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Homework Help: Integrals from definition

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to know if its possible to count undefined and defined integral from definition. When I have derivative of function, ex. [tex]x^3[/tex], I can count this from this: [tex]\left(x^n\right)^{\prime}=nx^{n-1}[/tex], but can also use definition
    [tex]\lim_{h\to0}\frac{\left(x+h\right)^3-x^3}{h}[/tex] and its the same. But when I have integral, ex. [tex]\int x^3\mbox{d}x[/tex], I dont have any definition to count this, I must use [tex]\int x^n\mbox{d}x=\frac{x^{n+1}}{n+1}[/tex]. What can I do, if I didnt know this formule? is there any definition to count undefined integral, or I can use only formules based on assumption that we know what function has this derivative?
    3. The attempt at a solution
    Trying to do something with derivative definition, but it didnt help me. Thanks for answer;]
     
    Last edited: Dec 20, 2009
  2. jcsd
  3. Dec 20, 2009 #2
    By "count", do you mean something like "evaluate"? In that case, the formula for the derivative you listed can be gotten by direct calculation by using the definition of the derivative. A function is differentiable at a if
    [tex]\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}[/tex]
    exists. If you compare this to the limit you wrote down for the derivative of x^3, you'll notice your limit is off in the f(a+h) term. Using this definition is often tedious, and thus the generalization (x^n)' = nx^(n-1) is useful.

    With integration, it's sort of the same idea. The integration formula you have is NOT the definition of the integral. The formula is a consequence of the fundamental theorem of calculus, which guarantees that a continuous function such as x^3 has an antiderivative. The typical definition for an integral encountered in an elementary calculus course involves Riemann sums. The basic idea is to approximate the area under a curve via rectangles. You could obtain the formula for that type of integral using this method, but it is not a pleasant exercise. If you want to learn more, visit

    http://en.wikipedia.org/wiki/Riemann_sum
    http://en.wikipedia.org/wiki/Darboux_sum

    (these two definitions are equivalent, but I find the later easier to use if you understand sups and infs)
     
  4. Dec 20, 2009 #3

    Landau

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    I don't understand what you mean by 'count' in this context, I guess you mean 'compute' or 'calculate'.

    The limit approach for calculating a derivative is only practical for 'easy' functions. For an integral, you could pick some partition of your interval [a,b] and actually compute the upper and lower sums, this may lead you to the answer. This also is not very practical.
    Or you could use some kind of numerical integration. I'm not sure what you're looking for though.
     
  5. Dec 20, 2009 #4
    I think you meant [tex]
    \lim_{h\to0}\frac{(x+h)^3-x^3}{h}
    [/tex]

    Anyway, suppose you know that (I'll leave out the integration constants)

    [tex]\int 1 dx = x[/tex]

    Then by using partial integration (and knowing how to calculate derivatives) you can get:

    [tex]f' = 1; g = x^n;

    \int f'g = fg - \int fg' \Rightarrow \int x^n = x^{n+1} - n \int x^n \Leftrightarrow \int x^n = \frac{x^{n+1}}{n+1}[/tex]

    You could also do a substitution instead. The point is that you can calculate unknown integrals by transforming them into known integrals. I'm not sure you can do it through a definition though since, if I recall correctly, an indefinite integral is defined as an anti-derivative so you have to know some derivative in the end. I could be wrong of course, I'm not a mathematician. You could try using the definition of the integral as a limit of a Riemann sum or something but I don't know whether it would work very well.
     
  6. Dec 20, 2009 #5
    well, thanks for answers, I am going think about this:)
     
  7. Dec 20, 2009 #6

    HallsofIvy

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    Science Advisor

    In general "inverse" problems are much harder than "direct" problems. For example, if you are given [itex]f(x)= 4x^5- 3x^3+ 2x^2+ x+ 5[/itex] and are asked to find f(1), that is a "direct" problem- you are given a formula so you can just use the formula to calculate. [itex]4(1)^5- 3(1)^3+ 2(1)^2+ (1)+ 5= 9.

    But the "inverse" problem is "find x such that f(x)= 9" and that involves solving a fourh degree equation. Because we know that f(1)= 9, we can say that x= 1 is a solution but we don't know if there are other solutions or what they are.

    The same is true of derivatives- we are given a formula and can, theoretically, differentiate any (differentiable) function using that formula. But "anti-derivatives" are defined only as the "inverse" of differentiation. That is, the anti-derivative of the function f is the function F such that F'= f. Since we know that if [itex]F(x)= x^3[/math] then [itex]F'(x)= 3x^2[/itex] we know immediately that an anti derivative of [itex]3x^2[/itex] is [itex]x^3[/itex]. But we need more theory to know that there are other anti-derivatives of the same function and, in fact, need Rolle's theorem to know that they all differ by a constant.
     
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