State the mode of X .

Determine the value of a .

Find E(X ) .

0     A1

[1 mark]

\(\int_0^1 {f(x)dx = 1} \)     (M1)

\( \Rightarrow a = \frac{1}{{\int_0^1 {{e^{ - x}}dx} }}\)

\( \Rightarrow a = \frac{1}{{\left[ { - {e^{ - x}}} \right]_0^1}}\)

\( \Rightarrow a = \frac{e}{{e - 1}}\) (or equivalent)     A1

Note: Award first A1 for correct integration of \(\int {{e^{ - x}}dx} \) .

This A1 is independent of previous M mark.

[3 marks]

\({\text{E}}(X) = \int_0^1 {xf(x)dx\left( { = a\int_0^1 {x{e^{ - x}}dx} } \right)} \)     M1

attempt to integrate by parts     M1

\( = a\left[ { - x{e^{ - x}} - {e^{ - x}}} \right]_0^1\)     (A1)

\( = a\left( {\frac{{e - 2}}{e}} \right)\)

\( = \frac{{e - 2}}{{e - 1}}\) (or equivalent)     A1

[4 marks]

A range of answers were seen to part a), though many more could have gained the mark had they taken time to understand the shape of the function. Part b) was done well, as was part c). In c), a number of candidates integrated by parts, but found the incorrect expression \( - x{e^{ - x}} + {e^{ - x}}\).

A range of answers were seen to part a), though many more could have gained the mark had they taken time to understand the shape of the function. Part b) was done well, as was part c). In c), a number of candidates integrated by parts, but found the incorrect expression \( - x{e^{ - x}} + {e^{ - x}}\).

A range of answers were seen to part a), though many more could have gained the mark had they taken time to understand the shape of the function. Part b) was done well, as was part c). In c), a number of candidates integrated by parts, but found the incorrect expression \( - x{e^{ - x}} + {e^{ - x}}\).