- #1

- 516

- 11

## Main Question or Discussion Point

So I've been spending a lot of time lately trying to figure out why an integral will give you the area under the curve. I asked the forum and got some great answers, but all were in terms of infinite sums, dx, and infinite rectangles. I think I've come upon a more fundamental answer that I haven't heard from anybody yet (although I'm sure its obvious to most people.) :)

[itex]y(x)[/itex] is a curve on a graph. Now as [itex]y(x)[/itex] gets larger, the area under the curve [itex]A(x)[/itex] gets larger. In fact, say [itex]y(x)=3[/itex]. Then the area at [itex]x=1[/itex] would be 3. Its rate of change at that instant would be 3.

Which means the rate of change of the area under the curve is equal to y(x): [itex]A'(x)=y(x)[/itex].

That being said, the actual area under the curve would equal the anti-deriviative of y(x): [itex]A(x)=∫y(x)[/itex].

Is this correct? If so, why do we use the confusing dx when it's really not even a number that you could multiply by to get the area of a rectangle (other than showing the integral is with respect to x). I guess I just don't like the [itex]\frac{dy}{dx}[/itex] notation.

Thanks

[itex]y(x)[/itex] is a curve on a graph. Now as [itex]y(x)[/itex] gets larger, the area under the curve [itex]A(x)[/itex] gets larger. In fact, say [itex]y(x)=3[/itex]. Then the area at [itex]x=1[/itex] would be 3. Its rate of change at that instant would be 3.

Which means the rate of change of the area under the curve is equal to y(x): [itex]A'(x)=y(x)[/itex].

That being said, the actual area under the curve would equal the anti-deriviative of y(x): [itex]A(x)=∫y(x)[/itex].

Is this correct? If so, why do we use the confusing dx when it's really not even a number that you could multiply by to get the area of a rectangle (other than showing the integral is with respect to x). I guess I just don't like the [itex]\frac{dy}{dx}[/itex] notation.

Thanks