Originally Posted by

**dokrbb** I'm not sure I understand really how it should be done, do I really need in this case to evaluate it as

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \left[ \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] = $

$\displaystyle \left[ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{bi}{n}\right)^2} \cdot \frac{b}{n} - \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{ai}{n}\right)^2} \cdot \frac{a}{n} \right] = $ and so on, without using the actual values of $\displaystyle b$ and $\displaystyle a$ ?

Hmmm? I'm not quite sure what you're doing...

Your nth Riemann sum is:

$\displaystyle R_n = \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$.

And they're asking you to evaluate $\displaystyle \lim_{n \to \infty} R_n$.

We proceed with the definition that:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = \int_a^b f(x) \ dx$.

Thus:

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n} = \int_0^2 \sqrt{4 - x^2} \ dx$.

All that's left to do is compute the integral on the RHS and you're done.