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Area of y=x^2 at x=1

  1. May 27, 2009 #1
    My first post here. So be gentle, my math cherry isn't popped yet.

    Currently reading "Essential Mathematics for Economic Analysis" by Sydsaeter and Hammond. I have a question from the book (page 313 if anyone cares).

    Assume [tex]f(x)=x^{2}[/tex]

    Also assume that [tex] F(x)=\frac{1}{3}x^{3}[/tex] is an indefinite inegral of [tex]f(x)[/tex].

    Now I want to calculate the area under f(x) in interval [0,1]. I do this by calculating [F(1)-F(0)] and get 1/3 as answer. The authors then say that 1/3 is a "reasonable" answer. From this comment I assume that the area is only approximated by 1/3.

    Why is this? Why does the F(1)-F(0) not give me exact answer for the area under f(x)=x^2.

    I have two versions as an answer.

    First is that any indefinite integral that we reconstruct from f(x) will always be just an approximation. I also think that in case of linear functions the indefinite integral will give the precise area. Correct?

    Second is that there does exist an indefinite integral of f(x) that gives the perfect true area under f(x)-x^2, but it has more terms involving x that we cannot reconstruct from f(x).
     
  2. jcsd
  3. May 27, 2009 #2
    I would test your answer as follows. Consider the intervals (0,1/2) and (1/2,1) separately and partition the area from 0 to 1 into 3 areas. the area under the curve from 0 to 1/2 the rectangle 1/4 high from 1/2 to 1 and the area from 1/2 to 1 that is above the rectangle and under the curve. If these sum to 1/3 then I would say your answer is probably exact. Another answer would probably be because you need to add an unknown constant to your answer to be more precise with respect to the general equation. Check that out when you determine the area under the second half of the curve.
     
    Last edited: May 27, 2009
  4. May 27, 2009 #3

    epenguin

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    It is an exact answer.:uhh: I think they mean to say, especially in a book for non-practiced users of maths: get into the habit of thinking about what is a reasonable answer, because of your high chances of error the extra guide is essential and is also part of the understanding and numeracy you are learning.

    For example here you might make a slip, If you got 2 that is not a reasonable answer. Do a picture of you function f inside a square of side 1 and you will see 1/2 is not a reasonable answer.
     
  5. May 27, 2009 #4
    After 2 hours of calculating (my brain is a Pentium 1) the area of the third area (the second half of the curve in [0,1]) I did eventually get 8/24 as answer for the area of three regions which you proposed, which is 1/3 indeed. So I probably have the exact answer for the area.

    What do you mean with general equation? The F(x) does not need a constant because we are calculat area from 0 to x by F(x) + C -[F(0) + C], so the constant cancels out and we need not worry about it. Is that what you meant?
     
  6. May 27, 2009 #5
    Hmm, are you implying that if I have an integrable derivative that I will always be able to calculate the area under the graph of basic functions EXACTLY?
     
  7. May 27, 2009 #6
    Have you learned the fundamental theorem of calculus yet? It seems that you're trying to use tools which you don't fully understand. Calculating the exact area under a curve used to a feat which could only be accomplished by geniuses so it's understandable that you have a hard time figuring out how to on your own but if you understand the FTC the task is reduced to a simple computation.

    For this particular problem you can even start with the definition of a definite integral and work it out without the FTC.

    [tex]\int_a^b f(x) \, \text{d}x = \lim_{x \to \infty}{\sum_{i=1}^n{f(x_i) \Delta x}[/tex]
     
  8. May 27, 2009 #7
    After a fashion, yes. The indefinite integral of a single-variable real function f over an interval [a, b] in which the function is positive and integrable is usually defined to be the area bounded by the curves y = 0, y = f(x), x = a, and x = b on the Cartesian graph of the function. This definition is motivated by the fact that the integral is usually the same as trying to measure the area under the curve by using sums of the areas of smaller and smaller rectangles (or other appropriate shapes) that cover the region, and noting that the sums seem to be approaching a fixed quantity (that is, they don't oscillate continuously, and they don't increase or decrease without bound; they are definitely bounded above or below). Thus, there must be some unique number that is the least of those bounds of all possible sums of covering shapes, and that number is called the integral of f over that interval, and it is reasonable to call that unique number the area as well. It is not considered an approximation. :)
    It is common to use the more manageable smaller and smaller rectangles defined over the interval and find the behavior of the sum as the size of the base of the rectangles decreases without bound. This type of sum is called a Riemann sum and the limiting procedure is referred to as Riemann integration. There is more rigour added to this description in any book on basic real analysis or even a good calculus text (Apostol, Spivak, Courant). For example, there are other types of integrals extending or replacing Riemann integration that allow integration of functions that "should have" an integral but don't have a Riemann one, and to support other types of "area" and objects like distributions, usually studied in measure theory/statistics.
    The older method of approximating the area of shapes bounded by smooth curves with summing smaller and smaller shapes of known area was made popular by Archimedes and his method of exhaustion. The Fundamental Theorem of Calculus connected these exhausting methods of calculating area with the simpler method of calculating derivatives.
     
    Last edited: May 27, 2009
  9. May 28, 2009 #8

    I have never seen a definition of definite inegral like your formula. It does not even make sense to me. As x approaches infinity? And how much is n? And isn't the [tex]\Delta x[/tex] way way bigger than dx?
     
