Illustrate graphically the inequality, \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \).

Use the inequality in part (a) to find a lower and upper bound for \(\pi \).

Show that \(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}} = \frac{{1 + {{( - 1)}^{n - 1}}{x^{2n}}}}{{1 + {x^2}}}} \).

Hence show that \(\pi  = 4\left( {\sum\limits_{r = 0}^{n - 1} {\frac{{{{( - 1)}^r}}}{{2r + 1}} - {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\).

     A1A1A1

A1 for upper rectangles, A1 for lower rectangles, A1 for curve in between with \(0 \le x \le 1\)

hence \(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } } \)     AG

[3 marks]

attempting to integrate from \(0\) to \(1\)     (M1)

\(\int_0^1 {f(x){\text{d}}x = [\arctan x]_0^1} \)

\( = \frac{\pi }{4}\)     A1

attempt to evaluate either summation     (M1)

\(\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \frac{\pi }{4} < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)

hence \(\frac{4}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \pi  < \frac{4}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } \)

so \(2.93 < \pi  < 3.33\)     A1A1

Note:     Accept any answers that round to \(2.9\) and \(3.3\).

[5 marks]

EITHER

recognise \(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}}} \) as a geometric series with \(r =  - {x^2}\)     M1

sum of \(n\) terms is \(\frac{{1 - {{( - {x^2})}^n}}}{{1 -  - {x^2}}} = \frac{{1 + {{( - 1)}^{n - 1}}{x^{2n}}}}{{1 + {x^2}}}\)     M1AG

OR

\(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}(1 + {x^2}){x^{2r}} = (1 + {x^2}){x^0} - (1 + {x^2}){x^2} + (1 + {x^2}){x^4} +  \ldots } \)

\( + {( - 1)^{n - 1}}(1 + {x^2}){x^{2n - 2}}\)     M1

cancelling out middle terms     M1

\( = 1 + {( - 1)^{n - 1}}{x^{2n}}\)     AG

[2 marks]

\(\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}{x^{2r}} = \frac{1}{{1 + {x^2}}} + {{( - 1)}^{n - 1}}\frac{{{x^{2n}}}}{{1 + {x^2}}}} \)

integrating from \(0\) to \(1\)     M1

\(\left[ {\sum\limits_{r = 0}^{n - 1} {{{( - 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} } \right]_0^1 = \int_0^1 {f(x){\text{d}}x + {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } \)     A1A1

\(\int_0^1 {f(x){\text{d}}x = \frac{\pi }{4}} \)     A1

so \(\pi  = 4\left( {\sum\limits_{r = 0}^{n - 1} {\frac{{{{( - 1)}^r}}}{{2r + 1}} - {{( - 1)}^{n - 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)\)     AG

[4 marks]

Total [14 marks]