CAIE P3 2019 June — Question 2 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeDouble integration by parts
DifficultyStandard +0.3 This is a straightforward double integration by parts problem with standard functions (polynomial times trigonometric). While it requires two applications of integration by parts and careful bookkeeping with the limits, it follows a completely standard algorithm with no conceptual challenges. The definite integral evaluation is routine, making this slightly easier than average.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08i Integration by parts
2 Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x ^ { 2 } \cos 2 x \mathrm {~d} x = \frac { 1 } { 32 } \left( \pi ^ { 2 } - 8 \right)\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
Commence integration and reach \(ax^2\sin 2x + b\int x\sin 2x\, dx\)M1*
Obtain \(\frac{1}{2}x^2\sin 2x - \int x\sin 2x\, dx\), or equivalentA1
Complete the integration and obtain \(\frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x\), or equivalentA1
Use limits correctly, having integrated twiceDM1
Obtain given answer correctlyA1
Total5 
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Commence integration and reach $ax^2\sin 2x + b\int x\sin 2x\, dx$ | M1* | |
| Obtain $\frac{1}{2}x^2\sin 2x - \int x\sin 2x\, dx$, or equivalent | A1 | |
| Complete the integration and obtain $\frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x$, or equivalent | A1 | |
| Use limits correctly, having integrated twice | DM1 | |
| Obtain given answer correctly | A1 | |
| **Total** | **5** | |
2 Show that $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x ^ { 2 } \cos 2 x \mathrm {~d} x = \frac { 1 } { 32 } \left( \pi ^ { 2 } - 8 \right)$.\\

\hfill \mbox{\textit{CAIE P3 2019 Q2 [5]}}
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