(adsbygoogle = window.adsbygoogle || []).push({}); Easier way to get exact sum/avr? [SOLVED, thanks awvvu]

Greetings,

First I will explain what I am trying to do.

I am trying to find the 'average' distance from the bottom and right edges of this box:

http://img174.imageshack.us/img174/8934/okvi7.jpg [Broken]

Basically (using 11 approximations)

What I am after for is equal to

1 + sqrt (1 + 0.1^2) + sqrt (1 + 0.2^2) + sqrt (1 + 0.3^2)

+ sqrt (1 + 0.4^2) + sqrt (1 + 0.5^2) + sqrt (1 + 0.6^2)

+ sqrt (1 + 0.8^2) + sqrt (1 + 0.8^2) + sqrt (1 + 0.9^2) + sqrt(2)

And then that divided by 11

I've used a nice messy excel spreadsheet to get this to 1000 approximations (1000 little 'slices') to get an average of 1.148001 and also using 5000 slices I get 1.147835

Basically as a summation what I _think_ I am looking for is (1000 slices):

[tex]\frac{1}{n} \sum_{n=1}^{1000} \sqrt{1^2 + (0.001n)^2} [/tex]

(that right?)

And extended to an infinte amount of slices:

[tex]\lim_{k\rightarrow\infty} \frac{1}{k} \sum_{n=1}^{k} \sqrt{1^2 + (\frac{n}{k})^2} [/tex]

(is this right/possible?)

What I am looking for is, using integration, or if it is do-able to evaluate that sum, to know if it is possible to get an 'exact' answer for the 'average' distance?

I imagine it would be very similar to the 1.1478 answer above, but I'm looking for more accuracy (basically to whatever precision the infinite sum gives) or if it just happens to equal a nice fraction for me (8/7 which is 1.14285...) or you know.. something nice and round

Thanks for reading, please let me know if you need any more info, or if I have gone wrong somewhere, or any hints to get me on the right track, etc.

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# Easier way to get exact sum/avr?

**Physics Forums | Science Articles, Homework Help, Discussion**