Why does the antiderivative give us the area?

[bezgin]

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

 

[dextercioby]

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

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.

 

[HallsofIvy]

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).

 

[Ethereal]

Look for something called the Fundamental Theorem of Calculus in calc textbooks. This thread helps too:

https://www.physicsforums.com/showthread.php?t=17384