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**Glitch** **The question:**

Suppose that $\displaystyle f(x) = \frac{1}{\sqrt{1 - x^2}}$

i) Show that f is increasing on the interval [0, 1/2].

ii) Find the upper Riemann sum for f with respect to the partition

{ $\displaystyle \frac{0}{2n}, \frac{1}{2n}, \frac{2}{2n}, \frac{3}{2n}, ..., \frac{n}{2n}$ } of [0, 1/2]

iii) Hence evaluate limit n -> infinity ($\displaystyle \frac{1}{\sqrt{4n^2 - 1^2}} + \frac{1}{\sqrt{4n^2 - 2^2}} + \frac{1}{\sqrt{4n^2 - 3^2}} + ... + \frac{1}{\sqrt{4n^2 - n^2}}$)

**My attempt:**

i) I took the derivative of f, and found the stationary points by equating it to 0. I noticed that the only stationary point was at 0 itself, thus the graph may only change gradient at this point. I substituted for 1/2 and found that the gradient was positive, so the interval [0, 1/2] is increasing.

ii) I worked out the width of each partition is $\displaystyle \frac{1}{2n}$, and the height of a given partition 'k' is $\displaystyle f(\frac{k}{2n})$. My sum then looked something like this:

$\displaystyle \sum^n_{k=0}{\frac{1}{\sqrt{1 - \frac{k^2}{4n^2}}}$