CAIE FP1 2015 June — Question 7 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypePolynomial times trigonometric
DifficultyChallenging +1.2 This is a standard reduction formula question requiring integration by parts twice to establish the recurrence relation, then applying it recursively with given limits. While it involves multiple steps and careful algebraic manipulation, the technique is routine for Further Maths students and follows a well-practiced pattern with no novel insight required.
Spec1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively
7 Let \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x\), where \(n\) is a non-negative integer. Show that $$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2$$ Find the exact value of \(I _ { 4 }\).
Question 7:
AnswerMarks Guidance
WorkingMarks Notes
\(I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} nx^{n-1}\cos x \, dx\)M1A1 (LNR)
\(= 0 + \left[nx^{n-1}\sin x\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} n(n-1)x^{n-2}\sin x \, dx\)M1A1 (LR)
\(= n\left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}\)A1 (5) (AG)
\(I_0 = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx = \left[-\cos x\right]_0^{\frac{\pi}{2}} = 1\)B1
\(I_2 = \pi - 2\), using reduction formulaM1A1
\(I_4 = 4 \times \left(\frac{\pi}{2}\right)^3 - 12(\pi - 2) = \frac{1}{2}\pi^3 - 12\pi + 24\)A1 (4)
Total: 9If \(I_2\) found by integration M1A1 (if correct) then M1A1 for \(I_4\) from reduction formula 
## Question 7:

| Working | Marks | Notes |
|---------|-------|-------|
| $I_n = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} nx^{n-1}\cos x \, dx$ | M1A1 | (LNR) |
| $= 0 + \left[nx^{n-1}\sin x\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} n(n-1)x^{n-2}\sin x \, dx$ | M1A1 | (LR) |
| $= n\left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2}$ | A1 | (5) (AG) |
| $I_0 = \int_0^{\frac{\pi}{2}} x^n \sin x \, dx = \left[-\cos x\right]_0^{\frac{\pi}{2}} = 1$ | B1 | |
| $I_2 = \pi - 2$, using reduction formula | M1A1 | |
| $I_4 = 4 \times \left(\frac{\pi}{2}\right)^3 - 12(\pi - 2) = \frac{1}{2}\pi^3 - 12\pi + 24$ | A1 | (4) |
| | **Total: 9** | If $I_2$ found by integration M1A1 (if correct) then M1A1 for $I_4$ from reduction formula |

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7 Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x$, where $n$ is a non-negative integer. Show that

$$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2$$

Find the exact value of $I _ { 4 }$.

\hfill \mbox{\textit{CAIE FP1 2015 Q7 [9]}}
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