\sqrt{0.5}) + \sin(\frac{1}{2}\sqrt{0.75})\}\)M1 Attempt four rectangles of width 0.25, with height on left-hand side; no need to see \(\sin 0\); allow M1 if evaluated in degrees (0.00452)
Lower bound \(= 0.253\)A1 Obtain 0.253 or better; soi as lower bound
\(0.25(\sin(\frac{1}{2}\sqrt{0.25}) + \sin(\frac{1}{2}\sqrt{0.5}) + \sin(\frac{1}{2}\sqrt{0.75}) + \sin\frac{1}{2})\)M1 Attempt rectangles of width 0.25 with height on right-hand side; or subtract \(\sin 0\) from part (ii) and add \(\sin 0.5\); M0 if \(\sin 0\) explicitly included
Upper bound \(= 0.373\) or \(0.374\)A1 Obtain 0.373 or 0.374 (from rounding upper bound up), or better; soi as upper bound
[4]
Question 10 - Part (b)(i)
AnswerMarks Guidance
AnswerMarks Guidance
\(t^2 = x - 1\)M1 Attempt to link \(dt\) and \(dx\)
\(2t\,dt = dx\)A1 Obtain correct equation linking \(dt\) and \(dx\); allow \(dt = \frac{1}{2}(x-1)^{-\frac{1}{2}}dx\)
\(\int \sin(\frac{1}{2}\sqrt{x-1})\,dx = \int \sin(\frac{1}{2}t)\,2t\,dt = \int 2t\sin(\frac{1}{2}t)\,dt\) A.G.A1 Attempt integrand in terms of \(t\) to obtain given answer; award A1 once all elements correct, even if not in same order; BOD if no brackets
[3]
Question 10 - Part (b)(ii)
AnswerMarks Guidance
AnswerMarks Guidance
\(-4t\cos(\frac{1}{2}t) + \int 4\cos(\frac{1}{2}t)\,dt\)M1* Attempt integration by parts; correct parts; as far as first stage
\(-4t\cos(\frac{1}{2}t) + 8\sin(\frac{1}{2}t)\)A1 Correct integral
\((-4\cos\frac{1}{2} + 8\sin\frac{1}{2}) - (-0 + 0)\)M1d* Attempt use of limits; using either \(t\) or \(x\), but must be consistent; condone no clear use of lower limit for M1
\(8\sin\frac{1}{2} - 4\cos\frac{1}{2}\) AGA1 Obtain given answer; must see some indication that lower limit considered
[4]
## Question 10 - Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.25\{\sin 0 + \sin(\frac{1}{2}\sqrt{0.25}) + \sin(\frac{1}{2}\sqrt{0.5}) + \sin(\frac{1}{2}\sqrt{0.75})\}$ | M1 | Attempt four rectangles of width 0.25, with height on left-hand side; no need to see $\sin 0$; allow M1 if evaluated in degrees (0.00452) |
| Lower bound $= 0.253$ | A1 | Obtain 0.253 or better; soi as lower bound |
| $0.25(\sin(\frac{1}{2}\sqrt{0.25}) + \sin(\frac{1}{2}\sqrt{0.5}) + \sin(\frac{1}{2}\sqrt{0.75}) + \sin\frac{1}{2})$ | M1 | Attempt rectangles of width 0.25 with height on right-hand side; or subtract $\sin 0$ from part (ii) and add $\sin 0.5$; M0 if $\sin 0$ explicitly included |
| Upper bound $= 0.373$ or $0.374$ | A1 | Obtain 0.373 or 0.374 (from rounding upper bound up), or better; soi as upper bound |
| | **[4]** | |

## Question 10 - Part (b)(i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $t^2 = x - 1$ | M1 | Attempt to link $dt$ and $dx$ |
| $2t\,dt = dx$ | A1 | Obtain correct equation linking $dt$ and $dx$; allow $dt = \frac{1}{2}(x-1)^{-\frac{1}{2}}dx$ |
| $\int \sin(\frac{1}{2}\sqrt{x-1})\,dx = \int \sin(\frac{1}{2}t)\,2t\,dt = \int 2t\sin(\frac{1}{2}t)\,dt$ **A.G.** | A1 | Attempt integrand in terms of $t$ to obtain given answer; award A1 once all elements correct, even if not in same order; BOD if no brackets |
| | **[3]** | |

## Question 10 - Part (b)(ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-4t\cos(\frac{1}{2}t) + \int 4\cos(\frac{1}{2}t)\,dt$ | M1* | Attempt integration by parts; correct parts; as far as first stage |
| $-4t\cos(\frac{1}{2}t) + 8\sin(\frac{1}{2}t)$ | A1 | Correct integral |
| $(-4\cos\frac{1}{2} + 8\sin\frac{1}{2}) - (-0 + 0)$ | M1d* | Attempt use of limits; using either $t$ or $x$, but must be consistent; condone no clear use of lower limit for M1 |
| $8\sin\frac{1}{2} - 4\cos\frac{1}{2}$ **AG** | A1 | Obtain given answer; must see some indication that lower limit considered |
| | **[4]** | |

---
10\\
\includegraphics[max width=\textwidth, alt={}, center]{febe231d-200a-4957-b41b-de5b9be98b0a-7_352_545_258_239}

The diagram shows the curve $y = \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right)$, for $1 \leqslant x \leqslant 2$.
\begin{enumerate}[label=(\alph*)]
\item Use rectangles of width 0.25 to find upper and lower bounds for $\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x$. Give your answers correct to 3 significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Use the substitution $t = \sqrt { x - 1 }$ to show that $\int \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = \int 2 t \sin \left( \frac { 1 } { 2 } t \right) \mathrm { d } t$.
\item Hence show that $\int _ { 1 } ^ { 2 } \sin \left( \frac { 1 } { 2 } \sqrt { x - 1 } \right) \mathrm { d } x = 8 \sin \frac { 1 } { 2 } - 4 \cos \frac { 1 } { 2 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR H240/01 2020 Q10 [11]}}