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Math Help - Trapezoidal

  1. #1
    MHF Contributor Mathstud28's Avatar
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    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
    \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

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

    \frac{2-0}{16}\bigg[\sin(0^2)+2\sin\bigg(\frac{1}{16}\bigg)+2\sin\bigg  (\frac{1}{4}\bigg)+ 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

    \int\sin(x^2)dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    \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

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

    \frac{2-0}{16}\bigg[\sin(0^2)+2\sin\bigg(\frac{1}{16}\bigg)+2\sin\bigg  (\frac{1}{4}\bigg)+ 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

    \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
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  3. #3
    Moo
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    Quote Originally Posted by CaptainBlack View Post
    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... ^^
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