Hiya

I've got ${\int }_0^2\sqrt{4-x^2}\mathrm{dx}=\pi$ but how do you expand $\sqrt{4-x^2}$?

Well, (4-x² )^1/2 = 2 (1-x² /4)^1/2

=2 ( 1 + (1/2) (-x² /4) + (1/2)(-1/2)/2 (-x² /4)² + (1/2)(-1/2)(-3/2)/3! (-x² /4)³ ... )

=2( 1 - x² /8 - 1/128 x⁴ - 1/1024 x⁶ - 5/32768 x⁸ ...)

and so on. Now you can integrate each term of this as a polynomial, carefully forgetting to mention to your teacher that what you are doing is highly dodgy mathematically (it does work, but you need to prove carefully in higher maths why it works). This will give you a reasonable approximation to pi.

David

How would you prove it?

Well, if you are going to do this properly, there are two things you need to prove, and both of them are university-level maths.

First, you need to prove that the series expansion of (4-x² )^1/2 is valid. This alone is well beyond A-level standard (your teachers will probably pretend they can prove it, but they won't do it properly). It is more or less identical to question 11 from the 2000 Mathematics 1A exam paper at Cambridge (the exam I will be doing in 2 weeks time). The way to do it is to apply Taylor's theorem, and find an exact expression in terms of an integral for the difference between the first N terms of the series and the value of the function. You can then show that this remainder term tends to zero.

Secondly, you need to show that it is valid to integrate the series termwise. It seems sort of obvious that it should be, but the standard proof that you can differentiate polynomials only works for ones of finite degree, so there is some work to be done. Anyway, this is justified by something called uniform convergence, which is slightly technical but again is in the Cambridge 1st year maths course, and probably in equivalent courses at other universities.

David

Er... I'm pretty sure you will be sitting the 2002 paper

unfortunately, since 2000 seems to have been an easy year (compared to 1998 for example)

And also the fact you know what's on the 2000 paper...