Question

Compute Integral cos (x^3) dx from 0 to 1 to within an error of .00001. You can leave your approximation as a sum.

Answer 1

I know this has to do with Maclaurin series and expansion and thats the way it should be done.

Answer 2

I have the maclaurin series for it so do i just evaluate it ?

Answer 3

yes just integrate each terms of maclarian series and put limits

Answer 4

The Maclaurin series of cos (x^3) is
\[\sum_{n=0}^{\inf}(-1)^{n} (x^3)^{2n}/(2n)!\]

Answer 5

at what point do i check to see within the .00001 ?

Answer 6

for better accuracy you should use more terms.

Answer 7

Your error will be the same as the error for the Taylor series.
If Rn+1 is the remainder of the Taylor series of n terms Tn, then int from 0 to 1 of f=int 0 1 of Tn+Rn+1=int 0 1 Tn + int 0 1 Rn+1=<int from 0 to 1 of Tn +max(Rn+1)

Answer 8

\[\int\limits_{0}^{1}\sum_{n=0}^{\inf}(-1)^{n} (x ^{6n})/(2n)! = \sum_{n=0}^{\inf}(-1)^n/(2n)! * (x^{6n+1})/(6n+1) from 0 \to 1\]

Answer 9

Excuse me, Rn not Rn+1. \[|R_{n}|\le \max_x \left | \frac{|x|^{n+1}}{(n+1)!}\max_z |f^{(n+1)}(z)| \right |\]

Answer 10

And that is your error, I do believe.

Answer 11

\[x,z \in (0,1)\]