# Why does the antiderivative give us the area?

1. Dec 25, 2004

### bezgin

Why does the antiderivative give us the area? I can't really find it in textbooks.

2. Dec 25, 2004

### dextercioby

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
$$\int_{x_{1}}^{x_{2}} f(x) dx$$
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 $F(x)=:\int f(x) dx$.Then Leibniz & Newton asserted that:
$$\int_{x_{1}}^{x_{2}} f(x) dx =F(x_{2})-F(x_{1})$$

Daniel.

PS.And to your question "why" the answer could be found in any calculus book.And you can find it.

3. Dec 25, 2004

### HallsofIvy

Staff Emeritus
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).

4. Dec 25, 2004

### Ethereal

Last edited by a moderator: Dec 25, 2004