Why do we integrate a function to find the area under it?

[Juwane]

Why, when finding the area by definite integral, we have to find the indefinite integral first? As I understand, to find the area of under the curve, all we need is the equation of the curve. On the other hand, the indefinite integral helps us to find the original function from its derivative. So what does this have to do with finding the area?

 

[arildno]

Why, when finding the area by definite integral, we have to find the indefinite integral first?

You don't have to. If you have, in general, infinite time at your disposal.

As I understand, to find the area of under the curve, all we need is the equation of the curve. On the other hand, the indefinite integral helps us to find the original function from its derivative. So what does this have to do with finding the area?

That is the truly beautiful insight in the fundamental theorem of calculus:

To sum up the area beneath some curve, essentially an INFINITE process, can trivially be done by finding an anti-derivative to the defining curve.

 

[Juwane]

[...]

That is the truly beautiful insight in the fundamental theorem of calculus:

To sum up the area beneath some curve, essentially an INFINITE process, can trivially be done by finding an anti-derivative to the defining curve.

One of the things that the fundamental theorem of calculus tells us that the area under a curve is equal to the area under that curve's derivative, right?

 

[elibj123]

One of the things that the fundamental theorem of calculus tells us that the area under a curve is equal to the area under that curve's derivative, right?

No, it tells us that integration (which is defined in a way that has nothing to do with derivatives or anti-derivatives) is opposite to derivation.

Which helps us in calculating integrals more efficiently.

 

[Juwane]

But it is true that the area under a curve is equal to the area under that curve's derivative, right?

 

[nicksauce]

But it is true that the area under a curve is equal to the area under that curve's derivative, right?

No. The area under y = 1 from x = 0 to 1 is 1. The area under the derivative is 0.

 

[pbandjay]

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition

 

[Juwane]

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition

Excerpt from the article:

It can thus be shown, in an informal way, that ƒ(x) = A’(x). That is, the derivative of the area function A(x) is the original function ƒ(x); or, the area function is simply the antiderivative of the original function.

 

[rochfor1]

Juwane: That quote you gave was just a restatement of the Fundamental Theorem of Calculus, nicksauce already gave an excellent example of where your assertion is false.

 

[Bohrok]

You don't have to. If you have, in general, infinite time at your disposal.

Left/right/middle sums all take too long to calculate the area?

 

[Landau]

@Juwane: how do get your assertion from the wikipedia quote? If $$A_f(x)$$ denotes the area function of some function f(x), then:

wiki (i.e. the fundamental theorem) says $$A_f'(x)=f(x)$$;
you're saying that $$A_f(x)=A_{f'}(x)$$.