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anonymous

  • one year ago

Measuring Integrals .. Need a suggestion on how I should approach this one. Image Coming...

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  1. anonymous
    • one year ago
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  2. dan815
    • one year ago
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    yep 6 is right

  3. anonymous
    • one year ago
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    Lol, you think I should just tell them they are correct and move on?

  4. dan815
    • one year ago
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    http://prntscr.com/7tsrio

  5. dan815
    • one year ago
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    from 1 to 3, the most that curve is fluctuating* is between a change in range of 0.5

  6. anonymous
    • one year ago
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    I need a good way to say that the variance around y=2 equates to no more than 0.6 in error.

  7. dan815
    • one year ago
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    what a second that doesnt make sense actually, the area should be 4 not 6

  8. dan815
    • one year ago
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    if it was integral from 0 to 3 then it would be 6

  9. anonymous
    • one year ago
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    its in the interval 1 to 3, so it's about 3*2

  10. Empty
    • one year ago
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    See the area of that rectangle outside and below is AT LEAST the integral, while the rectangle around the function represents your error

  11. anonymous
    • one year ago
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    no? oh yeah..

  12. anonymous
    • one year ago
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    1 to 3 would be 2...

  13. anonymous
    • one year ago
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    damn them... lol

  14. Empty
    • one year ago
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    Yeahhhhhh this is bunk yo

  15. anonymous
    • one year ago
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    So I get the satisfaction of telling them that the error is in fact a LOT MORE THAN 0.6!!!

  16. anonymous
    • one year ago
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    Given that the curve shows properties of a sin wave, with 4 peaks over the interval 1 to 3, and has an amplitude of less than .2 .. Is there a better way to approximate the error?

  17. anonymous
    • one year ago
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    it appears that both the amplitude and the wavelength are changing

  18. anonymous
    • one year ago
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    Im going to ignore that they messed up the integral equation, and assume they meant from 1 to 4, as per the written portion.

  19. dan815
    • one year ago
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    http://prntscr.com/7tszig

  20. anonymous
    • one year ago
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    Thanks dan, that will work.. just integrate the min and max, then calc the difference. // maybe 1/2 the difference.

  21. dan815
    • one year ago
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    yeah they just want you to show its less than 0.6

  22. anonymous
    • one year ago
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    No integration required folks. The width of the area is 2 units (x from 1 to 3). The max height of the area is \[2+0.05(3)\sin \left( 3 \times 3^{2}\right)\] The min height of the area is \[2-0.05(3)\sin \left( 3 \times 3^{2}\right)\]Calculate the the difference between the two areas to show the error is <0.6

  23. anonymous
    • one year ago
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    Thats brilliant ospreytriple .. I like that. that's a real solution.. only problem is, they 're asking to "explain how the plot tells you ....". So I think I'm supposed to use the image.

  24. anonymous
    • one year ago
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    I think they want me to say. The plot is saying there is a region of about + / - 0.2 that is centered on 2. And if we measure that area from x=1 to x=4, (Which is what I think they actually want me to measure, and the range of 1 to 3 was an error.. then we get.. width = 3 * 0.2 =0.6.

  25. anonymous
    • one year ago
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    width =3 height = 0.2 width * height = 0.6

  26. anonymous
    • one year ago
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    That might work as an estimate. You came up with 0.2 by examining the graph?

  27. anonymous
    • one year ago
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    yes 0;2 by looking at the plot

  28. anonymous
    • one year ago
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    I think I would write it as +/- 0.6

  29. anonymous
    • one year ago
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    If that satisfies your teacher, then fine. If you need more precision, you can use the function as I showed above.

  30. anonymous
    • one year ago
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    Thanks, I like that option too. I might include it.

  31. anonymous
    • one year ago
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    If I was grading your work, I would accept your explanation form examining the graph. But every teacher is different.

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