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]