Why isn't it calculated like this?

[Deniz666]

Why do we multiply f(x) and dx to find the area under a function and get its infinite sum. If we just calculate the infinite sum of f(x) in a certain interval, wouldn't we get the same result?
Since there are infinitely many dx anyway, does dx have any significance. After all, dx will behave like a point. In another case, it will appear in the calculation of the surface area.According to me, we don't have to include the ‘Lk' of the disk we calculated into the calculation. When the calculation is taken to infinity, the truncated cone will become a cylinder. Therefore, it will be sufficient to multiply the length of the circle by dx.
In fact, instead of multiplying by dx, as I mentioned in my first question, it will be enough to just take the length of the circle to infinity.
I would appreciate if you explain where i am doing wrong.

 

[blamocur]

If we just calculate the infinite sum of f(x) in a certain interval, wouldn't we get the same result?

If you add up an infinite number of [imath]f(x)[/imath]'s you'll get [imath]\infty[/imath], won't you?

 

[topsquark]

Why do we multiply f(x) and dx to find the area under a function and get its infinite sum. If we just calculate the infinite sum of f(x) in a certain interval, wouldn't we get the same result?

When we integrate we are looking at the "area under the curve." f(x) is height, not an area.

Another way to look at it is that the form of the integral comes from a sum: [imath]\lim_{\Delta x \to 0} \sum f(x) \Delta x[/imath]. So if the dx isn't there then you don't have the limit of the correct sum.

-Dan

 

[Deniz666]

When we integrate we are looking at the "area under the curve." f(x) is height, not an area.

Another way to look at it is that the form of the integral comes from a sum: [imath]\lim_{\Delta x \to 0} \sum f(x) \Delta x[/imath]. So if the dx isn't there then you don't have the limit of the correct sum.

-Dan

Multiplying f(xk) infinite times causes it not to be in a certain range, you argue that it must be multiplied by dx to be in a certain range, right?

 

[Steven G]

As \(\displaystyle \Delta x\rightarrow 0, n\rightarrow\infty\)

 

[BigBeachBanana]

Multiplying f(xk) infinite times causes it not to be in a certain range, you argue that it must be multiplied by dx to be in a certain range, right?

You know that area of a rectangle is height*width.
Height or f(x) by itself has an area of 0. Adding infinitely many zeroes give you 0, so there's no area.
Thus, the [imath]\Delta x[/imath] or the width is necessary to compute the area of a rectangle.
There's an animation at the end of this page demonstrating the concept [imath]\Delta x\rightarrow 0, n\rightarrow\infty[/imath] StevenG mentioned.