Integration by substitution?

**Timbuk2**

I want to integrate

[latex]\displaystyle{\frac{2}{\pi}\int_{-1}^1x\sqrt{1-x^2}dx}[/latex]

So I made a simple u-substitution

[latex]u = 1-x^2[/latex]
[latex]du = -2xdx[/latex]

So I can rewrite the above integral as
[latex]\displaystyle{\frac{1}{\pi}\int_{1-(-1)^2}^{1-(1)^2} \sqrt{u} du[/latex]
[latex]=\displaystyle{\frac{1}{\pi}\int_{0}^{0} \sqrt{u} du[/latex]

What's happened here? Is an integral evaluated between 0 and 0 even defined?

Is there another way to do this, a different substitution perhaps?

ray giraffe

Nothing's wrong, if you finish it off the answer is zero as it should be.

The integral between k and k is always zero, no matter what number k is.

**Timbuk2**

Oh right, that's ok then!

I don't see that as one of the properties of integrals though - it makes sense intuitively though.

It's just, we had a probability tutorial recently, and such an integral came up, and the tutor did something I was unsure about - he said

for that u-substitution,
limits
@ x=1: u =0
@ x=0: u=1
@ x=-1: u=0

and thus our function u is not injective, so we needed to split the above integral as such: (excuse the sloppy notation)
[latex]\int_{x=-1}^{x=1} = \int_{u=0}^{u=1} + \int_{u=1}^{u=0}[/latex]
[latex]\int_{x=-1}^{x=1} = \int_{u=0}^{u=1} - \int_{u=0}^{u=1}[/latex]
=0

Which I'm sure is right, I just didn't think it was necessary. We had to do that for a homework question, it'll be interesting to see if I lose marks for saying that the integral between 0 and 0 = 0 - hopefully not!

Iderown

You may like to try a trig substitution x = sin u

The original integrand is anti-symmetric about x=0. Its graph looks like a sine wave between x = -1 and x = +1. Area under it is 0.

When faced with a difficult looking integral which features pi, I suspect that a trig function may be involved.

Mwalimu

Splitting the interval for the function is a useful way to approach problems involving the area enclosed between the function/graph and the x axis. This is not always the saem as the integral. The integral over an interval can be zero, but the area involved might not be zero. The area above the x axis may be equal to the area below the x axis over the interval. The 'nett' value of the area is thus zero.