1. ## integration question

Hi,

I'm stuck on the domain of integration change in the following problem:

I have a Uniform distribution, U(a,b).

I'm deriving the variance:

var = $\int^b_a \left(x-\frac{b+a}{2}\right)^2 \frac{1}{b-a} dx$

Use a change of variable substitution: $y =x - \frac{b+a}{2}$

Take the constant out

$var = \frac{1}{b-a} \int^\frac{b-a}{2}_{-\frac{b-a}{2}} y^2 dy$

I'm not sure why the domain of integration has changed from a to b, to $\frac{b-a}{2}$. I'm presuming it's to do with the y change of variable. Any ideas?

2. Originally Posted by notgoodatmath
Hi,

I'm stuck on the domain of integration change in the following problem:

I have a Uniform distribution, U(a,b).

I'm deriving the variance:

var = $\int^b_a \left(x-\frac{b+a}{2}\right)^2 \frac{1}{b-a} dx$

Use a change of variable substitution: $y =x - \frac{b+a}{2}$

Take the constant out

$var = \frac{1}{b-a} \int^\frac{b-a}{2}_{-\frac{b-a}{2}} y^2 dy$

I'm not sure why the domain of integration has changed from a to b, to $\frac{b-a}{2}$. I'm presuming it's to do with the y change of variable. Any ideas?
Remember your limits were originally in terms of x.

If $y(x)=x-\frac{b+a}{2}$, what is $y(a)$ and $y(b)$? These will give you the new limits of integration with respect to y.

3. Originally Posted by notgoodatmath
Hi,

I'm stuck on the domain of integration change in the following problem:

I have a Uniform distribution, U(a,b).

I'm deriving the variance:

var = $\int^b_a \left(x-\frac{b+a}{2}\right)^2 \frac{1}{b-a} dx$

Use a change of variable substitution: $y =x - \frac{b+a}{2}$

Take the constant out

$var = \frac{1}{b-a} \int^\frac{b-a}{2}_{-\frac{b-a}{2}} y^2 dy$

I'm not sure why the domain of integration has changed from a to b, to $\frac{b-a}{2}$. I'm presuming it's to do with the y change of variable. Any ideas?
When you change variables you have to plug the limits of integration into the expression that you use to change variables, so just plug in a for x in $y =x - \frac{b+a}{2}$ and also plug in b to get your lower and upper limit

4. got it, thanks
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