# Math Help - Integration using Riemann sums

1. ## Integration using Riemann sums

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

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
{ $\frac{0}{2n}, \frac{1}{2n}, \frac{2}{2n}, \frac{3}{2n}, ..., \frac{n}{2n}$ } of [0, 1/2]
iii) Hence evaluate limit n -> infinity ( $\frac{1}{\sqrt{4n^2 - 1^2}} + \frac{1}{\sqrt{4n^2 - 2^2}} + \frac{1}{\sqrt{4n^2 - 3^2}} + ... + \frac{1}{\sqrt{4n^2 - n^2}}$)

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

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

This looks different to what's in part iii). Is it incorrect? Also, how do I solve part iii)? Thanks.

2. Any ideas?

3. Originally Posted by Glitch
The question:
Suppose that $f(x) = \frac{1}{\sqrt{1 - x^2}}$

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
{ $\frac{0}{2n}, \frac{1}{2n}, \frac{2}{2n}, \frac{3}{2n}, ..., \frac{n}{2n}$ } of [0, 1/2]
iii) Hence evaluate limit n -> infinity ( $\frac{1}{\sqrt{4n^2 - 1^2}} + \frac{1}{\sqrt{4n^2 - 2^2}} + \frac{1}{\sqrt{4n^2 - 3^2}} + ... + \frac{1}{\sqrt{4n^2 - n^2}}$)

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

$\sum^n_{k=0}{\frac{1}{\sqrt{1 - \frac{k^2}{4n^2}}}$
But the integral is $\frac{1}{2n}\sum_{k=0}^n\frac{1}{\sqrt{1- \frac{k^2}{4n^2}}}= \sum_{k=0}^n\frac{1}{2n\sqrt{1-\frac{k^2}{4n^2}}}$ $= \sum_{k=0}^n\frac{1}{\sqrt{4n^2(1- \frac{k^2}{4n^2}})}$

This looks different to what's in part iii). Is it incorrect? Also, how do I solve part iii)? Thanks.

4. Oh! I see what you did! I must have forgotten to multiply by width! >_<

Now that I understand that bit, how do I calculate the limit?

5. Well, you have now that the Riemann sum is $\sum_{k=0}^n\frac{1}{\sqrt{4n^2(1- \frac{k^2}{4n^2}})}= \sum_{k=0}^n\frac{1}{\sqrt{4n^2- k^2}}$ and the limit of that, as n goes to infinity is the integral from 0 to 1/2.

But the limit as n goes to infinity is exactly the sum you want to find.

So, go ahead and integrate from 0 to 1/2: $\int_{0}^{1/2} \frac{dx}{\sqrt{1- x^2}}$.
I recommend a trignometric substitution.

6. Oh, so I don't have to take a limit, I just solve it as an integral?