Question

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

Answer 1

yep 6 is right

Answer 2

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

Answer 3

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Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

See the area of that rectangle outside and below is AT LEAST the integral, while the rectangle around the function represents your error

Answer 10

no? oh yeah..

Answer 11

1 to 3 would be 2...

Answer 12

damn them... lol

Answer 13

Yeahhhhhh this is bunk yo

Answer 14

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

Answer 15

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?

Answer 16

it appears that both the amplitude and the wavelength are changing

Answer 17

Im going to ignore that they messed up the integral equation, and assume they meant from 1 to 4, as per the written portion.

Answer 18

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Answer 19

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

Answer 20

yeah they just want you to show its less than 0.6

Answer 21

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

Answer 22

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.

Answer 23

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.

Answer 24

width =3
height = 0.2
width * height = 0.6

Answer 25

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

Answer 26

yes 0;2 by looking at the plot

Answer 27

I think I would write it as +/- 0.6

Answer 28

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

Answer 29

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

Answer 30

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