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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol, you think I should just tell them they are correct and move on?

dan815
 one year ago
Best ResponseYou've already chosen the best response.3from 1 to 3, the most that curve is fluctuating* is between a change in range of 0.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need a good way to say that the variance around y=2 equates to no more than 0.6 in error.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3what a second that doesnt make sense actually, the area should be 4 not 6

dan815
 one year ago
Best ResponseYou've already chosen the best response.3if it was integral from 0 to 3 then it would be 6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its in the interval 1 to 3, so it's about 3*2

Empty
 one year ago
Best ResponseYou've already chosen the best response.0See the area of that rectangle outside and below is AT LEAST the integral, while the rectangle around the function represents your error

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 to 3 would be 2...

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeahhhhhh this is bunk yo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I get the satisfaction of telling them that the error is in fact a LOT MORE THAN 0.6!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Given 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it appears that both the amplitude and the wavelength are changing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im going to ignore that they messed up the integral equation, and assume they meant from 1 to 4, as per the written portion.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks dan, that will work.. just integrate the min and max, then calc the difference. // maybe 1/2 the difference.

dan815
 one year ago
Best ResponseYou've already chosen the best response.3yeah they just want you to show its less than 0.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No 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 \[20.05(3)\sin \left( 3 \times 3^{2}\right)\]Calculate the the difference between the two areas to show the error is <0.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0width =3 height = 0.2 width * height = 0.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That might work as an estimate. You came up with 0.2 by examining the graph?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 0;2 by looking at the plot

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I would write it as +/ 0.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If that satisfies your teacher, then fine. If you need more precision, you can use the function as I showed above.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, I like that option too. I might include it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I was grading your work, I would accept your explanation form examining the graph. But every teacher is different.
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