# Integral calculus help

1. Jul 25, 2010

### Rajini

Hello all,

I wanted to know whether the following two relations are same.
$$\int x\;f(x)\;{\rm d}x = {\rm A}$$
and
$$\int (x-{\rm A})\;f(x)\;{\rm d}x = 0$$
are same?
'A' is some positive number, distribution function f(x) is normalized to 1. (i am trying to understand some details of first moment of f(x) and got this doubt.)
thanks.

2. Jul 25, 2010

### stevenb

Looks the same as long as the integration limits are such that the integration covers the entire range of normalization. I'm not a mathematician, but I'm bothered by the unspecified integration limits.

3. Jul 25, 2010

### Rajini

Integration covers entire range in normalization. And for both equations the same integration limits.
f(x) is a normalized function. In the above 2 equation the same f(x) is used.
thanks.
edit:Stevenb, I solved, both are same.

Last edited: Jul 25, 2010
4. Jul 26, 2010

### paulfr

Looks to me like the A in the 2nd [lower] equation needs to be at least d/dx A or the derivative of A.
Then by the Fundamental Theorem of Calculus, the integral of the derivative of A is = A.

Even then the algebra looks incorrect.

The second equation should be
Integral [ x f(x) ] - [ d/dx A ] dx NOT
Integral ( x - A ) f(x) dx

This is using the rule that the integral of a sum is the sum of the integrals due to
the linearity of integration.

Last edited: Jul 27, 2010
5. Jul 27, 2010

### Rajini

Hi paulfr,

do you mean this :
$$\int \left( x-\frac{\rm dA}{{\rm d}{x}}\right)\;f(x)\;{\rm d}x = 0$$

6. Jul 27, 2010

### HallsofIvy

If that is a definite integral and $\int_a^b f(x)dx= 1$ then it is true that
$$\int_a^b (x- A)f(x) dx= \int_a^b xf(x)dx- A\int f(x)dx= \int_a^b xf(x)dx- A= 0$$
so
$$\int_a^b xf(x) dx= A$$.

7. Jul 27, 2010

### Rajini

HallyofIvy,
I never thought that it will be very simple as you show. thanks.

8. Jul 27, 2010

### paulfr

No I did not mean that.

My original post was wrong.
The A constant is not multiplied by f(x).

9. Jul 27, 2010

### paulfr

Your first line here implies that the integral of f(x) dx = 1
How do you arrive at that ?

10. Jul 27, 2010

### Gib Z

HallsofIvy didn't specify but he intended that [a,b] is the entire range of distribution, and since f(x) is normalized, by definition $$\int_a^b f(x) dx = 1$$

11. Jul 27, 2010

### Rajini

Hi Paulfr,
I already informed that my function is normalized to 1. So integral of f(x)=1.