Assume that equals its Maclaurin series for all x.
Use the Maclaurin series for to evaluate the integral
I get .334 for an answer, but evidently I'm doing something wrong, any insight would be outstanding, thanks
Assume that equals its Maclaurin series for all x.
Use the Maclaurin series for to evaluate the integral
I get .334 for an answer, but evidently I'm doing something wrong, any insight would be outstanding, thanks
Since we know that $\displaystyle sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$...we adapt this to $\displaystyle \sin(7x^2)$
$\displaystyle \int\sin(7x^2)dx=\int\sum_{n=0}^{\infty}\frac{(-1)^{n}(7x^2)^{2n+1}}{(2n+1)!}dx=\int\sum_{n=0}^{\i nfty}\frac{(-1)^{n}7^{2n+1}x^{4n+2}}{(2n+1)!}dx$
Integrating we get $\displaystyle \int_0^{.64}\sin(7x^2)dx=\sum_{n=0}^{\infty}\frac{ (-1)^{n}7^{2n+1}x^{4n+3}}{(2n+1)!\cdot(4n+3)}\bigg|_ 0^{.64}=.334$
hmm no luck with that, are there any more decimals after that, for whatever stupid reason the program likes 10 decimals. And then how did you go about solving that sum (like how do you account for both variables?) thanks again for your help, and sorry if these are dumb questions haha
Well maybe he wants the $\displaystyle a_1,a_2$ terms which are the second and third temrs...not the first and second...their sum would be a negative number so I dont think its them....I would have to say either your proffessor is making a mistake or I am misunderstanding something ...sorry
haha no problem, I would agree theres prlly an error, the professor is pretty useless when it comes to helping (its an online course), thanks for all of your help though! and just for kicks and giggles, what was the negative number you got? thanks again