2014 AJC P1 Q9

The graph of \(y=\sqrt{x+40}\) is shown in the diagram below.

1. By considering the shaded rectangle, show that
   \(\sqrt{n+40}<\displaystyle\displaystyle\int\limits_{n}^{n+1}{\sqrt{x+40}\mathrm{ \; d}x}\).[1]
2. Deduce that
   \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{80}<\displaystyle\displaystyle\int\limits_{-40}^{41}{\sqrt{x+40}\mathrm{\; d}x}\).[2]
3. Show also that
   \(\sqrt{n+41}>\displaystyle\displaystyle\int\limits_{n}^{n+1}{\sqrt{x+40}\mathrm{\; d}x}\).[2]
4. Deduce that
   \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{81}>\displaystyle\displaystyle\int\limits_{-40}^{41}{\sqrt{x+40}\mathrm{\; d}x}\).[1]
5. Hence, deduce that the value \(a\), where \(a\in \mathbb{Z}\), that satisfies the following inequality,
   \(9a<\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{81}<9\left( a+1 \right)\).[2]