# How definite integration works?

by Psyguy22
Tags: definite, integration
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 Let y= f(x) be continuous function such that $f(x)\ge 0$ for all $a\le x$. Define F(x) to be the area under the graph from a to x. Then F(x+$\Delta$x) is the area under the graph from a to x+ $\Delta$x. F(x+ $\Delta$x)- F(x) is the area under the graph from x to x+ $\Delta$x. Because f is continuous (this is the "deep" part!), there exist some x', $x\le x'\le x+ \Delta x$, such that $f(x')\Delta x$ is equal to that area. That is, $F(x+ \Delta x)- F(x)= f(x')\Delta x$ so that $$\frac{F(x+ \Delta x)- F(x)}{\Delta x}= f(x')$$ Now, x' is always between x and x+ $\Delta$ x so if we take the limit as $\Delta$x goes to 0, f(x') goes to f(x). That is: $$\lim_{\Delta x\to 0}\frac{F(x+ \Delta x)- F(x)}{\Delta x}=\frac{dF}{dx}= f(x)$$ That is, the area really is given by an anti-derivative. The interesting thing is that to do this- and so be able to calculate the are of some very complicated sets- we don't have to give a very detailed definition of "area". All that is needed about area is 1) The area of a set is a non-negative number 2) If sets A and B are disjoint (except possibe on their boundaries) the area of A U B is the area of A plus the area of B 3) The area of a rectangle is "height times width".