1. ## 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.

2. 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
$\displaystyle sin(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}x^{2n + 1}$

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

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

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

I'll let you finish up from there.

-Dan

3. I have a couple questions; where do I plug in the 0.6 and how do I take into account the factorial.

4. 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 $\displaystyle \int_a^{b}f(x)dx=F(b)-F(a),F'(x)=f(x)$...so apply that with b=.6 and a=0(drops out no point in putting it in)

5. Oh yeah and to account for the factorial I suppose you will have to use the gamma function haha $\displaystyle \Gamma(x)=\int_0^{1}\bigg[ln\bigg(\frac{1}{t}\bigg)\bigg]^{x}dt$

6. 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

7. ## 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

8. i keep getting 1.07 something any help would be fantastic

9. Originally Posted by waite3
i keep getting 1.07 something any help would be fantastic
.199

10. that seems not to be working either

11. Originally Posted by waite3
that seems not to be working either
$\displaystyle \int_0^{.6}sin(3x^2)dx=.199=\sum_{n=0}^{\infty}\fr ac{(-1)^{n}3^{2n+1}x^{4n+3}}{(4n+3)(2n)!}\bigg|_0^{.6}$

12. are there any additional decimals because sometime the program needs quite a few. thanks for the help.

13. Originally Posted by waite3
are there any additional decimals because sometime the program needs quite a few. thanks for the help.
0.19865856135