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>area between two curves above x-axis is positive
>area between two curves below x-axis is negative
Makes sense
>area between two curves that covers areas above and below the x axis is only positive
Somebody explain this shit. I understand that all areas are non negative or positive, but computing the integral for an area above x-axis is positive, below x-axis is negative, how does the integral follow those rules but still solves areas that expand across both quadrants?
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>negative area
It must suck to live in a developing country with no-tier education
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>>8619381
No, the area between two curves is not affected by where they are on the Cartesian plane. Taking the integral of f(x)-g(x) instead of g(x)-f(x) is the only thing that changes the sign of the integral.
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>>8619394
The integral will come out negative if the area is only below the x-axis.
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>>8619399
No, it won't.
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>>8619427
Note the exact same answer.
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>>8619427
>dat Pixel phone
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>>8619381
The problem is that you should not think of the integral as area.
When you define the integral you are multiplying a distance (usually called delta x) with a "height" that will be the function evaluated at some point. Obviously if the function evaluates to some negative value then that multiplication will yield a negative value, as delta x is always positive.
Now, to make the integral into area then you would have to add some absolute values in the definition but then that would make the fundamental theorem of calculus invalid. That is when it clicked: Hey, the integral should not be at all about area. And if we want area then we should partition a function and then use absolute values outside of the integral, to not mess up our precious theorem.
Basically, learn some calculus and see other applications of the integral other than "lol calculate the area of this shape".
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>>8619438
>you should not think of the integral as area
i think you're wrong
the integral has a close relationship with geometry, and deeper relations are revealed when you generalize the integral to work on chains and forms and whatnot
the geometric aspect should not be disregarded
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>>8619479
>the integral has a close relationship with geometry
It does but this is simply good luck. We can use the integral to find algebra but it is not as simple.
If you have to find the area under a curve in an interval then to succesfully do that you would have to partition that area in uniquely positive and uniquely negative portions and then absolute value the negative portions. It is not so simple. The relationship of the integral to area is not obvious, it is simply good luck. If the integral was algebra then simply computing the integral over the entire interval would yield our area.
And even harsher, as I just said, if you were to slightly redefine the integral so that it did correspond to area, then the fundamental theorem of calculus would be incorrect. So you would have a weaker theory of analysis.
>and deeper relations are revealed when you generalize the integral to work on chains and forms and whatnot
Sure, but the integral is not area.
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>>8619485
>algebra
I meant area every time I said algebra. ffs.
I'm watching Wildberger's latest video and he says algebra like 5 times per minute. Never watch wildberger and post at the same time, kids.
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>>8619490
lol retart
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>>8619545
no u
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Area between functions [math]f(\cdot)[/math] and [math]g(\cdot)[/math] in an interval [math][a,b][/math] is A=[math]\int_a^b (|f(x)-g(x)|)dx[/math]. Note that one or both functions could be the zero function.
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Are those absolute value bars? I can't tell. If so my professor never told us that it was supposed to be the absolute value which led to confusion for me, thanks.
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>>8619599
>If so my professor never told us that it was supposed to be the absolute value which led to confusion for me, thanks.
The professor never tells you because it is impractical. If you work with an absolute value inside the integral then the only way you can properly find the antiderivative through normal methods is if the function is strictly positive or strictly negative (that way you can remove the absolute value).
But if you were taking the antiderivative of the absolute value of a function that went from negative to positive then you could not easily apply theorems. Try it. It is not practical. It is way better to teach the usual method with no absolute values and then teaching you to properly partition functions and then use absolute values after computing the integral.
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>>8619599
>>8619646
To illustrate my point, take a look at the shit you'd have to do if you wanted to use the fundamental theorem of calculus to calculate the area between f(x) = x^2 and g(x) = x using that guy's "method".
Antiderivative: (no result found in terms of standard mathematical functions)
Antiderivative assuming all variables are real: [math] (4 - 9 x^2 + 6 x^3 + 3 (3 - 2 x) x^2 Sign[x] + Sign[1 - x] (-4 + 9 x^2 - 6 x^3 + (3 - 2 x) x^2 Sign[x]))/24 [/math]
top fucking kek. And here's the kicker: Sign is a piecewise function.
Good luck applying "retardo's theorem" in a test.
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>>8619656
>"retardo's theorem"
kek
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>>8619646
>>8619656
Of course you can subdivide the problem to see where f-g has negative or positive sign.
"Retardo's theorem"
Haha. Well you can't say it isn't right. Too bad it is too complex and that you need to always think the integral as the antiderivative.
Kicker: |x^2-x| is continous. That means that it has an antiderivative. You are bad in mathematics. https://www.wolframalpha.com/input/?i=abs(x%5E2-x)
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>>8619838
>Of course you can subdivide the problem to see where f-g has negative or positive sign.
But then that is equivalent to what you would usually do in a calc class. You already have to partition the function to find the area of certain curves.
>Well you can't say it isn't right.
It is right, just very impractical and a little dumb.
> Too bad it is too complex and that you need to always think the integral as the antiderivative.
Hey, don't think of me like that. The fundamental theorem of calculus is the only thing that makes computing integrals viable, at least for theorem proving and for real human beings. Computing it through the riemann sum is crazy long. Even a simple function like f(x)=x would take various steps to find what the limit converges to.
>Kicker: |x^2-x| is continous. That means that it has an antiderivative.
It does, but it is not practical. I even posted what the antiderivative is:
[math] (4 - 9 x^2 + 6 x^3 + 3 (3 - 2 x) x^2 Sign[x] + Sign[1 - x] (-4 + 9 x^2 - 6 x^3 + (3 - 2 x) x^2 Sign[x]))/24 [/math]
I wouldn't want to use the FTC on that shit.
>You are bad in mathematics.
I'm not. I'm really not.
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>>8619381
Pretty sure it's just a consequence of how the equations worked out
The point is you get a number representing area
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>>8619978
>The point is you get a number representing area
The fact that if you get the integral of a function where half of it is under the x axis or left to the y axis you get 0 should be enough to realize: Hey, maybe the integral has jack shit to do about area.
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>>8619381
Hmmm not much of a mathfag but if you do int(F(x) +2) don't you get int(F(x)) + 2x ? Wouldn't that change some shit compared to just int(F(x))?
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