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Why does the antiderivative give us the area?

  1. Dec 25, 2004 #1
    Why does the antiderivative give us the area? I can't really find it in textbooks.
     
  2. jcsd
  3. Dec 25, 2004 #2

    dextercioby

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    No,it's not the antiderivative that gives us the area.I should ask you what area.I assume you're talking about the area delimited by te graph of the function f(x),two vertical "bars" x_{1} & x_{2} and the 0x axis.
    The quantity that gives us that area is denoted by
    [tex] \int_{x_{1}}^{x_{2}} f(x) dx [/tex]
    and it is called the definite integral of the function f(x) from x_{1} to x_{2}.The fundamental theorem of calculus (Leibniz-Newton) allows us to express this quantity in terms of two values of the antiderivative of 'f'.
    Let F(x) be [itex] F(x)=:\int f(x) dx [/itex].Then Leibniz & Newton asserted that:
    [tex] \int_{x_{1}}^{x_{2}} f(x) dx =F(x_{2})-F(x_{1}) [/tex]

    Daniel.

    PS.And to your question "why" the answer could be found in any calculus book.And you can find it.
     
  4. Dec 25, 2004 #3

    HallsofIvy

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    Suppose y= f(x)> 0 for all x> a. Let A(X) be the "area function": the area of the region (call it R1) bounded on the left by x= a, below by y= 0, on the right by x= X and above by y= f(x). Even if we don't have a specific formula for such an area we know it must obey the basic rules for area:
    1. If A is a subset of B, the area of A is less than or equal to the area of B.
    2. If A and B area disjoint, then the area of AUB is the area of A plus the area of B.
    3. If A is a rectangle of width w and height h, then the area of A is wh.

    If h is a small (positive) number then A(x+h) is the area of the region (call it R2) bounded on the left by x= a, on the bottom by y= 0, on the right by x= X+h, and above by y= f(x). Let R3 be the region bounded on the left by x= X, on the bottom by y= 0, on the right by x= X+h, and above by y= f(x). Then
    R2= R1UR3 and R1 and R3 are disjoint so A(X+h)= A(X)+ area of R3 (we can't use "A()" for that last since it does not have x=a as left side).
    Now, let M be the maximum value of f(x) on X<= x<= X+h and m be the minimum value of f(x) on that same interval (these exist if f is continuous on the interval). Let R4 be the rectangle bounded on the left by the line x= X, below by y= 0, on the left by x= X+h, and on the top by y= m. Let R5 be the rectangle bounded on the left by the line x= X, below by y= 0, on the left by x= X+ h, and on the top by y= M. Since R4 is a subset of R3 which is a subset of R5 we have area of R4<= area of R3<= area of R5 or, since R4 is a rectangle of width h and height m and R5 is a rectangle of width h and height M, hm<= area of R3<= hM.

    From A(X+h)= A(X)+ area of R3 we have A(X+h)- A(X)= area of R3 so hm<= A(X+h)- A(X)<= hM. Dividing through by h, m<= (A(X+h)- A(X))/h<= M. As h->0, m and M both go to f(X) so, by the "squeeze" theorem
    lim(x->h) (A(X+h)- A(X))/h= f(X). In other words, A(x) is differentiable and its derivative at any X is f(X): f(x) is the derivative of the area function so the area IS an anti-derivative of f(x).
     
  5. Dec 25, 2004 #4
    Last edited by a moderator: Dec 25, 2004
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