# Integrals and negative area

1. Jan 17, 2010

### LucasGB

Sometimes, integrals can take up negative values. In those cases, how can I assume they represent the area under the curve, since areas can never be negative?

PS.: They not only take up negative values when f(x)<0, but also when f(x)>0 and the direction of integration is towards the negative direction.

2. Jan 17, 2010

### cronxeh

Last edited by a moderator: Apr 24, 2017
3. Jan 17, 2010

### sutupidmath

In general a Riemann integral does not give us the area under a curve. However, if you wat to calculate the area under a curve, then you must take the sign of the function into considertation. i.e. instead of integrating simply f(x) over an interval I, you should integrate |f(x)| over that interval.

say f(x)>0 in [a,b], and f(x)<0 in [b,c], and you want to find the area from a to c. Then you should not integrate simply f(x) over [a,c] to find the total area, since the areas are going to subtract instead of adding. Instead, if you integrate |f(x)| over [a,c] you notice that

integ(a to c) |f(x)|dx =integ(a,b) f(x)dx-integ(b,c)f(x)dx=Total Area

4. Jan 17, 2010

### LucasGB

All right, thank you both for your help. But a derivative is always the inclination of a tangent, and if the integral is not always the area under a curve, then what is the integral?

5. Jan 17, 2010

### mathman

The integral is "area". However when the curve is below the x axis, the area is counted as negative.

6. Jan 17, 2010

### LucasGB

That's what I thought, but user HallsofIvy, in the thread referenced above, states that "There is no such thing as negative area. The integral as area is one simple application. When you integrate you are NOT calculating an area. Strictly speaking when you integrate you are NOT calculating an area unless that is the particular application you are using the integral for. The integral of a function, from a to b, say, is a number. What that number means depends on the application."

7. Jan 17, 2010

### cronxeh

You can find area between 2 curves, volume, probability, determine if series converges, lenght of an arc, find mechanical work done in physics, concentration changes in chemistry, etc. It pretty much all depends on what the function itself is, in what space the function is

8. Jan 17, 2010

### LucasGB

Yes, I understand that. But even when you're doing all that, it is not incorrect to say that you are also calculating the area under a certain curve, and that area also happens to be a volume, a probability, a length, etc. I think this is just a semantics issue, I think I understand it now.

9. Jan 17, 2010

### cronxeh

I think you will like http://www.intmath.com/Applications-integration/Applications-integrals-intro.php" [Broken]

When you are interested in an area, you are adding positive values, and substracting negative values. So for something like y=5-x, if you integrate from 0 to 5 you get an area of 25/2, but if you integrate from 0 to 10 you are also substracting the negative triangle which is from 5 to 10.. and you get an area of 0. If you consider that calculus is study of change, its like saying my head is in the oven and my feet are an iceblock.. my average temperature is normal.

Last edited by a moderator: May 4, 2017
10. Jan 18, 2010

### HallsofIvy

You can't. "Area under the curve" is a simple way of thinking about some simple integrals. It is not all there is to integrals and, sometimes, if you want to maintain that simplistic representation, you have to do strange things thinking of "area below the x- axis" as "negative area".

Yes, same thing. That is simply an attempt to maintain a simplistic "representation" as "area", that doesn't really work in this situation.

11. Jan 18, 2010

### wofsy

Your two examples have different explanations.

The integral of a function over a finite interval can be thought of as an average. The Riemann sum,

sum over i (b a)/n f(xi) is the length of the interval,)b-a, times the average of the values f(xi).

So 1/b-a times the Riemann sum is the average and in the limit the integral is the limits of the averages. In a way 1/b-a times the integral is the average value of f. It is the same average you would get if you could take an infinitely large random sample of values of f and take the average of the sample.

If the values of f are negative, then the average is negative. The remarkable thing is that the area under the curve when f is positive can be thought of as this average times the length of the interval.

But when f is negative, the integral can be thought of as the negative of the area.

