Assume that \sin(x) equals its Maclaurin series for all x.
Use the Maclaurin series for sin(6 x^2) to evaluate the integral
(integral [0, 0.78]) sin(6 x^2)dx
. Your answer will be an infinite series. Use the first two terms to estimate its value.

Question

Assume that \sin(x) equals its Maclaurin series for all x. 
Use the Maclaurin series for sin(6 x^2) to evaluate the integral
(integral [0, 0.78]) sin(6 x^2)dx
. Your answer will be an infinite series. Use the first two terms to estimate its value.

anonymous · Answer

$$\int\limits_{0}^{0.78}\sin(6x^2)dx$$

anonymous · Answer

@mathstudent55

irishboy123 · Answer

writing the Maclaurin sin expansion in terms of t:

$\large sin(t) = \Sigma_{0}^{\infty}  \frac{(-1)^n }{(2n+1)! }\ t^{2n+1}$

it seems they are happy for you to stuff $t = 6 x^2$ in and go from there

to use an actual Mac series for $sin(6 x^2)$, you'd need to go

$f(0) = 0$
$f'(x) = (sin(6 x^2))' = ...$ and $f'(0) = ....$ so $x \ f'(0) = ...$

anonymous · Answer

Just plug it in directly?

anonymous · Answer

What do you think @sourwing

irishboy123 · Answer

"Just plug it in directly?"

yes, that is what the question mandates

for #1, plug it in term by terms and come up with a nice summation with a $\Sigma$ in it!

for #2, generate a few terms and hope you get the first two quickly.

[if you have a crack at the proper Mac expansion for that function, you will be busy for a while :p]

anonymous · Answer

So basically $$\sum_{n = 0}^{\infty} \frac{ -1^n }{ (2n+1)! } * (6x^2)^{2n+1}$$, and solve for n =  0 and n=1?

irishboy123 · Answer

sounds good

anonymous · Answer

Well what do I do to estimate the value of the series?

anonymous · Answer

I ask because I still have an x term in my result

irishboy123 · Answer

if you have an 'x' term in your result, well that is a good thing.  :p

anonymous · Answer

Okay lol. Sorry I'm clueless on the procedure here. Should I be evaluating the limit at both of the solutions? They are N_0 = -6x^2 and N_1 = -216x^2

irishboy123 · Answer

now, forget about the $\Sigma$ and just evaluate the first 2 terms

you are not summing, you are just using the first 2 terms of the series

make sense?!?!

you need x terms because you have to integrate

irishboy123 · Answer

nothing to do with limits, either.

just knock out the first 2 terms of the [bogus] expansion

anonymous · Answer

So get the first two terms in the series, but how do I use that to estimate the value of the series?

irishboy123 · Answer

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