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Definite Integral = Area Under curve(s)

  1. Feb 16, 2009 #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!
  2. jcsd
  3. Feb 16, 2009 #2


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    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....
  4. Feb 16, 2009 #3


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    No. If you look closely at the definitions in your text book, 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.

    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 text book 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.

    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.

    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.

    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!
    Last edited by a moderator: Feb 16, 2009
  5. Feb 16, 2009 #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 doesnt 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]
  6. Feb 16, 2009 #5
    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.
  7. Feb 16, 2009 #6
    1. (i) [tex]v=u+at[/tex]
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