Question

use a maclaurin series representation of the integrand from 0 to 1 of cos(x^2) to approximate the integral with an error of no more than .0001

Answer 1

\[\cos x=1-\frac{ x ^{2} }{ 2! }+\frac{ x ^{4} }{ 4! }-\frac{ x ^{6} }{ 6! }+ ...\]

put in x^2 in place of x and integrate term bu term until the integral is less than the desired accuracy

Answer 2

hint: you'll only need to do 2 terms

Answer 3

ok im still confused though

Answer 4

about...

Answer 5

where do i go after integrating

Answer 6

they want you to approximate \[\int\limits\limits_{0}^{1}\cos x ^{2}dx\] to the nearest .0001 by switching out the integrand with its MacLaurin series represetation.
so we get
\[\int\limits\limits_{0}^{1}\cos x ^{2}dx=\int\limits\limits_{0}^{1}\left( 1-\frac{ x ^{4} }{ 2! }+\frac{ x ^{8} }{ 4! }- ... \right) dx\]
so integrate term by term until the numbers in the 4th decimal place stop changing

Answer 7

ok thank you for your help