“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