# Integral arising in estimation of discrete series

1. Apr 28, 2008

### CRGreathouse

I'm trying to solve
$$f(t;a,b)=\int_a^b\sqrt{t-x^3}dx$$
or find a good estimate for it. The problem is 'nice', and so various niceness assumptions apply: $0\le a\le b\le t$ -- and if other assumptions are needed, they probably hold. :D

An example of a bad estimate would be $(b-a)\sqrt{t-a^3}$ -- I'm looking for a better, or ideally a tractable closed-form version. Mathematica gives a complicated form that seems to be concerned mostly with the case that $x^3>t$ which will not be the case here.

If it helps, a will be 'close' to 0 and b will be close to t.

2. Apr 29, 2008

### CRGreathouse

Failing a closed solution, any thoughts on a good estimate? Any fans of systems other than Mathematica want to test this integral?

3. Apr 29, 2008

### CRGreathouse

Playing around with it a bit, it looks like

$$\int_0^{t^3}\sqrt{t^3-x^3}dx=\frac{\Gamma(4/3)}{\Gamma(11/6)}\sqrt{t^5\pi}$$

4. May 1, 2008

### Gib Z

Simpson's rule gives:

$$\int^b_0 \sqrt{b^3 - x^3} dx \approx \frac{b}{6} \left( \sqrt{b^3} + 4\sqrt{\frac{7b^3}{8} \right)$$. Should be reasonable? Which reminds me, I think you meant t instead of t^3 as your upper bound in your last post.