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/sci/ - Science & Math

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5125734 No.5125734 [Reply] [Original]

I have one question to ask thee:

If I calculate the two areas shown in the picture, won't I get at least one negative answer? Because when I do the calculations I get two positive answers, even though both spaces are below the x-axis. Why is that?

>> No.5125740

It doesn't make much sense to say something has negative area.

>> No.5125743

If you integrate it from -1 to 0 you will get a negative value, which is why you're supposed to use the absolute values of what you get.

>> No.5125747

>>5125743
Oh nevermind, didn't see y=-1

>> No.5125767

>>5125734

You should get one positive and one negative answer, you must be messing up somewhere.

>> No.5125848

How are you calculating these areas?
If you're just integrating from -1 to 0 and then 0 to 2, you're way off. Integration measures the area between the curve and the x axis.

>> No.5126033

>>5125740
> It doesn't make much sense to say something has negative area.

Yeah, actually it does. If I have 40 acres of land and I sell some of it so that I only have 30 acres left, then I have experienced a net gain of -10 acres of land.

There's a case where a negative area makes perfect sense.

You are correct that, geometrically, an area is never negative. However, this isn't geometry. This is analytic geometry, which brings the full power of algebra and the real number system (including negatives) to the measurement of geometric objects.

When a signed area is negative, then it means that we're looking at the area of the bottomside of the surface instead of the topside. Imagine a rectangle that's 5 tall by -3 wide. It has a signed area of -15. The fact that the signed area is negative means that two of the edges have traded places, effectively causing us to see the bottomside of the rectangle as if it had been flipped over in 3d space. That's what the sign of the signed area means: which face (topside or bottomside) you're measuring the area of.

>> No.5126054

>>5126033

While you're technically right, it doesn't make sense in the context of OP's problem.

>> No.5126086

Don't know how you're approaching this, but I just get the feeling that you're in an elementary calculus course, and if you're calculating areas by Riemman integrals, you're measuring the area between the x-axis and the curve below it in this case. Both areas from -1 to 0 and from 0 to 1 would be negative.

>> No.5126090

>>5126054
>While you're technically right, it doesn't make sense in the context of OP's problem.

Oh, of course it makes sense in OP's problem!

Let's say you're integrating from a to b. Then you flip it so that you're integrating from b to a. Well, that's like flipping the whole x axis over, so you're now measuring the area of the backside -- so of course it's going to change the sign of the result.

In general, the integral yields the signed area from the x axis up or down to the function line. If you go down instead of up then you're going to get the opposite sign.

This goes straight to the heart of the geometric interpretation of the integral. The integral does not yield the area!!! Instead, the integral yields the SIGNED area, where the sign itself has a meaningful geometric interpretation. If your instructor is just ignoring the sign of the result, then he's doing you a big disservice.

>> No.5126643

Let me clarify: there are two spaces that I'm supposed to calculate the areas of. The two functions are both below the x-axis and together they limit two spaces.

I know that the only reason one takes -(integral) is because an area under the x-axis is usually negative, so it's a form of compensation. However, when I calculate the two areas without that compensation, I get two positive values. Those values together (added) are the same as when I use Mathematica to calculate the integral between -1 to 2 with the absolute value of g(x)-y.

>> No.5126656

>>5125767

Shouldn't both be negative, since both areas are below the x-axis?

>> No.5126672

OP You're supposed to integrate with respect to y in this situation. y terms lie on its axis but your x terms if you chose to integrate along it lie on y=-1

>> No.5126778
File: 99 KB, 741x597, full_retard2.jpg [View same] [iqdb] [saucenao] [google]
5126778

>>5126643
>2012
>going full Romney

Here is how you do it. Try and think of it conceptually, insead of like a Romney. Math is your fucking friend!

1) Add +1 to your y(x). This isn't changing the shape of your function, just shifting it so that Region 1 is above the x axis, and region 2 is below the x axis. The areas are still the same though, when you do a basic shift like this, right. You apply the shit just so you intergral concetually works out. Make sense?

2) Find the integral of your new fuction from x=-1 to x=0, you should get a postive answer. This is region 1.

3) Find the integral of your new function from 0 to 2, this should give you a negative answer, but for the sake of argumnet, you will just call it positive. This will be region 2.

Any questions? I hope that made sense.

>> No.5126774

>>5126672

I am doing just that, but both results come out positive.

If anyone could make the calculation and show me how you got the values negative, I'll owe you one. I have the right answer in front of me, but I have no idea why the two areas don't turn up negative.

