Question

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!

Answer 1

the integral from -3 to 3 on that graph is the area of that graph from -3 to 3

Answer 2

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?

Answer 3

You need to be careful: the 'area' under the curve when it is in negative y-territory is negative.

Answer 4

Right, and I add the absolute value of both pieces because they both are in negative territory.

Answer 5

Have you heard of Simpson's Rule?

Answer 6

No

Answer 7

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).

Answer 8

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).

Answer 9

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, \[\int\limits_{}{} \approx A_1-A_2\]

Answer 10

To estimate A1, I tried to count the rectangles, but I could't make it work

Answer 11

Why couldn't you make it work?

Answer 12

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.

Answer 13

I get roughly 4.25 above the x-axis. This is an approximation.

Answer 14

You use all the squares the graph touches, even if it doesn't make a square?

Answer 15

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.

Answer 16

What's simpsons rule (the gist of it...I have looked it up on Wikipedia just a moment ago)

Answer 17

I just estimate what fraction of a square is under the curve if the graph cuts it. That's all.

Answer 18

\[\int\limits_{a}^{b}f(x)dx \approx \frac{b-a}{6}\left[ f(a)+4f(\frac{a+b}{2})+f(b) \right]\]

Answer 19

So four square that get mostly touched by the curve, plus the little bit in the corner?

Answer 20

This is the simpson's rule? Thanks

Answer 21

i got 3.5

Answer 22

what was your method?

Answer 23

dindatc probably took his/her estimations with more refinement.

Answer 24

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.

Answer 25

sorry i calculate it mistakenly, i use the (2/3) * base * height
and i got 5

Answer 26

in the -x area, I got approx. maybe 2.5, looking at the graph. What is wrong with this estimation?

Answer 27

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.

Answer 28

I get, overall, as a rough estimation, -3.25 as the integral value from x= -4 to 4.

Answer 29

I'm going to estimate using Simpson's Rule. You can use it to compare with your rectangular estimation.

Answer 30

okay

Answer 31

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

Answer 32

i am really interested in seeing how simpson's rule comes out

Answer 33

2/3 is from the formula
the formula is (2/3) x base x height
which 2.5? both area have 2.5

Answer 34

Yes, where do you estimate 2.5 for both

Answer 35

What rule are you using...I just understood the base part now!

Answer 36

the left side has a base of 2.5, and the right has a base of 4

Answer 37

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

Answer 38

okay, I think I am starting to see this...how did you know how to do this? Is it a certain rule?

Answer 39

my teacher taught me that formula, he said it's a faster way to estimate the area of a parabola

Answer 40

(2/3)*base*height?

Answer 41

yes

Answer 42

if you could find the equation of that graph, i think you could get a better estimation

Answer 43

I wonder why they didn't teach us that...would save much frustration!

Answer 44

you would approximate that in the -x area, the height is 1 or more than that?

Answer 45

i approximate the height is 1

Answer 46

I think it is more, with that small piece at the top of the box. How did you get one?

Answer 47

I had to finish something else.

Answer 48

i chose 1 to make it easier to calculate

Answer 49

I estimated -3.25 using rough estimation with rectangles (eyeballing) and this calculation doesn't disagree too much, given it's again, an estimation.

Answer 50

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

Answer 51

for some reason, the computer won't take either method's answers. :(

Answer 52

I understand what you are doing dindatc, and I looked at your attachment, lokisan. I think I understand it, based on the rules...

Answer 53

@lokisan why you also calculate the A0 and A3?

Answer 54

Sorry, ilovemath, I can't offer any more answers. This is the problem with online submission...I hate it.

Answer 55

Because you have to find the area under the curve, don't you?

Answer 56

but isn't interval from -3 to 3?

Answer 57

Yes, from the interval -3 to 3. I have used up all my tries but nothing has worked.

Answer 58

Oh, yes...ummm...I'm tired...

Answer 59

You can still take the A_2 estimation. Just have to calculate the estimation fro -3 to 0.

Answer 60

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.

Answer 61

I mean, -1 to 3 is okay. Have to sort out -3 to -1.

Answer 62

You're welcome.

Answer 63

if it's only A1 and A2, the area is 14/3

Answer 64

you're welcome

Answer 65

peace

Answer 66

I get -4.9333...

Answer 67

Area from -3 to -1 is approx 1.73, and the area from -1 to 3 is approx -6.67

Answer 68

yep, our answer is almost the same

Answer 69

Phew...

Answer 70

no dice. computer will not accept the approximation, which I believe will be 8.4? I don't know what to tell you...

Answer 71

Where'd you get 8.4?

Answer 72

The integral is 1.73-6.67=-4.94

Answer 73

do you have to add them, since area must be positive?

Answer 74

no, the area under the x-axis is negative

Answer 75

ohhhhhhhhhhhhhhhhhh. estimation proved correct. sorry

Answer 76

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.

Answer 77

seeee

Answer 78

so, the integral can be a negative. i use the estimations and subtract them because they are on different sides of the x axis

Answer 79

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

Answer 80

okay

Answer 81

but area will be the absolute value of this, right?

Answer 82

Yes.

Answer 83

the result -4.9 means the area is 4.9 because an area will always positive

Answer 84

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.

Answer 85

There's a difference between *integral* and *are*.

Answer 86

*area*.

Answer 87

thanks

Answer 88

np

Answer 89

\[f(x)\approx\frac{1}{7}(x-3)(x+1)(x+3.5)\quad,\quad\int\limits_{-3}^{3}f(x)dx\approx -5\frac{1}{7}\]

Answer 90

so this what my equation would look like if I wanted to create one to make estimation easier?

Answer 91

I have to leave now...I'm *very* tired. You got your answer in the end.

Answer 92

yes, and thanks very much!

Answer 93

you're welcome.

Answer 94

even if the graph splits through the first box, we still use it to total 7 boxes for the 1/7 (meaning base)?

Answer 95

wait...this doesn't give you the right graph...