integral by series

• April 17th 2008, 05:11 PM
waite3
integral by series
Thanks for any help.

Assume that sin(x) equals its Maclaurin series for all x.
Use the Maclaurin series for sin(3x^2) to evaluate the integral

∫sin(3x^2) from 0.6 to 0

Your answer will be an infinite series. Use the first two terms to estimate its value.
• April 17th 2008, 06:22 PM
topsquark
Quote:

Originally Posted by waite3
Thanks for any help.

Assume that sin(x) equals its Maclaurin series for all x.
Use the Maclaurin series for sin(3x^2) to evaluate the integral

∫sin(3x^2) from 0.6 to 0

Your answer will be an infinite series. Use the first two terms to estimate its value.

The Maclaurin series for sin(x) is
$sin(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}x^{2n + 1}$

So
$sin(3x^2) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}\left ( 3x^2 \right ) ^{2n + 1}$

Thus
$\int sin(3x^2)~dx = \int \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}\left ( 3x^2 \right ) ^{2n + 1}~dx$

$= \sum_{n = 0}^{\infty} \frac{(-1)^n3^{2n+ 1}}{(2n + 1)!(4n + 3)}x^{4n + 3}$

I'll let you finish up from there.

-Dan
• April 17th 2008, 07:10 PM
waite3
I have a couple questions; where do I plug in the 0.6 and how do I take into account the factorial.
• April 17th 2008, 07:18 PM
Mathstud28
Quote:

Originally Posted by waite3
I have a couple questions; where do I plug in the 0.6 and how do I take into account the factorial.

use the fundamental theoroem of calculus $\int_a^{b}f(x)dx=F(b)-F(a),F'(x)=f(x)$...so apply that with b=.6 and a=0(drops out no point in putting it in)
• April 17th 2008, 07:20 PM
Mathstud28
Oh yeah and to account for the factorial I suppose you will have to use the gamma function haha $\Gamma(x)=\int_0^{1}\bigg[ln\bigg(\frac{1}{t}\bigg)\bigg]^{x}dt$
• April 18th 2008, 08:39 AM
waite3
i understand how to integrate the 0.6 now into the equation, but is there an easier way the solve this problem with the factorial in there
• April 21st 2008, 07:35 PM
waite3
problem is driving me nuts
can anyone give me an exact answer cause the answers i keep getting are not right when i plug them into the program I'm using.

thanks for any help
• April 22nd 2008, 05:24 PM
waite3
i keep getting 1.07 something any help would be fantastic
• April 22nd 2008, 05:36 PM
Mathstud28
Quote:

Originally Posted by waite3
i keep getting 1.07 something any help would be fantastic

.199
• April 22nd 2008, 05:48 PM
waite3
that seems not to be working either
• April 22nd 2008, 05:51 PM
Mathstud28
Quote:

Originally Posted by waite3
that seems not to be working either

$\int_0^{.6}sin(3x^2)dx=.199=\sum_{n=0}^{\infty}\fr ac{(-1)^{n}3^{2n+1}x^{4n+3}}{(4n+3)(2n)!}\bigg|_0^{.6}$
• April 22nd 2008, 05:55 PM
waite3
are there any additional decimals because sometime the program needs quite a few. thanks for the help.
• April 22nd 2008, 05:59 PM
Mathstud28
Quote:

Originally Posted by waite3
are there any additional decimals because sometime the program needs quite a few. thanks for the help.

0.19865856135