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

1. anonymous

2. dan815

yep 6 is right

3. anonymous

Lol, you think I should just tell them they are correct and move on?

4. dan815
5. dan815

from 1 to 3, the most that curve is fluctuating* is between a change in range of 0.5

6. anonymous

I need a good way to say that the variance around y=2 equates to no more than 0.6 in error.

7. dan815

what a second that doesnt make sense actually, the area should be 4 not 6

8. dan815

if it was integral from 0 to 3 then it would be 6

9. anonymous

its in the interval 1 to 3, so it's about 3*2

10. Empty

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

no? oh yeah..

12. anonymous

1 to 3 would be 2...

13. anonymous

damn them... lol

14. Empty

Yeahhhhhh this is bunk yo

15. anonymous

So I get the satisfaction of telling them that the error is in fact a LOT MORE THAN 0.6!!!

16. 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?

17. anonymous

it appears that both the amplitude and the wavelength are changing

18. 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.

19. dan815
20. anonymous

Thanks dan, that will work.. just integrate the min and max, then calc the difference. // maybe 1/2 the difference.

21. dan815

yeah they just want you to show its less than 0.6

22. 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

23. 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.

24. 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.

25. anonymous

width =3 height = 0.2 width * height = 0.6

26. anonymous

That might work as an estimate. You came up with 0.2 by examining the graph?

27. anonymous

yes 0;2 by looking at the plot

28. anonymous

I think I would write it as +/- 0.6

29. anonymous

If that satisfies your teacher, then fine. If you need more precision, you can use the function as I showed above.

30. anonymous

Thanks, I like that option too. I might include it.

31. anonymous

If I was grading your work, I would accept your explanation form examining the graph. But every teacher is different.