1. integral and taylor.

part (iv) h is small

k part (i) find integral of (1)/(4x^2+9)^0.5 dx between limits 0 and 2 DONE
(ii) given f(x)=arsinh(4/3+x) find f'(x) f''(x) bladeblah..
(iii) find maclaurin series for arsinh(4/3+x) as far as x^2
(iv) hence show that, when h is small integral of xarsinh(4/3+x)dx
roughly = (2/3)(h)^3

i got part (iii) as; ln3 + (4+3x)/5 -(54x^2 +144x +96)/125 +....
presuming this is wright in (iv) i get some utter rubbish -ve no. i screwed it up in rage! grr ???

2. A taylor series centred about a is given by.

$\displaystyle f(a+x) = f(a) + x f'(a) + \frac{f''(a)}{2}x^2$

so we have $\displaystyle arcsinh \left( \frac{4}{3} + x \right) = arcsinh \left( \frac{4}{3} \right) + \frac{1}{\sqrt{\frac{4}{3}^2+1}} x - \frac{\frac{4}{3}}{\left( \frac{4}{3}^2+1 \right)^\frac{3}{2}} \frac{x^2}{2}$

so the simplified series is

$\displaystyle arcsinh \left( \frac{4}{3} + x \right) = \ln3 + \frac{3}{5} x - \frac{18}{125} x^2$

Your wording for part vi) makes little sense.

do you mean $\displaystyle \int_{0}^{h}{x\mbox{arcsinh}\left( \frac{4}{3}+x \right)dx} \approx \frac{2}{3}h^{3}$ ? for small h

Bobak

3. err yeh

yeh i did thnx.... and the last bit??......is the taylor series the same as maclaurin??? not supposed to use taylor yet!?

maclaurin is; f(x) = f'(0) + f'(0)x + (f''(0)x^2)/2! etc sorry about this,,,,,,,i think i must've gone wrong in thw maclaurin/taylir series when i did
(f'(o)(4/3+x)^2)/2! someone on another post for that q told me the latter was right ,,am i wrong,,well fro ur anwer i mustve been,,,erm k let me give it a go

4. got it cheers buddy!!!!

5. Originally Posted by i_zz_y_ill
you got $\displaystyle \int_{0}^{h}{x\mbox{arcsinh}\left( \frac{4}{3}+x \right)dx} \approx \frac{2}{3}h^{3}$ ?

I am fairly sure that result is incorrect. could you post your working.