2021 EJC P1 Q8

The graph of \(y=\mathrm{f}\left( x \right)\), where \(\mathrm{f}\left( x \right)={{x}^{2}}+2\), \(0\le x\le 2\), is shown. Let \(R\) be the region bounded by the \(x\)-axis, the curve\(y={{x}^{2}}+2\), and the lines \(x=0\) and \(x=2\). Let \(n\) rectangles of equal width \(\frac{2}{n}\) be used to estimate the area of \(R\), as shown in the diagram above.

1. Given that \(\sum\limits_{i=1}^{n}{{{i}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}\), show that the total area of \(n\) rectangles is \(\frac{4\left( n+1 \right)\left( 2n+1 \right)}{3{{n}^{2}}}+4\).[4]
2. Hence find the least value of \(n\) such that the total area of \(n\) rectangles differs from the area of \(R\) by less than \(0.1\).[2]

Instead, \(n\) trapeziums of equal width are used to estimate the area of \(R\).

For example, when \(n=2\), the .trapeziums look like this:

3. Find the difference between the total area of the trapeziums and the area of \(R\) when \(4\) trapeziums are used.[3]
4. Using your answers to part (ii) and part (iii), comment on the use of trapeziums over rectangles to estimate the value of \(R\).[1]