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I Question about definite and indefinite integrals

  1. Dec 23, 2016 #1
    First, just to check, I write what I think and let me know if I am wrong:

    The definite integral of a function gives us a number whose geometric meaning is the area under the curve between two limiting points.
    We can calculate this integral as the limit of the sum of the rectangles and the integral symbol means a sum over all rectangles.

    The indefinite integral of the function f(x) is the function primitive F(x) whose derivative is f(x), and here we are using the fundamental theorem of calculus. Why do we use the same symbol?

    My question is, How could we calculate the indefinite integral without the fundamental theorem?

    Could we try it using the definite integral between the limits 0 and x?

    From a geometric point of view, How could we talk about the indefinite integral?

    Last edited: Dec 23, 2016
  2. jcsd
  3. Dec 23, 2016 #2


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    Because basically it's the same thing. Compare using the same term ##\sin ## for the sine function and for the value of the sine function at some point ##x##.
    We could create a table of F values from the sum of rectangle areas, but we would still need to state where we start integrating (i.e. the table would list ##F(x)-F(x_{\rm start}## ). So: Yes to your
    Note: this doesn't give you a function in the from of a 'recipe', though.
    It tells you how the area under the curve varies with the independent variable
  4. Dec 23, 2016 #3


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    You seem to be largely answering your own questions here. The only thing I would add is that you could see ##F## as a function which gives you definite integral between two points:

    ##\int_{a}^{b}f(x)dx = F(b) - F(a)##

    And, it turns out that any ##F## where ##F' = f## does the job. And, you call any such ##F## an indefinite integral of ##f##. More precisely, the set of functions ##\lbrace F: F' = f \rbrace## is the indefinite integral.
  5. Dec 23, 2016 #4
    The indefinite integral gives you the area (if you substitute for particular values), not the rate of change of the area, which is the function itself right?

  6. Dec 23, 2016 #5


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    Example: constant function. Indefinite integral: x itself, linear function. Area changes linearly with the independent variable.
  7. Dec 23, 2016 #6
    I think there was a misinterpretation.

    When you talked about changed linearly, you talked about the diferent values a function primitive F(x) can take, I thought you meant its rate of change F'(x) , that is why I said the rate of change was the funcion f(x) itself.

    I guess!
  8. Dec 23, 2016 #7


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  9. Dec 24, 2016 #8


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    Just to add to the confusion: In many cases it is possible to evaluate the definite integral but not the indefinite...
  10. Dec 24, 2016 #9
    Very interesting yes.
    I understand that. Finding an area is different to finding a function whose differential is a given one.

    Have I told you anything confusing?
  11. Dec 24, 2016 #10


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    Just an expression. Means: "This information that was not asked for and may not be very relevant"
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