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

1. Jan 24, 2010

### 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?

2. Jan 24, 2010

### arildno

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

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.

3. Jan 24, 2010

### Juwane

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?

4. Jan 24, 2010

### elibj123

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.

5. Jan 24, 2010

### Juwane

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

6. Jan 24, 2010

### nicksauce

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

7. Jan 24, 2010

### pbandjay

8. Jan 24, 2010

### Juwane

9. Jan 24, 2010

### 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.

10. Jan 24, 2010

### Bohrok

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

11. Jan 25, 2010

### 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 $$A_f(x)=A_{f'}(x)$$.