# Cant understand this integral(probability)

1. Feb 17, 2010

### nhrock3

if we integrate by U
then if we take steady point on U axes we
encounter 1-v and 3-v lines
thus making our intervals for each steady U point
but why 0<v<1??
there is a red point in the photo in our area in which v=3

??

2. Feb 17, 2010

### rsa58

the blue line should be parallel to the horizontal axis

3. Feb 17, 2010

### rsa58

since you are integrating in terms of u. for v between 0 and 1, du will be between the two lines mentioned. for v between 1 and 3, u will vary between the v-axis and 3-v. to check it try performing the volume integral which shows that it is indeed a probability density.

4. Feb 17, 2010

### nhrock3

no if we go by du then we go by the u axes

5. Feb 17, 2010

### cronxeh

I don't understand the chicken scratch on the photo you posted. The first integral doesn't make sense. The area between 0 and v=1 and u=1 is 1/2*base*height = 1/2*1*1 = 1/2. The probability in this area is 0.

The second shaded area has area of 1/2*3*3-1/2 = 4. The total probability in this area is 1/4*4 = 1 which is your state space.

If you are integrating along u that means the area of a little strip is height*du, the height being delta v