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i hope u know the formulas ?

\[a0+\sum_{1}^{\infty}_{n}cosnx+_{n}sinnx\]

wow...
anyways, could u find a0 =... ?

an and bn exist before cosx and sinx

yea with oyler formula!

\(\large a_0=(1/\pi)\int \limits_{-\pi}^{\pi}(3x^2-2x)dx=....?\)

euler formula! bn and an are determined with euler formula too.

∫−ππ(3x2)d-∫−ππ(2x)d=...? yea?

i don't know how we use euler here, i would integrate by parts for an and bn

u go on,and tell what u finally get for a0=....?

yes, its 1/2 pi
i thought he wrote a0/2 in the formula, which i generally do.

for find the a0 , u should separate ∫(3x2−2x)d to two ∫(3x2)dx-∫2x)dx.

yeah, that was correct...what u got a0 finally ?

a0 = 2(pi)^2

\(\large a_0=(1/2\pi)\int \limits_{-\pi}^{\pi}(3x^2-2x)dx=(1/2\pi)[x^3-x^2]^{\pi}_{-\pi}\)

3x^2 is an even function. and other one is Individual function.

we should use this state i think!

to find an and bn, you can use integration by parts.
try it...if u get stuck, i'll help.

Some good questions being asked today :)

why don't u try by yourself ?

we'll correct u if u get it wrong..

i tried on paper, it is so long to type here. :D

final step u got ?

what u got a0=...?an=...?bn=...?

an for first integration is this: an=\[\int\limits_{0}^{\pi}3x^{2}cosnxdx \rightarrow......?\]

did u miss 1/pi in the beginning ?

2/pi i missed in beginning! sry

so u having trouble with integration by parts ?

Fractional integrals is needed to solve this?

fractional integers ? no...

so how can u find the that integration?!

didn't u try integration by parts ? u know what that is, right ?
\(\int uv = ... ?\)

here u=x^2 , v = cos nx

uv-∫vdu u mean this?

yes.
u = x^2
what will u take dv =... ?

actually dv=cosnx and v=-sinnx

yes, yes, go ahead and solve it...

you'll again need product rule when u solve for sin nx (2x)dx

yea thnx... i'll try and finally put the answer here to prove;)

ok.