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Thread: Show that limit is a definite integral

  1. #1
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    Show that limit is a definite integral

    Show that the given limit is a definite integral $\displaystyle \int_a^b f(x) \, dx$ for a suitable interval $\displaystyle [a,b]$ and function $\displaystyle f$


    The limit is:


    $\displaystyle \lim \limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n}{n^2 + i^2}$


    That's the problem I can't solve. Here is how far I've made it:


    Since a definite integral can be defined as


    $\displaystyle \int_a^b f(x) \, dx = \lim\limits_{n \to \infty} \sum\limits_{i=1}^n f\left(a + (b-a)\frac{i}{n}\right) \cdot \frac{b-a}{n}$


    Then:


    $\displaystyle f\left(a + (b-a)\frac{i}{n}\right) \cdot \frac{b-a}{n} = \frac{n}{n^2 + i^2}$


    which simplifies to:


    $\displaystyle f\left(a + (b-a)\frac{i}{n}\right)= \frac{1}{b-a} \frac{n^2}{n^2 + i^2}$


    I'm not sure how to solve, but, I can guess that $\displaystyle b-a=1$ which would simplify things to:


    $\displaystyle f\left(a + \frac{i}{n}\right)= \frac{n^2}{n^2 + i^2}$


    I can also guess that $\displaystyle a=1$ so that:


    $\displaystyle f\left(\frac{n + i}{n}\right)= \frac{n^2}{n^2 + i^2}$


    Now, I can't find a function $\displaystyle f$ that satisfies this. $\displaystyle f(x)=\frac{1}{x^2}$ almost works but not quite.
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  2. #2
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    Re: Show that limit is a definite integral

    Quote Originally Posted by VinceW View Post
    Show that the given limit is a definite integral $\displaystyle \int_a^b f(x) \, dx$ for a suitable interval $\displaystyle [a,b]$ and function $\displaystyle f$
    The limit is:
    $\displaystyle \lim \limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n}{n^2 + i^2}$
    Let $\displaystyle a=0,~b=1~\&~f(x)=\frac{1}{1+x^2}$.
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  3. #3
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    Re: Show that limit is a definite integral

    that works. I take it that was just intuition? Not sure how I would have figured that out by myself. thanks!
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