Thread: 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 ???

A taylor series centred about a is given by.



so we have

so the simplified series is



Your wording for part vi) makes little sense.

do you mean ? for small h

Bobak

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

Originally Posted by i_zz_y_ill

you got ?

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