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

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

Mathematics
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yep 6 is right
Lol, you think I should just tell them they are correct and move on?

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from 1 to 3, the most that curve is fluctuating* is between a change in range of 0.5
I need a good way to say that the variance around y=2 equates to no more than 0.6 in error.
what a second that doesnt make sense actually, the area should be 4 not 6
if it was integral from 0 to 3 then it would be 6
its in the interval 1 to 3, so it's about 3*2
See the area of that rectangle outside and below is AT LEAST the integral, while the rectangle around the function represents your error
no? oh yeah..
1 to 3 would be 2...
damn them... lol
Yeahhhhhh this is bunk yo
So I get the satisfaction of telling them that the error is in fact a LOT MORE THAN 0.6!!!
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?
it appears that both the amplitude and the wavelength are changing
Im going to ignore that they messed up the integral equation, and assume they meant from 1 to 4, as per the written portion.
http://prntscr.com/7tszig
Thanks dan, that will work.. just integrate the min and max, then calc the difference. // maybe 1/2 the difference.
yeah they just want you to show its less than 0.6
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
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.
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.
width =3 height = 0.2 width * height = 0.6
That might work as an estimate. You came up with 0.2 by examining the graph?
yes 0;2 by looking at the plot
I think I would write it as +/- 0.6
If that satisfies your teacher, then fine. If you need more precision, you can use the function as I showed above.
Thanks, I like that option too. I might include it.
If I was grading your work, I would accept your explanation form examining the graph. But every teacher is different.

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