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A lower level question of curiosity.

  1. Mar 19, 2007 #1
    Hello, I am a high school student, and I am a new member to this forum.

    I was wondering how to find the area under the graph of one sin arc without using calculus. The problem comes from something one of my friends gave me. The question shows a picture with some rectangles in the space underneath the curve. I was thinking if i drew a large, scaled version and filled it with many rectangles i would be able to approximate the area. the key point to my question is that calculus is not involved.

    Are there any methods that are less demeaning than drawing tons of little rectangles? What are they?
     
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  3. Mar 19, 2007 #2

    JasonRox

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    Definitely. You can get great approximations without even using calculus.

    To get a good approximation you might want to look into Riemann Sums and once you get that look at the Trapezoidal Rule, which is just an improved version. There is also a Simpsons Rule, but I don't quite remember the method. I do believe it was the most accurate of the three though.

    In the end, I suggest you go with the Riemann Sums. It's the easiest one to follow, it's very straight forward. In fact, you probably already used Riemann Sums without even knowing it. (It involves rectangles.) Also, what's neat about it is that you can increase the approximation with relative ease compared to an actual drawing.

    Then, you will see that the value converges to a number as you approximate it even more and even more. Say, for example the answer is 3, then what I mean by converge is that, let's say the first try you get 2.5, the second 2.7, the third 2.9, then 2.95, then 2.96, and so on.

    The answer isn't 3. I decided not to use the answer so that you discover it yourself.
     
  4. Mar 19, 2007 #3

    JasonRox

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    Oh, and just search on Google for Riemann Sums. Look for a website with nice diagrams and such.
     
  5. Mar 19, 2007 #4
    Ok thanks, and for anyone else who wouldnt mind helping, I am in precalculus right now so I dont even know calculus to begin with.
     
  6. Mar 19, 2007 #5

    JasonRox

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    You don't need to know calculus for this. As long as you know how to draw a graph, and evaluate a function at a point there you're ready to go.

    i.e. Evaluate f(x)=sin(x) at Pi. f(Pi) = sin(Pi) = ?
     
  7. Mar 20, 2007 #6
  8. Mar 20, 2007 #7
    Thanks so far everyone, but the only site that explains how to use the Riemann Sums on the calculator is a little confusing, could anyone help with this? I have a Ti- 84 SI+
     
  9. Mar 20, 2007 #8

    matt grime

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    Since when were Riemann sums not calculus/analysis?
     
  10. Mar 20, 2007 #9

    JasonRox

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    I meant in a way where it doesn't use derivatives and such.
     
  11. Mar 20, 2007 #10

    JasonRox

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  12. Mar 20, 2007 #11

    matt grime

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    It's nothing to do with your calculator. Just draw some rectangles or parallelograms under the curve to estimate the area. Add 'em by hand if you want. You won't get the actual integral, just a numerical estimate of it. The real integral is defined as the limit of the estimates as the width of your rectangles tends to zero, so you cannot, by definition, avoid analysis.
     
  13. Mar 20, 2007 #12
    OK thanks you guys, so ive drawn from your responses that the only non-calc/analysis method to do this is with the little rectangles. Looks to be fun time!! (10)
     
  14. Mar 20, 2007 #13

    JasonRox

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    That's what Riemann Sums does. It draws little rectangles for you.

    Check out the website I linked about. It gives pictures explaining everything.
     
  15. Mar 20, 2007 #14
    O yea i know thats what it does but everyone else said that its considered analysis, so i thought i should stay away from it?
     
  16. Mar 20, 2007 #15
    Alright I took some time to draw the rectangles and such and i got an answer first of 1.87, then 1.89, then 1.92, and most recently 1.95, would it be safe to assume, following logical order that the area is approaching 2? Is this the correct answer? Thanks very much!!
     
  17. Mar 20, 2007 #16

    uart

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    No it wouldn't be safe to conclude that based just on the data you have. In general (series approximating) integrals do not necessarily converge to nice round numbers, so assuming "2" for that reason would be fallacy.

    In this case however (assuming you are integrating sin(x) from x = zero to Pi) then as it turns out yes the exact answer is 2. It's just that it wouldn't be safe to conclude that from your data alone.
     
  18. Mar 20, 2007 #17

    JasonRox

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    Oh really.

    Are you sure about that?
     
  19. Mar 21, 2007 #18

    matt grime

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    yes, Jason. He is. And he is correct. There is nothing posted by dtl42 that implies that that sequence of numbers he wrote converges to 2. It could be 2-sqrt(3)/e^4000.
     
    Last edited: Mar 21, 2007
  20. Mar 21, 2007 #19

    Gib Z

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    We'll, You could probably tell theres no e in it, seeing as its a sin curve :P Unless someone can prove pi and e are not algebaricly independant...

    But I agree with the point, we can't say the data implies a value of 2, after more terms we could say 2 may be a good approximation.

    EDIT: "We'll, You could probably tell theres no e in it, seeing as its a sin curve :P Unless someone can prove pi and e are not algebaricly independant..."

    If the bounds aren't designed to be fitting to that solution as well lol >.< I doubt this dudes doing that though.
     
  21. Mar 21, 2007 #20

    matt grime

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    what has that got to do with anything? If we're going to throw in some off topic comments, then, note that sin(x) = (expix - exp-ix)/2i, so maybe e and i will appear in the answer!
     
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