area of a region bounded(linear algebra)

Question

When we find the area of a region bounded by an upper curve f(x) and a Lower curve g(x), we integrate f(x)-g(x). Analytically, how do we know that this integral is positive and so actually represents the area. 

Answer 1

Hi Ana A.,

I would argue as follows. At every point x the value of f(x) is greater than the value of g(x) (the essence of "upper" and "lower"). Therefore for each interval dx, proceding in the direction from smaller x to larger x, dx is positive, and f(x) - g(x) is positive. Therefore, the limit of a series consisting of positive (product) increments is calculated for the integral, and thus that limit must be positive. What did you come up with as a rationale?

-- Cheers, --Mr. d.

Answer 2

Hi Ana A.,

I would argue as follows. At every point x the value of f(x) is greater than the value of g(x) (the essence of "upper" and "lower"). Therefore for each interval dx, proceding in the direction from smaller x to larger x, dx is positive, and f(x) - g(x) is positive. Therefore, the limit of a series consisting of positive (product) increments is calculated for the integral, and thus that limit must be positive. What did you come up with as a rationale?

-- Cheers, --Mr. d.

Answer 3

I think you have actually asked several questions rolled into one (2).

And I do not think there is an analytic way to know the answer.

Clearly if f(x)>0 and the integral converges, it represents the area...although not strictly true, for elementary purposes you can think of this situation as defining area.

When you consider the difference of 2 functions and the area between them, the situation is less clear and more complicated. Even if f(x)-g(x)>0, you do not know analytically what the integral represents (think about the area between 0 and 1 between y=x and y=-1/10); of course you do know that the integral is positive and is the ALGEBRAIC area...but not the actual area.

This is far from a complete answer to your question...and a complete answer requires higher math (higher than what I know)!