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

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

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

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

Why couldn't you make it work?

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?

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.

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

okay

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

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?

I had to finish something else.

i chose 1 to make it easier to calculate

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

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.

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...

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

seeee

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

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

*area*.

thanks

np

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.

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

oops. nevermind

nope. still wrong answer with the equation. Thanks though