Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration and differentiation

  1. Mar 18, 2012 #1
    Why is antiderivative and area under the curve the same thing? Its not at all intuitive to me

    Derivative is the slope at a point and its opposite is area?? Can someone just explain me why when we are finding an antiderivative, we are actually finding area under the curve

    i dont buy the fact that slope of a curve and area under the curve ae opposite of each other
     
  2. jcsd
  3. Mar 18, 2012 #2
    Oh my! Right now you're undergoing your very first mathematical crisis! I've been there some years ago and trust me, eventually you will understand it.

    I recommend you read a book by William Dunham, "The mathematical universe" chapters D, K and L.

    It's something you must discover by your own. For now I can just say that you are confusing "antiderivative" with "integral", which are quite different things. The first is a function, while the later is a real number.
     
  4. Mar 18, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The standard proof is this- look at the graph of y= f(x), a continuous. Let the area under the curve, above y= 0, and between x= a and x, be F(x). Let [itex]x^*[/itex] be a value x at which F takes its maximum, [itex]x_*[/itex] a value at which F takes its minimum on [a, x]. Then we must have [itex]f(x_*)(x- a)\le F(x)\le f(x^*)(x- a)[/itex]. Then
    [tex]\frac{F(x_*)}{x- a}\le f(x)\le\frac{F(x^*)}{x- a}[/tex]
    Because both x* and [itex]x_*[/itex] are between a and x, if we take the limit as x goes to a, x* and [itex]x_*[/itex] will also both go to a. But then,
    [tex]\lim_{x\to a}\frac{F(x_*)}{x-a}= \lim_{x\to a}\frac{F(x^*)}{x- a}= \frac{dF}{dx}[/tex] and we have
    [tex]\frac{dF}{dx}\le f(x)\le \frac{dF}{dx}[/tex]
     
    Last edited by a moderator: Mar 18, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration and differentiation
Loading...