anonymous
  • anonymous
Measuring Integrals .. Need a suggestion on how I should approach this one. Image Coming...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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dan815
  • dan815
yep 6 is right
anonymous
  • anonymous
Lol, you think I should just tell them they are correct and move on?

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dan815
  • dan815
http://prntscr.com/7tsrio
dan815
  • dan815
from 1 to 3, the most that curve is fluctuating* is between a change in range of 0.5
anonymous
  • anonymous
I need a good way to say that the variance around y=2 equates to no more than 0.6 in error.
dan815
  • dan815
what a second that doesnt make sense actually, the area should be 4 not 6
dan815
  • dan815
if it was integral from 0 to 3 then it would be 6
anonymous
  • anonymous
its in the interval 1 to 3, so it's about 3*2
Empty
  • Empty
See the area of that rectangle outside and below is AT LEAST the integral, while the rectangle around the function represents your error
anonymous
  • anonymous
no? oh yeah..
anonymous
  • anonymous
1 to 3 would be 2...
anonymous
  • anonymous
damn them... lol
Empty
  • Empty
Yeahhhhhh this is bunk yo
anonymous
  • anonymous
So I get the satisfaction of telling them that the error is in fact a LOT MORE THAN 0.6!!!
anonymous
  • anonymous
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?
anonymous
  • anonymous
it appears that both the amplitude and the wavelength are changing
anonymous
  • anonymous
Im going to ignore that they messed up the integral equation, and assume they meant from 1 to 4, as per the written portion.
dan815
  • dan815
http://prntscr.com/7tszig
anonymous
  • anonymous
Thanks dan, that will work.. just integrate the min and max, then calc the difference. // maybe 1/2 the difference.
dan815
  • dan815
yeah they just want you to show its less than 0.6
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
width =3 height = 0.2 width * height = 0.6
anonymous
  • anonymous
That might work as an estimate. You came up with 0.2 by examining the graph?
anonymous
  • anonymous
yes 0;2 by looking at the plot
anonymous
  • anonymous
I think I would write it as +/- 0.6
anonymous
  • anonymous
If that satisfies your teacher, then fine. If you need more precision, you can use the function as I showed above.
anonymous
  • anonymous
Thanks, I like that option too. I might include it.
anonymous
  • anonymous
If I was grading your work, I would accept your explanation form examining the graph. But every teacher is different.

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