Definite Integral = Area Under curve(s)

In summary: For example, the area between the graphs of y= x and y= x^2, both from x= 0 to x= 1 is \int_0^1 (x- x^2)dx. That is fine because x^2\ge x for x between 0 and 1. Now suppose that we want to find the area between the graphs of y= x^2- 4 and y= x+ 3, both from x= 0 to x= 3. x^2- 4= (x+ 2)(x- 2) so that graph crosses the x-axis at -
  • #1
karkas
132
1
I know the title can be a bit misleading, yet it's really close to what I want to ask.

Today I came upon an integral at school. The really easy one :[itex]\int_{-1}^{1}x^3 dx[/itex]. Of course, calculating the integral, you get 0. Yet, as far as I know, the way to get the area under the curve is by integrating the function inside the limits of which the area above or under x you want to get.

Yet [itex]f(x)=x^3[/itex] occupies some space from -1 to 0, and the same space from 0 to 1. Why then does the integral consider the one being positive space and the other one being negative, sum them up and calculate the area as zero? This is a question that I've had before, and by looking I saw that there are restrictions or,lack of a better word, modifications that must be applied to the integral when calculating for the area under the curve. Could you please introduce me to them?

Thanks in advance!
 
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  • #2
Well, you should be able to get an answer yourself, just by going back to the definition of the definite integral, and reviewing the argument that indicates how it relates to the area under a curve.



There is a deeper idea that areas in a plane, or volumes in 3-space can be oriented, just like segments in a line... and reflections reverse orientation. When you're thinking of the area under a curve, what you're really thinking is the area between the x-axis and your curve... and you can orient that direction: you can think of it being from the x-axis to the curve. When paired with an orientation on the x-axis (we integrate from left to right), that gives us a two-dimensional orientation on the area we're integrating.

And you'll notice that when the curve negative, the direction 'from' the x-axis 'to' the curve is opposite than when the curve is positive...
 
  • #3
karkas said:
I know the title can be a bit misleading, yet it's really close to what I want to ask.

Today I came upon an integral at school. The really easy one :[itex]\int_{-1}^{1}x^3 dx[/itex]. Of course, calculating the integral, you get 0. Yet, as far as I know, the way to get the area under the curve is by integrating the function inside the limits of which the area above or under x you want to get.
No. If you look closely at the definitions in your textbook, that is for functions with positive value only. With your example, there is no "area under the curve" because you have not specified a lower limit. If f(x)> 0, we can assume y= 0 is the lower limit but if f(x)< 0, that does not make sense.

Yet [itex]f(x)=x^3[/itex] occupies some space from -1 to 0, and the same space from 0 to 1.
No, as a matter of fact, it doesn't! [tex]f(x)= x^3[/tex] is not a geometric object and has nothing to do with "space". If you were to write [tex]y= x^3[/tex], then I would be inclined to interpret it as a question about the graph and have another objection: it is two-dimensiona and so has no "area' which is what you mean "space" here. In order to have area you must have a two-dimensional area. Often we state curves as boundaries of regions: "find the area bounded by [tex]y= x^3[/tex], x= 1, and y= 0". If a problem said "find the area of the region under the graph of [tex]y= x^3[/tex] between x= 0 and x=1, I would have to determine a lower boundary and would probably pick y=0. That really should be said. I think you will find that your textbook does state at the beginning of the introductory chapter on integration that it is talking about regions bounded on the bottom at y= 0 throughout the chapter and so doesn't say it again. But later it will give lower bounds, perhaps y= 0, perhaps other lines or curves, explicitely.

Why then does the integral consider the one being positive space and the other one being negative, sum them up and calculate the area as zero?
It doesn't. There is no such thing as negative area (and "area" is the correct word here, not "space"). The integral as area is one simple application. When you integrate [tex]f(x)= x^3[/tex] from -1 to 1, you are NOT calculating an area. Strictly speaking when you integrate any [tex]\int_a^b f(t)dt[/tex] 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.

