# Thread: Double integration: limits

1. ## Double integration: limits

I am given the following probability density function:
$f (x,y) = \frac{2x + 2 - y}{4}$ for 0<x<1 and 0<y<2
$f (x,y) = 0$ otherwise

I am to find P[X + Y > 1].

My attempts at the double integration failed.

The answer begins:
$\int_0^1 \int_{1-x}^2 \frac{2x + 2 - y}{4} dydx$

How did they pick which integral was inner?

How did they choose 1-x as the lower limit on the inner integral?

2. Here is a graph of the region

They solved x+y=1 for y to get y=1-x

3. Originally Posted by Boris B
I am given the following probability density function:
$f (x,y) = \frac{2x + 2 - y}{4}$ for 0<x<1 and 0<y<2
$f (x,y) = 0$ otherwise

I am to find P[X + Y > 1].

My attempts at the double integration failed.

The answer begins:
$\int_0^1 \int_{1-x}^2 \frac{2x + 2 - y}{4} dydx$

How did they pick which integral was inner?

How did they choose 1-x as the lower limit on the inner integral?
Draw the line x + y = 1 inside the rectangle defined by 0 < x < 1 and 0 < y < 2. This line divides the rectangle into two regions: The region above the line (upper region) and the region below the line (lower region). Note that in the upper region x + y > 1. So you need to set up a double integral of f(x, y) over the upper region .......