When f is mixed positive and negative then the integral becomes a difference of two areas -
but it is still the average value.

the original motivation for the Riemann integral, came from Physics. If one wants to know the total amount of a Physical quantity one takes its average value in small areas or volumes and adds them up. For instance if you want to know how much fluid is flowing across a surface you divide the surface up into small squares, multiply the area of each square by the average amount of fluid per unit area crossing that square and add over all of the squares.

In practice though you may not know the average amount of fluid per unit area in each square and would like to just take any value and hope that if the squares are really small that you get a good approximation. This approximation is a Riemann sum.

In my mind, the idea of averages of physical quantities as the idea behind Riemann sums is key and areas secondary. The interesting question you are really asking is when does the average times the length of the interval equal the area under the curve?

2. When one integrates from right to left rather than from left to right you are not multiplying by the length of the inteval (b - a)/n but by its negative. So in this case you really don't have a true average but its negative.

12. Jan 18, 2010

### LucasGB

Thank you all for your replies, but one question still remains. Geometrically, a derivative of a function can always be thought of as the inclination of a tangent to the curve defined by that function. Always. Is there an analogous to integrals? Geometrically, what IS an integral?

13. Jan 18, 2010

### elibj123

Mathematically, or should I say, in calculus, integration is a weighted summation of infinite number of values spreaded continuously on some "interval" (as opposed to discrete summation where the you have a countable set of values), with the weighting factor being a differential of a variable of integration (which acts as an indexer of the summation).

In analytic geomtry, an integral may be used as a tool to calculate an area, or volume, or even length, but you cannot restrict yourself to identify integration with calculating an area. When I perform a Fourier Transformation, or Laplace Transformation, or just calculate the fourier series coefficient of a function, I don't assosciate it with anything from geomtry.

14. Jan 18, 2010

### wofsy

You were given the answer. Geometrically,the integral is a signed area. The area can be signed if you function has negative values or if you integrate from right to left.

I was trying to emphasize that this is a restricted idea that does not apply generally and the idea of averages is what is really useful.

A derivative is a special geometrical idea. The integral less so. Wild functions that are not always differentiable can still be integrated. try finding the area under the Devil's Staircase.

15. Jan 19, 2010

### Tac-Tics

The integral is not defined by areas. It's defined as a limit of a Riemann sum. To understand where this "signed area" comes from, you need to understand how Riemann sums work.

Essentially, you break the interval of integration into a billion tiny intervals of length dx. Then, the integral is approximately equal to f(a) dx + f(a + dx) dx + f(a + 2dx) dx + f(a + 3dx) dx + ... + f(b - dx) dx + f(b) dx. The actual value of the integral is the limit of this sum as the length, dx, shrinks to zero (and the number of terms in the sum approaches infinity).

Take a single term from a Riemann sum, say f(x) dx. Look familiar? It's the area of a rectangle with a height of f(x) and a width of dx. An naturally, as you take the limit of the sum, the integral keeps that property, and the result is the integral is closely related to the area.

There's a caveat, though! The area formula, length x width, only works when both the length and width greater or equal to zero! That's not necessarily the case for an arbitrary function f. If f(x) = -x^2 - 1, then every value of f(x) is negative. So it's NOT the area when f(x) < 0.

It is still, however, closely related. If f(x) < 0, then f(x) dx is going to be the opposite of the area. You can simply flip the sign and get the correct answer. If you integrate the ABSOLUTE VALUE of f(x), you get the total area between f and the x axis.

Also related, by convention, the bounds of the integral determine the sign. In other words, the integral of f from a to b is the OPPOSITE of the integral of f from b to a. In the Riemann sums, basically, the dx's can be negative, too.

It doesn't make a lot of sense from a geometric standpoint, but algebraically, it makes a lot of sense. It adds tons of symmetry to our equations and the math all works out. But you have to live with the fact that the integral is not exactly a measure area. The two concepts are, however, intimately related.

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