(a)     Using the Maclaurin series for the function ${{\text{e}}^x}$, write down the first four terms of the Maclaurin series for ${{\text{e}}^{ - \frac{{{x^2}}}{2}}}$.

(b)     Hence find the first four terms of the series for $\int_0^x {{{\text{e}}^{ - \frac{{{u^2}}}{2}}}} {\text{d}}u$.

(c)     Use the result from part (b) to find an approximate value for $\frac{1}{{\sqrt {2\pi } }}\int_0^1 {{{\text{e}}^{ - \frac{{{x^2}}}{2}}}{\text{d}}x}$.

(a)     ${{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots$

putting $x = \frac{{ - {x^2}}}{2}$     (M1)

${{\text{e}}^{ - \frac{{{x^2}}}{2}}} \approx 1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{{2^2} \times 2!}} - \frac{{{x^6}}}{{{2^3} \times 3!}} \approx \left( {1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} - \frac{{{x^6}}}{{48}}} \right)$     A2

[3 marks]

(b)     $\int_0^x {{{\text{e}}^{ - \frac{{{u^2}}}{2}}}{\text{d}}u \approx \left[ {u - \frac{{{u^3}}}{{3 \times 2}} + \frac{{{u^5}}}{{5 \times {2^2} \times 2!}} - \frac{{{u^7}}}{{7 \times {2^3} \times 3!}}} \right]_0^x}$     M1(A1)

$= x - \frac{{{x^3}}}{{3 \times 2}} + \frac{{{x^5}}}{{5 \times {2^2} \times 2!}} - \frac{{{x^7}}}{{7 \times {2^3} \times 3!}}$     A1

$\left( { = x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}}} \right)$

[3 marks]

(c)     putting x = 1 in part (b) gives $\int_0^1 {{{\text{e}}^{ - \frac{{{x^2}}}{2}}}{\text{d}}x \approx } 0.85535 \ldots$     (M1)(A1)

$\frac{1}{{\sqrt {2\pi } }}\int_0^1 {{{\text{e}}^{ - \frac{{{x^2}}}{2}}}{\text{d}}x \approx } 0.341$     A1

[3 marks]

Total [9 marks]

This was one of the most successfully answered questions. Some candidates however failed to use the data booklet for the expansion of the series, thereby wasting valuable time.