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Math Help - Use Riemann sums and a limit to compute the exact area under the curve.

  1. #1
    JDS
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    Use Riemann sums and a limit to compute the exact area under the curve.

    Use Riemann sums and a limit to compute the exact area under the curve.

    Y = 4x2 x on (a) [0, 1]; (b) [-1, 1] ; (c) [1,3]

    I have already done a and verified it is correct, however b and c I am unsure of. This is as far as I get on b, then I am not sure what to do next....

    (b)
    ∆x = 2/n, We use right endpoints as evaluation points, xi = -1 + 2i/n

    (my small amount of work is shown in the picture I attached.)

    Also, I worked this problem out on paper, but the end result seemed to way off, so I started over and only managed to confuse myself.

    I guess I am really getting confused in the part where xi = -1 + 2i/n, where usually it is a single term such as i/n. Anyways any help on this matter is greatly appreciated!
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    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    I think what I would do here, is instead of working out 3 separate sums, I would use the interval [a,b], to derive a formula for any interval. Then we can plug in the intervals we are given.

    We will have:

    \Delta x=\frac{b-a}{n}

    x_k=a+k\Delta x

    Hence:

    A_n=\Delta x\sum_{k=0}^{n-1}\left(4x_k^2-x_k \right)=\Delta x\sum_{k=0}^{n-1}\left(4(a+k\Delta x)^2-(a+k\Delta x) \right)=

    \Delta x\sum_{k=0}^{n-1}\left(4a^2+8a\Delta xk+4(\Delta x)^2k^2-a-\Delta xk \right)=\Delta x\sum_{k=0}^{n-1}\left(a(4a-1)+\Delta x(8a-1)k+4(\Delta x)^2k^2 \right)=

    \Delta x\left(a(4a-1)n+\Delta x(8a-1)\frac{(n-1)n}{2}+2(\Delta x)^2\frac{(n-1)n(2n-1)}{3} \right)=

    (b-a)\left(a(4a-1)+\frac{(b-a)(8a-1)(n-1)}{2n}+\frac{2(b-a)^2(n-1)(2n-1)}{3n^2} \right)

    Thus, we find:

    \lim_{n\to\infty}A_n=(b-a)\left(a(4a-1)+\frac{(b-a)(8a-1)}{2}+\frac{4(b-a)^2}{3} \right)=

    (b-a)\left(\frac{6a(4a-1)+3(b-a)(8a-1)+8(b-a)^2}{6} \right)=

    (b-a)\left(\frac{8(a^2+ab+b^2)-3(a+b)}{6} \right)=\frac{4(b^3-a^3)}{3}+\frac{a^2-b^2}{2}

    Now it's simply a matter of using the given values to answer parts a), b) and c).
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Thank you MarkFL2!! I just 'tested' your derived formula with my known correct answer for a and it was spot on! You are a life saver! Thank you!!! I will post my answers for b and c in a moment to see if I did them correctly. Thanks again!!
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    (b)

    (4(1^3 - (-1)^3)/3) + (((-1)^2 - 1^2)/2)
    = 8/3 + 0/2.....but doesnt this create an undefined answer or null answer because of 0/2?

    or is it simply 8/3?
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    (c)

    (4(3^3 - 1^3)/3) + ((1^2 - 3^2)/2)

    = 104/3 - 8/2

    = 92/3
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    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Yes, b) is 8/3 and c) is 92/3.
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Quote Originally Posted by MarkFL2 View Post
    Thus, we find:


    (b-a)\left(\frac{6a(4a-1)+3(b-a)(8a-1)+8(b-a)^2}{6} \right)=

    (b-a)\left(\frac{8(a^2+ab+b^2)-3(a+b)}{6} \right)=\frac{4(b^3-a^3)}{3}+\frac{a^2-b^2}{2}
    I am trying to use your method for a different formula, but once I reach the point above, I get stuck.

    How are you coming up with that whole last line I quoted. I just don't get how that all factors down to the final equation. I mean it definitely gave me the correct answer for the other problems. But when trying to mimic this formula for a different (but similar equation) I just do not understand how to factor out this last part. Would you be so kind as to explain that last part a bit further? Thank you so much (in advance!) for your taking the time to help me!
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Ah, nevermind....I think I see what mistake I have made....In my calculations I forgot to factor in the (b-a) at the beginning. Makes quite a bit of difference!
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    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Hey, I am glad to see you taking the generalized approach! Not only is it less work in the end, but it forces you to use more algebra up front.

    edit: I will be glad to explain any step I made that is not clear.
    Last edited by MarkFL; December 5th 2012 at 09:15 PM.
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    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    If I were to look at my set of assigned problems and see several functions of the same type (linear, quadratic, etc.) I might be tempted to generalize not only for the intervals, but for the functions as well.
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Quote Originally Posted by MarkFL2 View Post
    Hey, I am glad to see you taking the generalized approach! Not only is it less work in the end, but it forces you to use more algebra up front.

    edit: I will be glad to explain any step I made that is not clear.
    Well, regardless of the "Eureka" moment I had , I think I may have to take you up on the offer to explain a bit further.