This is a question that I've had before, and by looking I saw that there are restrictions or,lack of a better word, modifications that must be applied to the integral when calculating for the area under the curve. Could you please introduce me to them?
Certainly. If you are using an integral to find the area of a region, you first have to specify the entire region. That is why "the area under the graph of [tex]y= x^3[/tex], from x=-1 to x= 1" makes no sense- it has no lower boundary. If you are asked (as I am sure you will be soon!) to find "the area between the graphs of y= f(x) and y= g(x), for x between a and b" (where [tex]f(x)\le g(x)[/tex] for all x) you have an upper boundary, the graph of y= f(x), a lower boundary, the graph of y= g(x), and left and right boundaries, the lines x= a and x= b. In that case the area is given by [tex]\int_a^b (f(x)- g(x))dx[/tex]. The assumption that [tex]f(x)\ge g(x)[/itex] for all x, means that f(x)- g(x) is never negative and so the area is not negative.

A little more "sophisticated" problem would be to find the area between the graphs of y= f(x) and y= g(x) when the graphs cross one or more times. In that case you would have to integrate |f(x)- g(x)| so that, no matter which graph was the upper bound, the integral would be positive.

Thanks in advance!

The integral has many applications. "Area" is one of them, perhaps the simplest. Please don't think that, just because the integral is introduced in terms of area, it must represent some area!
 
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  • #4
Nice quoting the "Thanks in advance" there :)

To the point now. First of all, I understand that there is no such thing as positive or negative space and length, but that was my initial thought on what the integral "thought it should be doing", if you understand what I am trying to say.

I see in my textbook that for functions that also have negative values, we use [itex]\int_{a}^{b}|f(x)|dx[/itex], instead of [itex]\int_{b}^{a}f(x)dx [/itex]. But doesn't that mean we are still going to have [itex]\int_{-1}^{1}|f(x)|dx[/itex] , or are we going to take [itex]\int_{1}^{-1}x^3dx = [x^4/4]^{1}_{-1}=1/2[/itex], or [itex]\int_{1}^{-1}|x^3|dx = |[x^4/4]^{1}_{-1}|=?[/itex](1/2 again??)

Basically, I am sort of confused. You mentioned something about a lower limit. Does this mean I should be calculating the integral (asking for area between curve and x'x axxis) by cutting the initial integral into two separate integrals? Like this? : [itex]\int_{-1}^{1}x^3dx=\int_{-1}^{0}x^3dx + \int_{0}^{1}x^3dx?[/itex]
 
  • #5
karkas said:
To the point now. First of all, I understand that there is no such thing as positive or negative space and length, but that was my initial thought on what the integral "thought it should be doing", if you understand what I am trying to say.

The fact that the integral corresponds *roughly* to area under a curve (taking sign into account) is mostly just a convenient coincidence for calculus teachers. In fact, other equally legitimate ways of graphing a function would NOT produce this. A polar graph of a function or a cartesian graph with non-linear coordinates, such as using a logarithmic scale for the y-axis (common in sound engineering), lead to areas which do NOT correspond to the integral of the function.

If you want the actual area instead of a signed area, you must take the integral of the absolute value of the function you want. The absolute value flips all the negative parts of the graph so that the negative areas magically turn into positive ones.
 
  • #6
1. (i) [tex]v=u+at[/tex]
 

Related to Definite Integral = Area Under curve(s)

1. What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve. It is used to find the total accumulation of a quantity over a given interval.

2. How is a definite integral calculated?

A definite integral is calculated by dividing the area under the curve into small rectangles, finding the area of each rectangle, and then summing these areas together. As the rectangles get smaller and smaller, the sum becomes more accurate and approaches the exact value of the integral.

3. What is the significance of the area under a curve?

The area under a curve represents the total amount of a quantity over a given interval. This can be useful in many applications, such as finding the total distance traveled by an object, the total amount of water in a reservoir, or the total sales of a company over a certain period of time.

4. Can definite integrals be negative?

Yes, definite integrals can be negative. This occurs when the area under the curve is below the x-axis, meaning that the quantity being measured is decreasing over the given interval.

5. What is the relationship between definite integrals and derivatives?

Definite integrals and derivatives are inverse operations. The derivative of a function represents its rate of change at a specific point, while the definite integral represents the total change in the function over a given interval. This means that the definite integral can be used to find the original function when its derivative is known.

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