Cant understand this integral(probability)

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The discussion centers on integrating probability density functions with respect to the variable U, specifically addressing the integration limits for variable V. Participants clarify that for V values between 0 and 1, the integration results in a probability density area of 1/2, while for V values between 1 and 3, the area calculates to 1, confirming the total probability within the defined state space. The conversation emphasizes the importance of understanding the geometric interpretation of the integration process and the significance of the area under the curve in probability calculations.

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nhrock3
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2ngr2uo.jpg


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

??
 
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the blue line should be parallel to the horizontal axis
 
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.
 
no if we go by du then we go by the u axes
 
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
 

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