>> No.5126788
File: 48 KB, 600x400, romney-pinocchio-romnocchio.jpg [View same] [iqdb] [saucenao] [google]
5126788

>>5126672
>integrate with respect to y

>> No.5126791

>>5126778
\thread

>> No.5126802
File: 20 KB, 300x480, 258Troll_spray.jpg [View same] [iqdb] [saucenao] [google]
5126802

>>5126672

>> No.5126888

>>5126778

I agree that it should give me a negative answer, but it does not. Since y is the upper function in the second area, I do the following subtraction: y-g(x), inside the integral. This makes the result positive, but I fail to see why I shouldn't do this subtraction when it is what I did with the first area.

>> No.5127092
File: 110 KB, 320x480, 01e8bee6fb3dfe147857728210073413.jpg [View same] [iqdb] [saucenao] [google]
5127092

>>5126888
WTF are you doing? Calculus will give you a negative answer for the second region.

Post your steps, or someshit, so we can see where you are failing so fucking hard. I have a feeling you can''t even do basic integration.

>> No.5127103
File: 38 KB, 400x399, WTFAMIREADING.jpg [View same] [iqdb] [saucenao] [google]
5127103

>>5126888
>upperfucntion
>subtracting

STOP BEING A FUCKING FAGOOT!

Just add +1 to y(x) and itegrate form 0 to 2.

It gives you a negative fucking answer, HOW FUCKING HARD IS THAT?

>> No.5127176
File: 31 KB, 1261x506, Explanation.png [View same] [iqdb] [saucenao] [google]
5127176

>>5127092
>>5127103

No need to get irate.

>> No.5127250

Okay... Let's see.

You have f(x) = -1 and g(x) = x^3 - x^2 - 2x - 1.
You want to find the shaded area in the picture.

We will be needing the area of f(x) = -1 to be able to find our desired area.
I think it's pretty safe to say from geometric common sense that the area of f(x) from -1 to 0 will be -1 and from 0 to 1 will be -1. We're really only interested in magnitude of these areas, so let's go ahead and disregard the negative signs.

Now, integrate g(x) from -1 to 0 and it'll give you -7/12. This area corresponds to the two segments that aren't shaded below the x-axis and above y = -1 from -1 to 0. Once again, magnitude is what we're looking for, so let's omit the negative.

We know that the integral of f(x) from -1 to 0 was 1. Subtracting (7/12) from 1 and we have (5/12). Now, do we assign this a negative value or leave it as is? Well, the area we've just found is above your desired axis of integration (I don't know how else to explain it) so we'll leave it positive as is convention.

From 0 to 1, the integral of g(x) is -25/12. Omit negatives. Notice that (25/12) is larger than 1, the area of f(x) in the interval from 0 to 1, and therefore must extend further down past y = -1 in the interval. The area you're looking for in this interval is simply the difference between 25/12 and 1, 13/12. Since this is below the axis you're using as a standard, we'll assign this a negative value.

The net area from -1 to 1 is the sum of the two areas, (-13/12) + (5/12) and gives us with a net area of (-7/12). This makes sense, as the curve goes steeper down into the negative numbers from 0 to 1 than it goes up from -1 to 0.

That's how I would approach the problem.

>> No.5127265

OP i didn't read the rest of the thread, but if you integrate -1 --> 0, and subtract 0 --> 2 you receive the total displacement of the function from -1 --> 2. The total area is the absolute value of -1 --> 0 + absolute value of 0 --> 2.

Just apply yourself and try to use calculus to receive meaningful information

>> No.5127335

>>5127250

>above your desired axis of integration

This is more than enough, thank you.

The second area should be between 0 and 2, but the crux for me is why this value suddenly becomes negative. I know that it should, but it doesn't with the technique of integration I've been taught (integral of upper function minus the integral of the lower function). Maybe this technique only applies to functions that are above the x-axis?

>> No.5127361

>>5127335

Huh, yeah, could have sworn that was 1 and not a 2.

Well, one observation:

If f(x) = -1 and g(x) x^3 - x^2 - 2x - 1,

from (-1, 0) notice that g(x) is larger than f(x), but the roles are switched for the interval (0,2).

To find the area between two curves, you have to subtract the smallest integral from the biggest.

So, denoting A1 to be the area for (-1,0) and A2 to be the area for (0,2), you'd find A1 by the integral of g(x) - f(x) and A2 by f(x) - g(x).

Also I'm sure the method works if the functions are below x-axis.