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A Quick Question about Integrals

  1. Sep 2, 2012 #1
    Hello, I recently learned about integrals in my high school calculus class. Yes, I know you add 1 to the exponent and divide the coefficient by that number, and the integral of a velocity over time graph gives the total distance traveled, but I have a different question: What is a quick and easy way to define an integral, as if explaining it to a non-mathematically minded individual?

    Derivative of f(x): Rate at which the function is changing at point x
    Integral of f(x): ?
     
  2. jcsd
  3. Sep 2, 2012 #2

    pwsnafu

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    Let ##a## and ##b## be real numbers with ##a<b##. Let ##f : [a,b]\rightarrow \mathbb{R}## be a positive real function. Then ##\int_a^b f(x) \, dx## is the area of the region bounded by 1) the x-axis, 2) the vertical line ##x=a##, 3) the vertical line ##x=b##, and 4) ##f(x)## for ##a \leq x \leq b##.

    Notice this a real number. To obtain a function, you fix one of the terminals and allow the other vary.

    Edit: Just a note: we do not define the integral using area. We interpret the integral this way. But it is sufficient for someone who isn't mathematically minded.
     
    Last edited: Sep 2, 2012
  4. Sep 2, 2012 #3

    Simon Bridge

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    [itex]\int_a^b f(x)dx[/itex] is the area beween the x-axis and y=f(x) between x=a and x=b.

    eg. the area under the v-t graph is the displacement (note: NOT the total distance traveled.)

    if g(x)=f'(x) then [itex]\int g(x)dx = f(x)+c[/itex]
    It is the antiderivative ... i.e. given the equation of the slope of the tangent to a point on a graph, what is the graph?

    In actual fact it is a summation - where the interval of the sum is very small.
    We just use it to find areas.
     
  5. Sep 2, 2012 #4

    Bacle2

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    Sorry to nitpick, Simon, but there are exceptions to ∫f'dx =f ,like that of the Cantor

    function C(x),for which C'(x)=0 a.e., so that ∫C'(x)dx=∫0dx=0 ≠ C(x).
     
  6. Sep 2, 2012 #5

    Simon Bridge

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    I'm sorry - I did not claim that ∫f'dx =f ... please read more carefully. I also did not claim that there were no exceptions or that I was making a definition - complete or otherwise.
    That is not correct - [itex]\int 0dx = c[/itex] ... a constant.

    Note: the cantor function has zero derivative almost everywhere - but not everywhere. The actual complete derivative function will look quite special - which gives you a clue how an antiderivative will work.

    There are functions with zero derivative everywhere ... constant functions.

    Cantor functions are usually considered stuff for advanced students - is that the level that the question is being asked?

    However, having put in the niggle - your challenge is to improve my description in the context of post #1 and relevant to OPs level of understanding :)
     
  7. Sep 2, 2012 #6

    Bacle2

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    Sorry if I misread your post, Simon.

    To the OP:

    You may find this interesting:

    Look up the case of the Cantor function:

    http://en.wikipedia.org/wiki/Cantor_function

    Note that the Cantor function is an exception to:

    ∫f'= f .

    In order to have ∫f'=f we need f to be absolutely continuous.
     
  8. Sep 2, 2012 #7

    Simon Bridge

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    When you differentiate a function, you lose some information ... so any reconstruction by integration will be accurate only to an arbitrary constant. It is possible to lose information so badly that you cannot reconstruct the original function by integration ... but usually you just have to be careful to keep track.

    eg. if [tex]u(x)=\left\{
    \begin{array}{lr}
    1 & : x > 0\\
    0 & : x \leq 0
    \end{array}
    \right.[/tex]... you can see that the derivative is 0 everywhere except at the origin.
    We need to be able to distinguish this situation from a constant function. This is what the Dirac Delta function is for.

    But OP has only just learned to integrate polynomials ... all these special functions and such will be so many octarine words hanging in the air.

    I'll stick by the conclusion post #2: the integration is really a sum.
     
  9. Sep 2, 2012 #8
    So let's say I have a graph showing the price of something over time, and let's say it's modeled by the equation f(x) = ln(x). The derivative, f'(x) = 1/x, can be used to determine the price's rate of change at a particular point. However, what would the integral of this function, F(x) = xln(x) - x, tell us?
     
  10. Sep 2, 2012 #9

    Bacle2

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    This may not be too satisfying of an answer , but one of the things the integral

    tells you is the functions whose rate of change is ln(x). Like Simon said, it is

    difficult to generalize without a context.
     
  11. Sep 2, 2012 #10
    Oh that is interesting I haven't thought of it like that. Anyway, in my price example, would integrating the function serve any purpose?
     
  12. Sep 2, 2012 #11

    Bacle2

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    Let's see: what is the other variable: pricing against, or as a function of what:

    time, money supply, etc.?

    ( you can tell I am not much of an investor )
     
  13. Sep 2, 2012 #12
    Just a graph of $ over time
     
  14. Sep 2, 2012 #13

    Bacle2

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    I'm sorry, I need to head out right away. If no one has answered, I'll get back to you.
     
  15. Sep 2, 2012 #14

    AlephZero

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    I don't think it tells you anything with an obvious meaning. Just because you can calculate something, that doesn't mean calculating it is meaningful.

    But if you divide the integral by the time interval (end time - start time), that would tell you the average priice during the time interval.
     
  16. Sep 2, 2012 #15
    So let's say the time is in days, and the interval goes on for 10 days. Would [itex]\frac{10ln(10)-10}{10}[/itex] give the average price for the 10 days?

    Or do you mean [itex]\int_{0}^{10}\frac{xln(x)-x}{10}[/itex]
     
  17. Sep 2, 2012 #16

    AlephZero

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    I mean $$\frac{\int_{t_1}^{t_2} f(t)\,dt} { t_2 - t_1 }$$
     
  18. Sep 3, 2012 #17

    Simon Bridge

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    Imagine the price of $P does not change over time ... then the integral over a time period T (days) would be would be PT ... units are dollar-days. It is not clear what this would mean. It would probably be called the dollar-displacement for the system.

    In your example (post #16) dividing through by T would just get the price back.
    You are dividing the area by the length of the base.

    If the price started at Q and ended at P, linearly, then the integral is (P+Q)T/2. Dividing through by T gets you (P+Q)/2 ... the average price.

    That should be general to any p(t) ... the integral of p divided by the time period is the time-average of p.
     
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