1. ## Trapezoidal

hello i have a big problem i can solve this problem pleae help
Use the trapezoidal rule with n-8 to approximate the definite integral of sin(x^2)dx when a=0 and b=2
$\displaystyle \frac{b-a}{2n}\bigg[f(x_0)+\sum_{m=1}^{n-1}2f(x_m)+f(x_n)\bigg]$

So we have for this case

$\displaystyle \Delta{x}=\frac{2-0}{8}=\frac{1}{4}$

$\displaystyle \frac{2-0}{16}\bigg[\sin(0^2)+2\sin\bigg(\frac{1}{16}\bigg)+2\sin\bigg (\frac{1}{4}\bigg)+$$\displaystyle 2\sin\bigg(\frac{9}{16}\bigg)+2\sin(1)+\sin\bigg(\ frac{25}{16}\bigg)+2\sin\bigg(\frac{9}{4}\bigg)+2\ sin\bigg(\frac{49}{16}\bigg)+\sin(4)\bigg]\approx\int_0^{2}\sin(x^2)dx I dont know how high you are in but if you are interested \displaystyle \int\sin(x^2)dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!} 2. Originally Posted by Mathstud28 \displaystyle \frac{b-a}{2n}\bigg[f(x_0)+\sum_{m=1}^{n-1}2f(x_m)+f(x_n)\bigg] So we have for this case \displaystyle \Delta{x}=\frac{2-0}{8}=\frac{1}{4} \displaystyle \frac{2-0}{16}\bigg[\sin(0^2)+2\sin\bigg(\frac{1}{16}\bigg)+2\sin\bigg (\frac{1}{4}\bigg)+$$\displaystyle 2\sin\bigg(\frac{9}{16}\bigg)+2\sin(1)+\sin\bigg(\ frac{25}{16}\bigg)+2\sin\bigg(\frac{9}{4}\bigg)+2\ sin\bigg(\frac{49}{16}\bigg)+\sin(4)\bigg]\approx\int_0^{2}\sin(x^2)dx$

I dont know how high you are in but if you are interested

$\displaystyle \int\sin(x^2)dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!}$
Presumably someone PM this to you. It might be an idea to say so and who did. As it is the question is sort of floating with no context.

RonL

3. Originally Posted by CaptainBlack
Presumably someone PM this to you. It might be an idea to say so and who did. As it is the question is sort of floating with no context.

RonL
http://www.mathhelpforum.com/math-he...bers/bega.html

I had a PM from him/her too
And he/she posted his/her question in the thread of someone else... ^^