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Math Help - Approximate the area under the curve using n rectangles and the evaluation rules

  1. #1
    JDS
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    Approximate the area under the curve using n rectangles and the evaluation rules

    (NOTE: I am using an example in my book, that I have the answer to already, because I want to understand how they arrived at the answer, that way, I can do the other problems on my own)

    Okay, so here is the problem:
    Approximate the area under the curve on the given interval using n rectangles and the evaluation rules (a) left endpoint, (b) midpoint, and (c) right endpoint.

    y = x2 + 1 on [0, 1] and n=16


    So, this is how the book solved this.....

    (a) ci = i∆x where i is from 0 to 15

    A16 = ∆x (I do not know how to input the correct symbol here, but its 15 on top and i=0 on bottom) f(ci)

    = 1/16 (that symbol again) [(i/16 + 1/16)2+ 1]

    (approximately =) 1.3652

    Now, I sort of get how they get up to the last part, but I do not understand how they came to the conclusion of 1.3652. Any clarification on this would be greatly appreciated.
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  2. #2
    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    Im attaching a Screen Shot of a similar problem that I attempted, feedback/ideas/suggestions are welcome.
    Attached Thumbnails Attached Thumbnails Approximate the area under the curve using n rectangles and the evaluation rules-calculus-problem.jpg  
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    MHF Contributor MarkFL's Avatar
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    1.) We are given f(x)=x^2+1 on the interval [0,1] with n=16.

    \Delta x=\frac{1-0}{16}=\frac{1}{16}

    x_k=x_0+k\Delta x=k\Delta x

    a) left end-points:

    A\approx\left(\sum_{k=0}^{15}f(x_k) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=0}^{15}\left(x_k^2+1 \right)=\frac{1}{16}\left((\Delta x)^2\sum_{k=0}^{15}k^2+\sum_{k=0}^{15}(1) \right)=

    \frac{1}{16}\left(\frac{1}{256}\cdot\frac{15(15+1)  (2\cdot15+1)}{6}+16 \right)=\frac{1}{16}\left(\frac{1}{256}\cdot\frac{  15\cdot16\cdot31}{6}+16 \right)=

    \frac{1}{16}\left(\frac{1}{16}\cdot\frac{5\cdot31}  {2}+16 \right)=\frac{1}{16}\left(\frac{155}{32}+16 \right)=\frac{1}{16}\cdot\frac{667}{32}=\frac{667}  {512}=1.302734375

    Since the function is increasing on the interval, we should expect the left-point approximation to be less than the true value of 4/3. Your book gives a value greater than this, and I notice a problem with the formula used, specifically with the value of x_k.

    b) midpoints:

    A\approx\left(\sum_{k=0}^{15}f\left(\frac{x_k+x_{k  +1}}{2} \right) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=0}^{15}\left(\left(\frac{x_k  +x_{k+1}}{2} \right)^2+1 \right)=

    \frac{1}{16}\sum_{k=0}^{15}\left(\left(\frac{k \Delta x+(k+1)\Delta x}{2} \right)^2+1 \right)=\frac{1}{16}\sum_{k=0}^{15}\left(\left( \frac{ \Delta x}{2} \right)^2(2k+1)^2+1 \right)=

    \frac{1}{16}\left(\left( \frac{ \Delta x}{2} \right)^2\sum_{k=0}^{15}(4k^2+4k+1)+\sum_{k=0}^{15  }(1) \right) \right)=\frac{1}{16}\left(\frac{1}{1024}\left(4 \sum_{k=0}^{15}(k^2)+4\sum_{k=0}^{15}(k)+\sum_{k=0  }^{15}(1) \right)+16 \right) \right)=

    \frac{1}{16}\left(\frac{1}{1024}\left(4\cdot\frac{  15\cdot16\cdot31}{6}+4\cdot\frac{15\cdot16}{2}+16 \right)+16 \right) \right)=\frac{1}{16}\left(\frac{5456}{1024}+16 \right)=

    \frac{1}{16}\cdot\frac{1365}{64}=\frac{1365}{1024}  = 1.3330078125

    c) right end-points:

    A\approx\left(\sum_{k=1}^{16}f(x_k) \right)\cdot\Delta x=\frac{1}{16}\sum_{k=1}^{16}\left(x_k^2+1 \right)=\frac{1}{16}\left((\Delta x)^2\sum_{k=1}^{16}k^2+\sum_{k=1}^{16}(1) \right)=

    \frac{1}{16}\left(\frac{1}{256}\cdot\frac{16(16+1)  (2\cdot16+1)}{6}+16 \right)=\frac{1}{16}\left(\frac{1}{256}\cdot\frac{  16\cdot17\cdot33}{6}+16 \right)=

    \frac{1}{16}\left(\frac{1}{16}\cdot\frac{17\cdot11  }{2}+16 \right)=\frac{1}{16}\left(\frac{187}{32}+16 \right)=\frac{1}{16}\cdot\frac{699}{32}=\frac{699}  {512}=1.365234375

    Okay, now I see that what you have as part a) in your first post is part c). If you have any questions about anything I posted here, please don't hesitate to ask.
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    ah, yes my apologies about the initial mix up.....Thank you for showing all the calculations, I think that I keep forgetting to apply the rules for n(n+1)/2 for values of i and the rule for values of i^2 being n(n+1)(2n+1)/6.....with that in mind I think I may have to go back and do the other problems over again (which I posted in the picture file I attached above)
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    I am redoing a, b and c, for the other problem i posted here, I will post back here with each part when I finish to allow you to critique my work, if you do not mind!
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    Okay so I finished part a of the other problem, using (I hope) the correct rules this time. If you do not mind, please take a look and let me know if I made any mistakes. Thanks in advance


    NOTE: I gott get some rest, but will take a look here tomorrow, and try to do b and c for the other problem and post my answers, thanks!!
    Attached Thumbnails Attached Thumbnails Approximate the area under the curve using n rectangles and the evaluation rules-calculus-problem2.jpg  
    Last edited by JDS; December 3rd 2012 at 08:55 PM. Reason: just an added note at the end
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    MHF Contributor MarkFL's Avatar
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    Looks good!
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    okay, I just finished part B (midpoints) and am attaching a pic of the work I did. Once again, if you do not mind, give it a look and let me know if I have made any mistakes. Thanks!!
    Attached Thumbnails Attached Thumbnails Approximate the area under the curve using n rectangles and the evaluation rules-calculus-problem2b.jpg  
    Last edited by JDS; December 4th 2012 at 07:26 AM. Reason: forgot to attach the pic initially!
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    and now I am done with Part C, please let me know if I have made any mistakes, As Always....THANK YOU!!!!
    Attached Thumbnails Attached Thumbnails Approximate the area under the curve using n rectangles and the evaluation rules-calculus-problem2c.jpg  
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    MHF Contributor MarkFL's Avatar
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    Both results look good, the only quibble I have is in part c); your lower index of summation should be 1, not 2.
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    JDS
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    ah, yeah, got ya. Thank you!
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    Re: Approximate the area under the curve using n rectangles and the evaluation rules

    Quote Originally Posted by JDS View Post
    ah, yeah, got ya. Thank you!
    In future, DO NOT BUMP THREADS.
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