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anonymous
 5 years ago
I need some help estimating the integral from 3 to 3 on this graph:
https://instruct.math.lsa.umich.edu/webwork2_course_files/ma115173w11/tmp/gif/mcdotreb4414setChap5Sec3prob7image1.png
How do I estimate the integral with only a graph? I have tried rectangles, but I am not getting the right answer. Please Help!
anonymous
 5 years ago
I need some help estimating the integral from 3 to 3 on this graph: https://instruct.math.lsa.umich.edu/webwork2_course_files/ma115173w11/tmp/gif/mcdotreb4414setChap5Sec3prob7image1.png How do I estimate the integral with only a graph? I have tried rectangles, but I am not getting the right answer. Please Help!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the integral from 3 to 3 on that graph is the area of that graph from 3 to 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I need to find the area under the curve between the curve and the xaxis. But how, when there is not concrete numbers on the graph to work with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to be careful: the 'area' under the curve when it is in negative yterritory is negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, and I add the absolute value of both pieces because they both are in negative territory.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you heard of Simpson's Rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so it sounds like they only want you estimating with rectangles to get you used to the Riemann sum (which is essentially what this kind of integration is).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We don't have any other instructions in this section of my math book except to count squares, do the lefthand and righthand estimation, and use the intergral (but not the antidrevative).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would: (1) estimate the area under the curve *above* the xaxis (2) estimate the area under the curve *below* the xaxis Subtract (2) from (1): that is, \[\int\limits_{}{} \approx A_1A_2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To estimate A1, I tried to count the rectangles, but I could't make it work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why couldn't you make it work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is one full square with and area of one, but then there are two peices of a square that can't be made to fit together to make another square. So the are for 3 to 0 is 1. something, I think.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I get roughly 4.25 above the xaxis. This is an approximation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You use all the squares the graph touches, even if it doesn't make a square?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, you don't have to be perfect. There isn't enough information. That's why I asked if you'd heard of Simpson's Rule because you could get this done and dusted in a couple of minutes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's simpsons rule (the gist of it...I have looked it up on Wikipedia just a moment ago)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just estimate what fraction of a square is under the curve if the graph cuts it. That's all.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{a}^{b}f(x)dx \approx \frac{ba}{6}\left[ f(a)+4f(\frac{a+b}{2})+f(b) \right]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So four square that get mostly touched by the curve, plus the little bit in the corner?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the simpson's rule? Thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what was your method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dindatc probably took his/her estimations with more refinement.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the main point of the exercise is to teach you the geometrical motivation for what is going on, and to realise the difference between 'integral' and 'area'  i.e. the integral in negative regions is negative, whereas area is always positive. If you haven't used Simpson's Rule in class, stick to rectangular estimation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i calculate it mistakenly, i use the (2/3) * base * height and i got 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the x area, I got approx. maybe 2.5, looking at the graph. What is wrong with this estimation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know it is wrong, but I don't see how to do the rectangle estimation, I suppose. This problem doesn't come with an equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I get, overall, as a rough estimation, 3.25 as the integral value from x= 4 to 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to estimate using Simpson's Rule. You can use it to compare with your rectangular estimation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks dindatc. I am looking at the graph right now...where do you get 2.5 from? Below the graph, I see where you got 4 from, but not 2/3 or 2.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am really interested in seeing how simpson's rule comes out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02/3 is from the formula the formula is (2/3) x base x height which 2.5? both area have 2.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, where do you estimate 2.5 for both

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What rule are you using...I just understood the base part now!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the left side has a base of 2.5, and the right has a base of 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on the left one, the graph intersect xaxis on x=1 and x = 3.5(approximately) so the base would be 1(3.5) = 2.5 on the right one, 2.5 is the height of that shape, i marked it with green line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, I think I am starting to see this...how did you know how to do this? Is it a certain rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my teacher taught me that formula, he said it's a faster way to estimate the area of a parabola

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you could find the equation of that graph, i think you could get a better estimation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wonder why they didn't teach us that...would save much frustration!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you would approximate that in the x area, the height is 1 or more than that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i approximate the height is 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is more, with that small piece at the top of the box. How did you get one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I had to finish something else.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i chose 1 to make it easier to calculate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I estimated 3.25 using rough estimation with rectangles (eyeballing) and this calculation doesn't disagree too much, given it's again, an estimation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the actual height is more than one, so the differences of the two areas will be smaller, therefore the answer will be smaller than 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for some reason, the computer won't take either method's answers. :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understand what you are doing dindatc, and I looked at your attachment, lokisan. I think I understand it, based on the rules...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@lokisan why you also calculate the A0 and A3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, ilovemath, I can't offer any more answers. This is the problem with online submission...I hate it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because you have to find the area under the curve, don't you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but isn't interval from 3 to 3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, from the interval 3 to 3. I have used up all my tries but nothing has worked.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, yes...ummm...I'm tired...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can still take the A_2 estimation. Just have to calculate the estimation fro 3 to 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks so much for helping me, Lokisan and dindatc! I will continue to work on it, and bring it to the mathlab to see what they say. I wish all we had was written homework.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean, 1 to 3 is okay. Have to sort out 3 to 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it's only A1 and A2, the area is 14/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Area from 3 to 1 is approx 1.73, and the area from 1 to 3 is approx 6.67

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep, our answer is almost the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no dice. computer will not accept the approximation, which I believe will be 8.4? I don't know what to tell you...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral is 1.736.67=4.94

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have to add them, since area must be positive?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, the area under the xaxis is negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhhhhhhhhhhhhhhh. estimation proved correct. sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you're looking for area, then you add the magnitudes, but if it's the integral, that part *under* the xaxis will be negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, the integral can be a negative. i use the estimations and subtract them because they are on different sides of the x axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All the 'area' under the xaxis is negative. This comes from the definition of the integral You have to remember that it's taking function values for the 'height' of the rectangle. When the curve is UNDER the xaxis, the function values are NEGATIVE, so the height is NEGATIVE which means \[ \delta x \times f(x)\]will be negative also

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but area will be the absolute value of this, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the result 4.9 means the area is 4.9 because an area will always positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it asks for area, that is positive. So in that instance, you'd have to be careful with a function and not just integrate from one end to the other if part of the function hits negative territory. You'd have to integrate up to the point where the function cuts the xaxis, record the value, keep going during the interval where the integral is negative, take note of that value, and then continue where it's positive.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a difference between *integral* and *are*.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)\approx\frac{1}{7}(x3)(x+1)(x+3.5)\quad,\quad\int\limits_{3}^{3}f(x)dx\approx 5\frac{1}{7}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this what my equation would look like if I wanted to create one to make estimation easier?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to leave now...I'm *very* tired. You got your answer in the end.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, and thanks very much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even if the graph splits through the first box, we still use it to total 7 boxes for the 1/7 (meaning base)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait...this doesn't give you the right graph...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope. still wrong answer with the equation. Thanks though
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