Finding P(X+Y < 1) for joint distribution f(x,y) = x² + xy³

[Gauss M.D.]

Homework Statement

f(x,y) = x² + xy³ for 0 < x < 1, 0 < y < 2
and 0 otherwise.

Calculate P(X+Y < 1)

Homework Equations

The Attempt at a Solution

P(X+Y < 1) = P(X < 1-Y) which means y is now bounded by 0:1 instead of 0:2 and x is bounded by 0:y.

So we get ∫_[0-1]∫_[0-y] x² + xy³ dx dy

Integrating twice I get the answer P(X < 1-Y) = 1/8, which is incorrect.

What am I doing wrong?

 

[Gauss M.D.]

Oops.

 

[Gauss M.D.]

I.e. nevermind.