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Math Help - Evaluating limit of an area under the graph of a continuous function.

  1. #1
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    Evaluating limit of an area under the graph of a continuous function.

    The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:
    (a) Use this definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit


    A= lim n-->infinity ( f(x1)1/n +f(x2)1/n + ...+f(xn)1/n )

    x1= x1+deltax = 1*deltax
    x2= 2deltax
    x3= 3deltax

    xi=ideltax= i(1/n)

    My answer =



    (b) Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a).

    I am not sure at all how to solve this part.

    Any help greatly appreciated!!
    Last edited by Steelers72; February 9th 2013 at 07:01 PM.
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    Re: Evaluating limit of an area under the graph of a continuous function.

    Starting from your expression \lim_{n\to\infty} \sum_{i=1}^n \left(\frac{i}{n}\right)^3 \frac{1}{n}, you can factor \frac{1}{n^4} out of the sum, since it doesn't depend on i:

    \lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3

    and now the sum is exactly 1^3+2^3+\dots+n^3, so you can use the formula you were given.

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    Re: Evaluating limit of an area under the graph of a continuous function.

    Thanks Hollywood! Since the numbers are infinite, how do I know where to stop adding? It's asking for an exact numerical answer some I'm unsure of the exact amount.
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    Re: Evaluating limit of an area under the graph of a continuous function.

    You first evaluate the *finite* sum to get a function of n, then take the limit of this function as n goes to infinity.

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    Re: Evaluating limit of an area under the graph of a continuous function.

    Ok, after plugging in 1^3 and 2^3 into ( n(n+1)/2 )^2

    I got 1+1296 which equals 1297.

    Now I use the 1/n^4 summation i^3 as n-->infinity ?

    Would the answer be 0 because 1/large = approaches zero?
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    Re: Evaluating limit of an area under the graph of a continuous function.

    0 wasn't right; scratch that thought
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    Re: Evaluating limit of an area under the graph of a continuous function.

    \lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3 =

    \lim_{n\to\infty}\frac{1}{n^4}\left(\frac{n(n+1)}{  2}\right)^2 =

    \lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{  4}

    Can you take it from there?


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    Re: Evaluating limit of an area under the graph of a continuous function.

    Thanks for your help hollywood; this calculus is new to me so I'm slow at learning it. thanks for the patience. Do we have to plug in 1^3 and 2^3 into that formula?

    Or if we are solving for n here is my work:

    1/n^4 is zero since a small # over an infinitely large number = 0. However, the n^4+2n^3 +n^2/4 part would be infinity since a very large # over a small number goes to infinity. 0*infinity would be zero though? Am I overlooking something? Sorry for my naivety.
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  9. #9
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    Re: Evaluating limit of an area under the graph of a continuous function.

    Yes, both the numerator and denominator go to infinity as n goes to infinity. But you can simplify the expression to something that you can take the limit of:

    \lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{ 4}=

    \lim_{n\to\infty}\frac{n^4+2n^3+n^2}{ 4n^4}=

    \lim_{n\to\infty}\frac{1}{4}+\frac{1}{2n}+\frac{1}  {4n^2}

    and now you can see that \frac{1}{2n} and \frac{1}{4n^2} go to zero as n goes to infinity, and of course \frac{1}{4} stays \frac{1}{4}, so the answer is \frac{1}{4}.

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