The graph of $y=\sqrt{x+40}$ is shown in the diagram below. By considering the shaded rectangle, sho

Question

The graph of $y=\sqrt{x+40}$ is shown in the diagram below. By considering the shaded rectangle, show that $\sqrt{n+40}[1] Deduce that $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{80}[2] Show also that $\sqrt{n+41}>\displaystyle\int\limits_{n}^{n+1}{\sqrt{x+40}\mathrm{ d}x}$. [2] Deduce that $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{81}>\displaystyle\int\limits_{-40}^{41}{\sqrt{x+40}\mathrm{ d}x}$. [1] Hence, deduce that the value $a$, where $a\in \mathbb{Z}$, that satisfies the following inequality, $9a[2]

Accepted Answer

$a = 54$

2014 AJC P1 Q9

2014 AJC P1 Q9

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