# Dx in a function to be intigrated

1. Jan 22, 2009

### dE_logics

Suppose in an integrated function f(x), its a necessity to have dx, why is it that all x in that function is not replaced by dx?...or is it?

2. Jan 22, 2009

### Gib Z

Sorry but I really can't understand what you are trying to say? Could you illustrate it with some examples to show us what you mean?

3. Jan 22, 2009

### dE_logics

I've seen this in many derivations.

Suppose the original function is (sin x)/x...and we're suppose to figure out the area.

So...the function to be integrated should be (sin dx)/dx...i.e replace all x with dx so they return results with infinite accuracy when the infinitely small sections are summed up.

But instead why do we take (sin dx)/x (I'm not sure if this is right, I've arbitrarily converted one of the x to dx).

4. Jan 22, 2009

### nicksauce

You have it wrong, you don't replace any x's with dx's. Imagine moving your finger along the x-axis... x is the value of your finger is pointing to, f(x) is the value of the function at that point, and dx is a small increment in x. The small unit of area is thus f(x) dx, and you integrate f(x) dx.

Example: The area under the curve f(x) = sin(x) / x from 0 to 5 is
$$A=\int_0^5 \frac{\sin{x}}{x}dx$$

5. Jan 22, 2009

### dE_logics

Humm...I see thanks!

6. Jan 22, 2009

### Feldoh

$$\int_a^b f(x) dx$$ is an operator in itself, but think of as taking an infinitesimal length on the independent axis (this is dx), and multiplying it by the height of some function f(x) so you get the area under the curve and then summing it up from a to b.

7. Jan 23, 2009

### dE_logics

Yep...as a result infinite accuracy.