CAIE FP1 2008 November — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionNovember
Marks6
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TopicIntegration with Partial Fractions
TypeBasic partial fractions then integrate
DifficultyStandard +0.8 This question requires understanding the subtle conceptual distinction between 'mean value with respect to x' versus 'mean value with respect to y', then correctly setting up and evaluating integrals involving exponentials. The second part requires changing variables (x = ln y, dx = dy/y) and manipulating the resulting integral, which goes beyond routine integration practice and tests deeper understanding of the mean value concept.
Spec4.08e Mean value of function: using integral
2 Let \(y = \mathrm { e } ^ { x }\). Find the mean value of \(y\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant 2\). Show that the mean value of \(x\) with respect to \(y\) over the interval \(1 \leqslant y \leqslant \mathrm { e } ^ { 2 }\) is \(\frac { \mathrm { e } ^ { 2 } + 1 } { \mathrm { e } ^ { 2 } - 1 }\).
AnswerMarks Guidance
MV (\(y\) wrt \(x\)) over \([0,2] = (1/2) \int_0^e e^x dx = (1/2)[e^x]_0^e = (e^2 - 1)/2 = 3.19\)M1A1
\(\int_1^{e^2} \frac{\ln y \, dy}{e^2 - 1}\)M1
\(= \frac{[y\ln y - y]_1^{e^2}}{e^2 - 1}\)M1A1 M1A1 for integration of \(\ln y\) can be earned independently
\(= \left[\frac{2e^2 - e^2}{e^2 - 1}\right] - \left[\frac{-1}{e^2-1}\right]\) (oew)
\(= \frac{e^2 + 1}{e^2 - 1}\)A1 (AG) 
MV ($y$ wrt $x$) over $[0,2] = (1/2) \int_0^e e^x dx = (1/2)[e^x]_0^e = (e^2 - 1)/2 = 3.19$ | M1A1 |

$\int_1^{e^2} \frac{\ln y \, dy}{e^2 - 1}$ | M1 |

$= \frac{[y\ln y - y]_1^{e^2}}{e^2 - 1}$ | M1A1 | M1A1 for integration of $\ln y$ can be earned independently |

$= \left[\frac{2e^2 - e^2}{e^2 - 1}\right] - \left[\frac{-1}{e^2-1}\right]$ (oew) | |

$= \frac{e^2 + 1}{e^2 - 1}$ | A1 | (AG) |
2 Let $y = \mathrm { e } ^ { x }$. Find the mean value of $y$ with respect to $x$ over the interval $0 \leqslant x \leqslant 2$.

Show that the mean value of $x$ with respect to $y$ over the interval $1 \leqslant y \leqslant \mathrm { e } ^ { 2 }$ is $\frac { \mathrm { e } ^ { 2 } + 1 } { \mathrm { e } ^ { 2 } - 1 }$.

\hfill \mbox{\textit{CAIE FP1 2008 Q2 [6]}}
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