  10. May 28, 2009 #9

    cristo

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    There is a typo above, it should read:

    [tex] \int_a^b f(x) \, \text{d}x = \lim_{n \to \infty}{\sum_{i=1}^n{f(x_i) \Delta x}[/tex]
     
  11. May 28, 2009 #10
    Why are we taking every value of x instead of only those values in [a,b]? I can only make the sense of the formula if I rewrite it:

    [tex] \int_{-\infty}^\infty f(x) \, \text{d}x = \lim_{n \to \infty}{\sum_{i=1}^n{f(x_i) \Delta x}[/tex]
     
  12. May 28, 2009 #11

    cristo

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    I'm not sure what you mean; we're not taking "every value of x." Basically, the idea comes from approximating the area under a graph by areas of rectangles. Divide the domain into a number of pieces, and multiply the average value of the function in that piece (f(x_i)) with the length of the piece (\Delta x). This gives the area under the graph for each piece; summing them up gives an approximation of the total area.

    Then, the exact value is found by using infinitely many pieces, or by making the size of each division infinitely small.

    Edit:

    Perhaps this is where your confusion lies. \Delta x is the width of the pieces you divide the interval into. When you then take the limit of the expression above, you are taking infinitely many pieces or, as I mentioned before, making the size of each piece infinitely small. Thus, in this limit \Delta x becomes the infinitesimal dx.
     
    Last edited: May 28, 2009
  13. May 28, 2009 #12
    Yes, I understand that. However, how is it apparent from the right-hand side formula that we are taking rectangles only from interval [a,b]? It is not apparent. So we are taking EVERY rectangle. Therefore we are calculating the improper integral with minus infinity and infinity as boundaries.
     
  14. May 28, 2009 #13

    cristo

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    The interval must be finite, since \Delta x is defined as (b-a)/n, in the simple example of taking areas of rectangles.
     
  15. May 28, 2009 #14
    I thought that

    [tex]\Delta x = x_{i+1} - x_1[/tex]
     
  16. May 28, 2009 #15

    No we're not. Think of it this way. Assume f is a continuous function. The length of [a,b] is given by b-a. When we approximate the area under the curve by rectangles, we split the interval into n even parts (we don't have to, but it makes this explanation simpler), so the width of each rectangle is given by (b-a)/n. Let's assign this value to h. So we have h = (b-a)/n. Now, the end points of our rectangles are a, a + h, a + 2h, a + 3h, ... , a + nh = b.

    We will take the right end point of each rectangle to approximate our area. The right end point of the first rectangle is at x = a + h, for the second rectangle the right end point is at x = a + 2h and remembering we have n rectangles, the right end point of the last rectangle is at x = a + nh = b. So we see that the area of the i'th rectangle is given by [tex] hf(a + ih) [/tex]. We can now write [tex] \int_a^b f(x) dx != h[f(a + h) + f(a + 2h) + f(a + 3h) + ... + f(a + nh)] = \sum_{i = 1}^n h(f(a + ih)) [/tex] where the '!=' is used to mean 'approximately equal to'.

    We know that to approximate the area under the curve more and more precisely we have to take more and more rectangles, which means that the width of each rectangle is shrinking. So we must take the limit as h (the width of each rectangle) approaches 0 of our approximation expression, and so we define [tex] \int_a^b f(x) dx = \lim_{h \rightarrow 0} \sum_{i = 1}^n h(f(x + ih)) [/tex].

    Remember that h = (b-a)/n, implying that n = (b-a)/h. As h tends to 0, n increases beyond all bounds. We will also say that x_i = a + ih (x_i is the right end point of the i'th rectangle) and so we can write [tex] \int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n h(f(x_i)) [/tex], and letting [tex] \Delta x = h [/tex] we can also write [tex] \int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n \Delta x(f(x_i)) [/tex], the same formulation you were arguing about.
     
  17. May 28, 2009 #16

    Remember that (b-a)/n is just the width of each rectangle, and obviously [tex] x_{i + 1} - x_i [/tex] is also the width of each rectangle if you take for the rectangles end points [tex]x = x_{i+1} [/tex] and [tex] x = x_i [/tex], so [tex] (b-a)/n = x_{i+1} - x_i = \Delta x [/tex]
     
  18. May 28, 2009 #17
    JG89, your post is very very clear. I wish all posts on this forum were like yours.

    I get it now. The (b-a)/a is cleverly hidden in [tex]\Delta x[/tex].

    I mean I knew that [tex]\Delta x = x_2 - x_1[/tex]

    But then I did not think it throught that over the whole graph in [a,b]
    [tex]\Delta x = b - a[/tex] which is the width of single rectangle. Then we take n rectangles in that area and get:

    [tex]\Delta x = (b - a) / n[/tex]

    Awesome.
     
  19. May 28, 2009 #18
    Just to see if I grasp it comletely, if we assume

    [tex] \int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n \Delta x(f(x_i)) [/tex],

    then from it follows that [tex]x_0 = a[/tex] right? :shy:
    No, it doesn't "follow" from that. The endpoints of the interval must be given initially and both [itex]\Delta x[/itex] and all [itex]x_i[/itex] from them.
     
    Last edited by a moderator: May 28, 2009
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