CAIE P2 2022 June — Question 7 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas Between Curves
TypeArea with Trigonometric Curves
DifficultyStandard +0.8 This question requires finding the area between two curves where one involves a square root and the other sin²x. Students must set up the correct integral, apply the identity sin²x = (1-cos2x)/2, integrate √(2π-2x) using substitution, and combine results. The multi-step integration with both algebraic and trigonometric components, plus careful handling of limits and exact form, makes this moderately challenging but still within standard P2 scope.
Spec1.02b Surds: manipulation and rationalising denominators1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08f Area between two curves: using integration
7 \includegraphics[max width=\textwidth, alt={}, center]{1b9c6b41-69dd-4132-92c7-9507cbd741dd-10_551_657_274_735} The diagram shows the curves \(y = \sqrt { 2 \pi - 2 x }\) and \(y = \sin ^ { 2 } x\) for \(0 \leqslant x \leqslant \pi\). The shaded region is bounded by the two curves and the line \(x = 0\). Find the exact area of the shaded region.
Question 7:
AnswerMarks Guidance
AnswerMark Guidance
Integrate \(\sqrt{2\pi - 2x}\) to obtain form \(k_1(2\pi - 2x)^{\frac{3}{2}}\)M1 any constant \(k_1\)
Obtain correct \(-\frac{1}{3}(2\pi - 2x)^{\frac{3}{2}}\)A1 OE
Apply limits to obtain \(\frac{1}{3}(2\pi)^{\frac{3}{2}}\) or exact equivalentA1 at this stage or later
Use identity \(-\sin^2 x = -\frac{1}{2} + \frac{1}{2}\cos 2x\)B1 OE
Integrate to obtain form \(k_2 x + k_3 \sin 2x\)M1 any constants \(k_2\), \(k_3\)
Obtain correct \(-\frac{1}{2}x + \frac{1}{4}\sin 2x\)A1 OE if area taken as positive
Substituting correct limits into *their* \(\frac{1}{2}x - \frac{1}{4}\sin 2x\)M1 Can be SOI by \(\pm\frac{1}{2}\pi\); Allow one slip
Obtain \(\frac{1}{3}(2\pi)^{\frac{3}{2}} - \frac{1}{2}\pi\) or exact equivalentA1 CWO 
## Question 7:

| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate $\sqrt{2\pi - 2x}$ to obtain form $k_1(2\pi - 2x)^{\frac{3}{2}}$ | M1 | any constant $k_1$ |
| Obtain correct $-\frac{1}{3}(2\pi - 2x)^{\frac{3}{2}}$ | A1 | OE |
| Apply limits to obtain $\frac{1}{3}(2\pi)^{\frac{3}{2}}$ or exact equivalent | A1 | at this stage or later |
| Use identity $-\sin^2 x = -\frac{1}{2} + \frac{1}{2}\cos 2x$ | B1 | OE |
| Integrate to obtain form $k_2 x + k_3 \sin 2x$ | M1 | any constants $k_2$, $k_3$ |
| Obtain correct $-\frac{1}{2}x + \frac{1}{4}\sin 2x$ | A1 | OE if area taken as positive |
| Substituting correct limits into *their* $\frac{1}{2}x - \frac{1}{4}\sin 2x$ | M1 | Can be SOI by $\pm\frac{1}{2}\pi$; Allow one slip |
| Obtain $\frac{1}{3}(2\pi)^{\frac{3}{2}} - \frac{1}{2}\pi$ or exact equivalent | A1 | CWO |

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7\\
\includegraphics[max width=\textwidth, alt={}, center]{1b9c6b41-69dd-4132-92c7-9507cbd741dd-10_551_657_274_735}

The diagram shows the curves $y = \sqrt { 2 \pi - 2 x }$ and $y = \sin ^ { 2 } x$ for $0 \leqslant x \leqslant \pi$. The shaded region is bounded by the two curves and the line $x = 0$.

Find the exact area of the shaded region.\\

\hfill \mbox{\textit{CAIE P2 2022 Q7 [8]}}
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