The probability density function of the continuous random variable X is given by

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k{2^{\frac{1}{x}}},}&{1 \leqslant x \leqslant 2} \\
  {0,}&{{\text{otherwise}}}
\end{array}} \right.\]

where k is a constant. Find the expected value of X .

\(k\int_1^2 {{2^{\frac{1}{x}}}{\text{d}}x = 1 \Rightarrow k = \frac{1}{{\int_1^2 {{2^{\frac{1}{x}}}{\text{d}}x} }}{\text{ }}( = 0.61556...)} \)     (M1)(A1)

\({\text{E}}(X) = k\int_1^2 {x{2^{\frac{1}{x}}}{\text{d}}x = 2.39....k{\text{ or 1.47}}} \)     M1A1

 Note: Condone missing dx in any part of the question.

[4 marks]

This question was well attempted by most candidates. However many were not alert for the necessity of using GDC to calculate the definite integrals and wasted time trying to obtain these values using standard calculus methods without success.