A continuous random variable X has the probability density function f given by

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

(a)     Determine c.

(b)     Find ${\text{E}}(X)$.

(a)     the total area under the graph of the pdf is unity     (A1)

area $= c\int_0^1 {x - {x^2}{\text{d}}x}$

$= c\left[ {\frac{1}{2}{x^2} - \frac{1}{3}{x^3}} \right]_0^1$     A1

$= \frac{c}{6}$

$\Rightarrow c = 6$     A1

(b)     ${\text{E}}(X) = 6\int_0^1 {{x^2} - {x^3}{\text{d}}x}$     (M1)

$= 6\left( {\frac{1}{3} - \frac{1}{4}} \right) = \frac{1}{2}$     A1

Note: Allow an answer obtained by a symmetry argument.

[5 marks]

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