# Q regarding Integration

1. Aug 10, 2004

### ReyChiquito

Im trying to understand a "simple" identity.

I have the problem $$\ddot{y}+g(t,y)=0$$
with $$y(0)=y_{0}$$ and $$\dot{y}(0)=z_{0}$$

What i need to prove is that the solution can be expresed in the form
$$y(t)=y_{0}+tz_{0}-\int_{0}^{t}(t-s)g(s,y(s))ds$$

Integrating twice is clear that
$$y(t)=y_{0}+tz_{0}-\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds$$

Now the book goes
$$\int_{0}^{t}\{\int_{0}^{s}g(\tau,y(\tau))d\tau\}ds=\int_{0}^{t}\{\int_{\tau}^{t}ds\}g(\tau,y(\tau))d\tau$$

???????????????????? How do i prove that? should i do a change of variable? use fubini?

Im lost, pls help.

2. Aug 10, 2004

### mathman

It is not any change of variables. If you look at the domain of integration in the (s,tau) plane, you will see it is a triangle. All the book is describing is how the triangle is expressed when you switch the order of integration. Fubini's theorem says you can do the switch.

3. Aug 10, 2004

### ReyChiquito

Ohhhhhhhhhhhhh.... i know i should have named the tread "stupid Q about integration".
Thx a lot dood. I sure need to review my calc notes.