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Fundamental Theorem of Calculus

  1. Nov 23, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    The derivative of an integral with a constant as its lower bound and a function as its upper bound is the function at its upper bound multiplied by the derivative of the upper bound.

    3. The attempt at a solution

    How come the constant term has no bearing? I understand the derivative is un-doing the integral, but how come the constant term disappears?

    Is this why?

    https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1459734_10201135696496253_598181064_n.jpg?oh=a9ce27391aab214e4545d0bddf547aa5&oe=5292F89D
     
    Last edited: Nov 23, 2013
  2. jcsd
  3. Nov 23, 2013 #2

    haruspex

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    Having x as the variable of integration and in the range is a 'pun'. I.e. they're different variables really, so to make it clearer let's give them different names: ##\frac d{dx}\int_{t=1}^{x^2}t.dt##
    Increasing x extends the integral at the right-hand end, but doesn't change the left-hand end. So the change in value of the integral does not depend on where the left-hand end is. Draw a picture.
     
  4. Nov 23, 2013 #3

    Qube

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    Yes, it's the dummy variable of integration.


    I love geometric approaches to calculus. Indeed, what you say is true.
     
  5. Nov 23, 2013 #4

    Ray Vickson

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    Besides what others have told you, you can apply the general formula:
    [tex] \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(t)|_{t=b(x)} \frac{db(x)}{dx} - f(t)|_{t=a(x)} \frac{d a(x)}{dx}
     
  6. Nov 23, 2013 #5

    Ray Vickson

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    Besides what others have told you, you can apply the general formula:
    [tex] \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(t)|_{t=b(x)} \frac{db(x)}{dx} - f(t)|_{t=a(x)} \frac{d a(x)}{dx}[/tex]
    For a constant lower limit we have da/dx = 0.
     
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