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

- katieb

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- anonymous

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

- anonymous

- IrishBoy123

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 \(x \ f'(0) = ...\)

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## More answers

- anonymous

Just plug it in directly?

- anonymous

What do you think @sourwing

- IrishBoy123

"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

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

- IrishBoy123

sounds good

- anonymous

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

- anonymous

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

- IrishBoy123

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

- anonymous

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

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

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

- anonymous

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

- IrishBoy123

|dw:1439852188969:dw|

- IrishBoy123

|dw:1439852553035:dw|

- anonymous

\[6x^2-x^5\]??

- IrishBoy123

|dw:1439852806038:dw|

- anonymous

-1/6 * 6x^5

- anonymous

-6x^5/6

- anonymous

-x^5

- IrishBoy123

you are *not* summing a series, or *valuing* a series
you are hacking an integral; by approximating the integrand to a hack of its Mac expansion

- anonymous

brb I have to go feed goats

- anonymous

Maclaurin series of \(sin(6x^2) \) is
\(sin(6x^2) =\sum_{n=0}^{\infty}\dfrac{(-1)^n (6x^2)^{2n+1}}{(2n+1)!}\)
\[\int_0^{0.78} sin(6x^2)dx=\int_0^{0.78}\sum_{n=0}^{\infty}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!}dx\\=\sum_{n=0}^{\infty}\int_0^{0.78}\dfrac{(-1)^n(6x^2)^{2n+1}}{(2n+1)!} dx\]
This integral is easy to take, right?
After take out the integral, you just calculate the first 3 term of it (approximate theorem) and you get the answer.

- anonymous

oh, your problem asks for the first 2 terms, not 3 as I suggested. That is easier.!!

- anonymous

I'm sorry but nothing about this seems easy to me. I hate to keep bothering you but is there any way you can force me to understand this? @Crazyandbeautiful

- anonymous

OH, I assumed you know the formula of Maclaurin series of sin (6x^2)
If you don't know it, you can derive it from Maclaurin series of sinx, ok?

- anonymous

\(sin x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}\) ok?

- anonymous

Okay

- anonymous

so, just replace x by 6x^2 to get Maclaurin series of sin(6x^2)

- anonymous

ok?

- anonymous

Alright

- anonymous

now, you want to take int of sin (6x^2) from 0 to 0.78 , right?
Just take integral both sides.

- anonymous

The left hand side is \(\int_0^{0.78} sin(6x^2)dx\)

- anonymous

The right hand side is integral of the sum (...) right?

- anonymous

Yes or NO? where don't you get?

- anonymous

I'm just writing this in my notes as we go, sorry

- anonymous

Yes

- anonymous

And we know that integral of sum = sum of integral, right?

- anonymous

Okay

- anonymous

hence the right hand side is
\[\sum_{n=0}^{\infty} \int_0^{0.78}(-1)^n\dfrac{(6x^2)^{2n+1}}{(2n+1)!}\] ok?

- anonymous

Okay, I'm with you

- anonymous

Now, calculate the integral. Can you?

- anonymous

I can try.

- anonymous

But what do I do with the fact that this has both n and x in it?

- anonymous

Don't forget \((6x^2)^{2n+1}= 6^{2n+1}*x^{4n+2}\)

- anonymous

hey, just x is the variable, n is a number!! nothing to do with n, take it as constant

- anonymous

you take integral w.r.t.x (dx) right? how n works as variable?

- anonymous

We will use n for the far left summation, not for integral.

- anonymous

Okay

- anonymous

It turns to
\[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!}\int_0^{0.78} x^{4n+2}dx\]

- anonymous

ok?? take integral, what do you get?

- anonymous

|dw:1439861021631:dw|

- anonymous

\[\frac{ x ^{4n+3} }{ 4n+3 }\]???

- anonymous

where are your limits?

- anonymous

Sorry, evaluated from 0 to 0.78

- anonymous

ok, =?

- anonymous

Just sub .78 and 0 for x and subtract?

- anonymous

0 for x =0, just plug x =0.8 in, ok?

- anonymous

oh, 0.78

- anonymous

So it's just \[\frac{ .78^{4n+3} }{ 4n+3 }\] ??

- anonymous

So, now you have
\[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\]

- anonymous

ok ?

- anonymous

Yes

- anonymous

You need just 2 terms!! easy!!
n =0, plug in,
n =1, plug in
add them together, done.

- anonymous

That's it? Just add the resulting two terms together?

- anonymous

Yes, dat sit. If you want, you can add you and me to get 4 terms. :)

- anonymous

show me your result, please.

- anonymous

Working on it now

- anonymous

-0.745188?

- anonymous

I got 0.0457318875

- anonymous

ok, let's do together

- anonymous

I'm absolutely sure that you're right, I just did mine mostly in my head...

- anonymous

\[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\]
\[(-1)^0\dfrac{6^{2*0+1}}{(2*0+1)!} *\dfrac{0.78^{4*0+3}}{4*0+3}+(-1)^1\dfrac{6^{2*1+1}}{(2*1+1)!} *\dfrac{0.78^{4*1+3}}{4*1+3}\]

- anonymous

\[-1^{0}*\frac{ 6^{2(0)+1} }{ (2(0)+1)! }*\frac{ 0.78^{4(0)+3} }{ 4(0)+3 }\]

- anonymous

Oh you're much faster at this than I am

- anonymous

hahaha... I just copy and paste the previous post, then change the numbers. hehehe

- anonymous

Well for n = 0, term by term, the first one would equal 1 * 6/1* .78^3/3 which I said totals 0.949104

- anonymous

and your first term = \(\dfrac{6}{1!}\dfrac{0.78^3}{3}\) right? = \(2*0.78^3=0.949104\) Yes,

- anonymous

The second term is ??

- anonymous

For n = 1 I said -1* 6^3/6 * .78^7/7 which I thought should be -0.90337211

- anonymous

\(-\dfrac{6^3}{3!}\dfrac{0.78^7}{7}=-0.9033721125\)
Yes, again,

- anonymous

add them together , you get??

- anonymous

0.045732

- anonymous

yup

- anonymous

So I did just mess something up in my head the first time.

- anonymous

easy, right?

- anonymous

Thank you so much Crazyandbeautiful, you are literally the most beautiful person in the world to me right now

- anonymous

haahaha... I'm not beautiful, I am handsome. hahaha.... jk

- anonymous

lol you're probably going to be the deciding factor in how well I do in my test next week actually, because I clearly had no idea what I was doing.

- anonymous

Handsome, beautiful, the actual envisionment of God, whatever you are dude you saved my life here

- anonymous

You got the method, right? don't scare me. I spent more than 1h to explain and you said you don't know what you are doing. it is a real scaring.

- anonymous

I had no idea BEFORE this chat string

- anonymous

Believe me, I was taking notes

- anonymous

ok, good luck

- anonymous

Thanks again

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