    I will post the new equation I am working on and my attempt to work out my derived formula, and the answer that gives.

    I will also show my original attempt to solve the problem (before I attempted your better method). As you might already guess, my two answers did not match, which tells me I did one or the other incorrectly!

    So here is the equation i have been trying to derive:

    y = x2 + 3x on [0, 1]

    I have attached my work in pic files and they are labeled accordingly.
    Attached Thumbnails Attached Thumbnails Use Riemann sums and a limit to compute the exact area under the curve.-original-attempt-solve-equation-using-riemann-sums-limit.jpg   Use Riemann sums and a limit to compute the exact area under the curve.-new-method-my-attempt-derive-formula.jpg  
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    In your first attempt, I am with you up to this point:

    A_n=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac  {3}{n^2}\cdot\frac{n(n+1)}{2}

    Now, I would simplify as follows:

    A_n=\frac{(n+1)(2n+1)}{6n^2}+\frac{3(n+1)}{2n}= \frac{(n+1)((2n+1)+9n)}{6n^2}=\frac{(n+1)(11n+1)}{  6n^2}

    Now, what is the limit of this sum as n\to\infty?

    I am going to look over your more generalized approach next.
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    MHF Contributor MarkFL's Avatar
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    In your second approach, you actually made the same error that you made in your first approach. I am with you up to this point:

    A_n=(b-a)\left(a(a+3)+\frac{(b-a)(2a+3)(n-1)}{2n}+\frac{(b-a)^2(n-1)(2n-1)}{6n^2} \right)

    Now, when you took the limit as n\to\infty you used a denominator of 6 for the last term within the parentheses, when it should be a 3, do you see why?

    I will let you rework this on your own once you understand why you need a 3 there. Everything else you did is correct except for that.

    edit: If you have a bunch of quadratic functions, you may want to generalize further and use f(x)=ax^2+bx+c on [d,e].
    Last edited by MarkFL; December 5th 2012 at 10:32 PM.
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Quote Originally Posted by MarkFL2 View Post
    In your first attempt, I am with you up to this point:

    A_n=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac  {3}{n^2}\cdot\frac{n(n+1)}{2}

    Now, I would simplify as follows:

    A_n=\frac{(n+1)(2n+1)}{6n^2}+\frac{3(n+1)}{2n}= \frac{(n+1)((2n+1)+9n)}{6n^2}=\frac{(n+1)(11n+1)}{  6n^2}

    Now, what is the limit of this sum as n\to\infty?

    I am going to look over your more generalized approach next.
    Ah, thank you for the clarification on that part. Simplifying it does help. After doing so, I come up with:

    A_n=\frac{11}{6}\ + \frac{2}{n}\ + \frac{1}{6n^2}
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    JDS
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    Re: Use Riemann sums and a limit to compute the exact area under the curve.

    Quote Originally Posted by MarkFL2 View Post
    In your second approach, you actually made the same error that you made in your first approach. I am with you up to this point:

    A_n=(b-a)\left(a(a+3)+\frac{(b-a)(2a+3)(n-1)}{2n}+\frac{(b-a)^2(n-1)(2n-1)}{6n^2} \right)

    Now, when you took the limit as n\to\infty you used a denominator of 6 for the last term within the parentheses, when it should be a 3, do you see why?

    I will let you rework this on your own once you understand why you need a 3 there. Everything else you did is correct except for that.

    edit: If you have a bunch of quadratic functions, you may want to generalize further and use f(x)=ax^2+bx+c on [d,e].
    I do not see why it should be a 3, at least not yet...I am now reworking the problem at this point and will let you know if I cannot figure out why it should be a 3...thanks for all your help and guidance thus far!


    Edit: So, yeah, I see it now....2/6 reduces to 1/3....Im gonna chalk that up to being sleep deprived!
    Last edited by JDS; December 5th 2012 at 10:38 PM. Reason: Figured it out!
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