Antiderivative to find Area

“Why should I apply Antiderivative to find area of a function ?”

It might seems a very silly question, but I was searching for a solution.
I got some idea from Wikipedia Fundamental theorem of calculus , but still I was looking for example.

At last, I have found answer from these:

1. Area function f(x) = 2 within interval [a,x] = [-1,x]

Let assume, the distance of the rectangle from 0 is x.
Total Length of the rectangle will be x+1 over x-axis and width is 2.

Area = length * width

A(x) = 2(x+1) = 2x + 2

Derivative of A(x) is:
A/(x) = 2 = f(x)

2. Area function f(x) =x+1 within interval [a,x] = [-1,x]

Triangle Area = 1/2 * base * height

A(x) = 1/2 (x+1)(x+1) = x2/2 + x + 1/2

Derivative of A(x) is:
A/(x) = x + 1 = f(x)

3. Area function f(x) = 2x+3 within interval [a,x] = [-1,x]

Trapezoid Area = 1/2 * h * (a+b)

h = heigh
a,b is the parallel sides lengths.

A(x) = 1/2 (x+1)((2x+3)+1) = x2 + 3x + 2

Derivative of A(x) is:
A/(x) = 2x + 3 = f(x)

Result:
This is not only valid for linear function, but also for continuous function.
To find the area of  a function we can simply apply antidifferentiation.

A(x)  representing the area
A/(x) representing the area function.

If we know the area function A/(x) we can get the area by antidifferentiation which will provide us A(x) (area of the function).

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Source:
Calculus – Early Transcendentals by Anton, Bivens, Davis.
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition
http://en.wikipedia.org/wiki/Rectangle
http://en.wikipedia.org/wiki/Triangle
http://en.wikipedia.org/wiki/Trapezoid

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