I need some help estimating the integral from -3 to 3 on this graph: https://instruct.math.lsa.umich.edu/webwork2_course_files/ma115-173-w11/tmp/gif/mcdotreb-4414-setChap5Sec3prob7image1.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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I need some help estimating the integral from -3 to 3 on this graph: https://instruct.math.lsa.umich.edu/webwork2_course_files/ma115-173-w11/tmp/gif/mcdotreb-4414-setChap5Sec3prob7image1.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!

Mathematics
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the integral from -3 to 3 on that graph is the area of that graph from -3 to 3
So I need to find the area under the curve between the curve and the x-axis. But how, when there is not concrete numbers on the graph to work with?
You need to be careful: the 'area' under the curve when it is in negative y-territory is negative.

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Right, and I add the absolute value of both pieces because they both are in negative territory.
Have you heard of Simpson's Rule?
No
Okay, 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).
We don't have any other instructions in this section of my math book except to count squares, do the left-hand and right-hand estimation, and use the intergral (but not the antidrevative).
I would: (1) estimate the area under the curve *above* the x-axis (2) estimate the area under the curve *below* the x-axis Subtract (2) from (1): that is, \[\int\limits_{}{} \approx A_1-A_2\]
To estimate A1, I tried to count the rectangles, but I could't make it work
Why couldn't you make it work?
There 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.
I get roughly 4.25 above the x-axis. This is an approximation.
You use all the squares the graph touches, even if it doesn't make a square?
Yeah, 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.
What's simpsons rule (the gist of it...I have looked it up on Wikipedia just a moment ago)
I just estimate what fraction of a square is under the curve if the graph cuts it. That's all.
\[\int\limits_{a}^{b}f(x)dx \approx \frac{b-a}{6}\left[ f(a)+4f(\frac{a+b}{2})+f(b) \right]\]
So four square that get mostly touched by the curve, plus the little bit in the corner?
This is the simpson's rule? Thanks
i got 3.5
what was your method?
dindatc probably took his/her estimations with more refinement.
I 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.
sorry i calculate it mistakenly, i use the (2/3) * base * height and i got 5
in the -x area, I got approx. maybe 2.5, looking at the graph. What is wrong with this estimation?
I 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.
I get, overall, as a rough estimation, -3.25 as the integral value from x= -4 to 4.
1 Attachment
I'm going to estimate using Simpson's Rule. You can use it to compare with your rectangular estimation.
okay
thanks 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
i am really interested in seeing how simpson's rule comes out
2/3 is from the formula the formula is (2/3) x base x height which 2.5? both area have 2.5
Yes, where do you estimate 2.5 for both
What rule are you using...I just understood the base part now!
the left side has a base of 2.5, and the right has a base of 4
on the left one, the graph intersect x-axis 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
okay, I think I am starting to see this...how did you know how to do this? Is it a certain rule?
my teacher taught me that formula, he said it's a faster way to estimate the area of a parabola
(2/3)*base*height?
yes
if you could find the equation of that graph, i think you could get a better estimation
I wonder why they didn't teach us that...would save much frustration!
you would approximate that in the -x area, the height is 1 or more than that?
i approximate the height is 1
I think it is more, with that small piece at the top of the box. How did you get one?
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I had to finish something else.
i chose 1 to make it easier to calculate
I estimated -3.25 using rough estimation with rectangles (eyeballing) and this calculation doesn't disagree too much, given it's again, an estimation.
but 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
for some reason, the computer won't take either method's answers. :(
I understand what you are doing dindatc, and I looked at your attachment, lokisan. I think I understand it, based on the rules...
@lokisan why you also calculate the A0 and A3?
Sorry, ilovemath, I can't offer any more answers. This is the problem with online submission...I hate it.
Because you have to find the area under the curve, don't you?
but isn't interval from -3 to 3?
Yes, from the interval -3 to 3. I have used up all my tries but nothing has worked.
Oh, yes...ummm...I'm tired...
You can still take the A_2 estimation. Just have to calculate the estimation fro -3 to 0.
Thanks 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.
I mean, -1 to 3 is okay. Have to sort out -3 to -1.
You're welcome.
if it's only A1 and A2, the area is 14/3
you're welcome
peace
I get -4.9333...
Area from -3 to -1 is approx 1.73, and the area from -1 to 3 is approx -6.67
yep, our answer is almost the same
Phew...
no dice. computer will not accept the approximation, which I believe will be 8.4? I don't know what to tell you...
Where'd you get 8.4?
The integral is 1.73-6.67=-4.94
do you have to add them, since area must be positive?
no, the area under the x-axis is negative
ohhhhhhhhhhhhhhhhhh. estimation proved correct. sorry
If you're looking for area, then you add the magnitudes, but if it's the integral, that part *under* the x-axis will be negative.
seeee
so, the integral can be a negative. i use the estimations and subtract them because they are on different sides of the x axis
All the 'area' under the x-axis 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 x-axis, the function values are NEGATIVE, so the height is NEGATIVE which means \[ \delta x \times f(x)\]will be negative also
okay
but area will be the absolute value of this, right?
Yes.
the result -4.9 means the area is 4.9 because an area will always positive
If 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 x-axis, 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.
There's a difference between *integral* and *are*.
*area*.
thanks
np
\[f(x)\approx\frac{1}{7}(x-3)(x+1)(x+3.5)\quad,\quad\int\limits_{-3}^{3}f(x)dx\approx -5\frac{1}{7}\]
so this what my equation would look like if I wanted to create one to make estimation easier?
I have to leave now...I'm *very* tired. You got your answer in the end.
yes, and thanks very much!
you're welcome.
even if the graph splits through the first box, we still use it to total 7 boxes for the 1/7 (meaning base)?
wait...this doesn't give you the right graph...
oops. nevermind
nope. still wrong answer with the equation. Thanks though

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