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What is an integral?

  1. May 2, 2012 #1
    I've this example of an integral.

    [itex]\int^{x}_{a} f(t)dt[/itex]

    What t and dt means? Is there a relation between t and dt?
     
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  3. May 2, 2012 #2

    HallsofIvy

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    Since you are talking about an integral I assume you have some idea of what an integral is, either taking a course in Calculus or are reading a book on Calculus.

    Every text book I have seen introduces the integral in terms of the Riemann sum in which you divide an area under the graph of y=f(x) into very thin rectangles- each having base [itex]\Delta x[/itex] and height [itex]f(x_i^*)[/itex] for [itex]x_i^*[/itex] being some x value within the base of the ith rectangle. The approximate area would be the sum of the areas of those rectangles: [itex]\sum f(x_i^*)\Delta x[/itex]. One can show that the exact area is the limit of that as the base of those rectangles goes to 0. Then in the form [itex]\int f(x)dx[/itex] or [itex]\int f(t)dt[/itex] x and t are the independent variables in the functions f(x) and f(t), the limits of [itex]f(x_i^*)[/itex] and [itex]f(t_i^*)[/itex], and dx and dt are "infinitesmal" sections of the axis.
     
  4. May 2, 2012 #3
    So what's the purpose of having dx in the expression? If I'm correct, dx means the derivative of x. And, in each point of x in the interval [itex]\Delta x[/itex], x will be a fixed value (is it a constant?), and the derivative of a constant is always 0.
     
  5. May 2, 2012 #4

    micromass

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    The "dx" is just a notation. It doesn't mean anything. It's just used to denote which variable we integrate.

    For example

    [tex]\int_0^1 (t+x)dx[/tex]

    means that we integrate with respect to x. While

    [tex]\int_0^1 (t+x)dt[/tex]

    means that we integrate with respect to t.
     
  6. May 2, 2012 #5
    As far as I understand, in the first case we will only integrate x, and in the second case, t.

    So, being f a function, f=(t+x)

    In the first case, what happens to x? Can you integrate this example, please?
     
  7. May 2, 2012 #6

    micromass

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    With the integral

    [tex]\int (t+x)dx[/tex]

    we have that t is a constant and x is the integration variable, so

    [tex]\int (t+x)dx = tx+\frac{x^2}{2}+C[/tex]

    While in

    [tex]\int (t+x)dt[/tex]

    we treat x as a constant. Thus

    [tex]\int (t+x)dx = \frac{t^2}{2}+tx+C[/tex]
     
  8. May 2, 2012 #7
    Thanks, now I got it.
     
  9. May 3, 2012 #8

    hunt_mat

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    I wrote some notes on integration, they should be available on the Math & Science Learning Materials section under notes on integration.
     
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