CAIE FP1 2011 June — Question 9 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyStandard +0.8 This is a two-part Further Maths question requiring (1) centroid calculation using integration formulas for area and moments, and (2) arc length integration with algebraic manipulation to reach an exact answer. Both parts require careful setup and computation beyond standard A-level, but follow established techniques without requiring novel insight.
Spec4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric
9 The curve \(C\) has equation \(y = x ^ { \frac { 3 } { 2 } }\). Find the coordinates of the centroid of the region bounded by \(C\), the lines \(x = 1 , x = 4\) and the \(x\)-axis. Show that the length of the arc of \(C\) from the point where \(x = 5\) to the point where \(x = 28\) is 139 .
Question 9:
Centroid coordinates:
AnswerMarks Guidance
\(\bar{x} = \frac{\int_1^4 x^{\frac{5}{2}}dx}{\int_1^4 x^{\frac{3}{2}}dx}\), \(\bar{y} = \frac{\frac{1}{2}\int_1^4 x^3 dx}{\int_1^4 x^{\frac{3}{2}}dx}\)M1M1
Numerators: \(\left[\frac{2}{7}x^{\frac{7}{2}}\right]_1^4 = \frac{254}{7}\), \(\left[\frac{1}{8}x^4\right]_1^4 = \frac{255}{8}\)A1A1
Denominator: \(\left[\frac{2}{5}x^{\frac{5}{2}}\right]_1^4 = \frac{62}{5}\)B1
\(\bar{x} = \frac{635}{217}\ (= 2.93)\), \(\bar{y} = \frac{1275}{496}\ (= 2.57)\)M1A1 Accept rational values; Part mark: 7
Arc length:
AnswerMarks Guidance
\(y = x^{\frac{3}{2}} \Rightarrow y' = \frac{3}{2}\sqrt{x} \Rightarrow (y')^2 = \frac{9x}{4}\)B1
\(s = \int_5^{28}\sqrt{1 + \frac{9x}{4}}\ dx\)B1
\(= \left[\frac{8}{27}\left(1 + \frac{9x}{4}\right)^{\frac{3}{2}}\right]_5^{28}\) or \(\left[\frac{1}{27}(4+9x)^{\frac{3}{2}}\right]_5^{28}\)M1A1
\(= \frac{8^4 - 7^3}{27}\) or \(\frac{16^3 - 7^3}{27} = 139\) (AG)A1 Part mark: 5
Total: [12]
## Question 9:

**Centroid coordinates:**

$\bar{x} = \frac{\int_1^4 x^{\frac{5}{2}}dx}{\int_1^4 x^{\frac{3}{2}}dx}$, $\bar{y} = \frac{\frac{1}{2}\int_1^4 x^3 dx}{\int_1^4 x^{\frac{3}{2}}dx}$ | M1M1 |

Numerators: $\left[\frac{2}{7}x^{\frac{7}{2}}\right]_1^4 = \frac{254}{7}$, $\left[\frac{1}{8}x^4\right]_1^4 = \frac{255}{8}$ | A1A1 |

Denominator: $\left[\frac{2}{5}x^{\frac{5}{2}}\right]_1^4 = \frac{62}{5}$ | B1 |

$\bar{x} = \frac{635}{217}\ (= 2.93)$, $\bar{y} = \frac{1275}{496}\ (= 2.57)$ | M1A1 | Accept rational values; Part mark: 7

**Arc length:**

$y = x^{\frac{3}{2}} \Rightarrow y' = \frac{3}{2}\sqrt{x} \Rightarrow (y')^2 = \frac{9x}{4}$ | B1 |

$s = \int_5^{28}\sqrt{1 + \frac{9x}{4}}\ dx$ | B1 |

$= \left[\frac{8}{27}\left(1 + \frac{9x}{4}\right)^{\frac{3}{2}}\right]_5^{28}$ or $\left[\frac{1}{27}(4+9x)^{\frac{3}{2}}\right]_5^{28}$ | M1A1 |

$= \frac{8^4 - 7^3}{27}$ or $\frac{16^3 - 7^3}{27} = 139$ (AG) | A1 | Part mark: 5

**Total: [12]**

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9 The curve $C$ has equation $y = x ^ { \frac { 3 } { 2 } }$. Find the coordinates of the centroid of the region bounded by $C$, the lines $x = 1 , x = 4$ and the $x$-axis.

Show that the length of the arc of $C$ from the point where $x = 5$ to the point where $x = 28$ is 139 .

\hfill \mbox{\textit{CAIE FP1 2011 Q9 [12]}}
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