calculus - power series ASAP please :)
posted by COFFEE on .
using power series, integrate & evaluate to 4 dec. places
integral from 0 to 1: sin x^2 dx
i'm REALLY stuck on this. and i need help asap..
what is the inverse of "sin x^2" so that i could have it in a fraction that will fit the power series equation? and that is: (A)/(1-R)
You have to use the series expansion of the sin function, not the series expansion of the geometric series.
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
So we have:
sin(x^2) = x^2 - x^6/3! + x^10/5! - x^14/7! + ...
Then integrate term by term:
Int sin(x^2) dx =
x^3/3 - x^7/42 + x^11/1320 - x^15/75600 + ...
and to solve to four decimal places... how do i work that? plugging x=0 and x=1?
Yes, you substitute x = 1 and subtract the value you get for x = 0 (but that's zero in this case). The error is of the order of the next term in the series we ignored (you an derive more rigorous error estimates using the Lagrange error formula).
You'll have 6 decimal figures accuracy using the terms up to x^15...
thanks again :)