Understanding the Relationship of Integral Calculus: A vs. (x-A)f(x) = 0

[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.

 

[stevenb]

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.
thanks.

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.

 

[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.

 

[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.

 

[Rajini]

Hi paulfr,

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

 

[HallsofIvy]

If that is a definite integral and \(\displaystyle \int_a^b f(x)dx= 1\) 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
$$\int_a^b xf(x) dx= A$$.

 

[Rajini]

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

 

[paulfr]

Hi paulfr,

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

No I did not mean that.

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

 

[paulfr]

If that is a definite integral and \(\displaystyle \int_a^b f(x)dx= 1\) 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
$$\int_a^b xf(x) dx= A$$.

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

 

[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 $$\int_a^b f(x) dx = 1$$

 

[Rajini]

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