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**Mathlv** Hello,

Would someone please help explain this to me?

Given:

$\displaystyle f(x,y) = {\rm{ \{ }}{e^{ - y}},{\rm{ }}0 < x < y$ and = 0 elsewhere

Find P(x>1, y>1)

So the set up happen to be as follow and then solve the integration

$\displaystyle \int_1^\infty {\int_1^y {{e^{ - y}}} dxdy} $

= $\displaystyle \int_1^\infty {{e^{ - y}}(y - 1)dy} $

Let u = y-1 $\displaystyle \Rightarrow $ du = dy

= $\displaystyle \int_0^\infty {{e^{ - (u + 1)}}udu = {e^{ - 1}}} \int_0^\infty {u{e^{ - u}}} du$

= $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$

What I do not understand is how my instructor comes to $\displaystyle {e^{ - 1}} \times 1 = {e^{ - 1}}$

How did he gets to 1?

Also I do not know how he changes from $\displaystyle \int_1^\infty {} $ to $\displaystyle \int_0^\infty {} $