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Indefinite Integration

  1. Jul 1, 2012 #1
    Presently, I am reading about computing definite integrals; and in one of the examples the authors provides, there is a statement made: "Recall that the point behind indefinite integration...is to determine what function we differentiated to get the integrand."

    I was wondering if someone could perhaps explain this to me?
  2. jcsd
  3. Jul 1, 2012 #2
    Here is an example:

    ∫xdx = [itex]\frac{x^{2}}{2}[/itex] + constant
    The reason for this is because [itex]\frac{d(\frac{x^{2}}{2} + constant)}{dx}[/itex] = 2x/2 + 0 = x.

    i.e. in an indefinite integration (like the above) we try to find the function, that when differentiated, will give what we are going to integrate.
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