1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the values of 'a' and 'b' for the following PDF

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A probability density function is defined by ##f\left(x\right)=\left|a\right|\left(x+1\right)-\left(ax-1\right)^2## where ##x∈[0,b]##. Determine the values of the constants ##a## and ##b##, given that the graph passes through the point ##(b,0)##.

    2. Relevant equations
    ##∫_a^bf(x)dx=1## where ##f(x)## is a probability density function

    3. The attempt at a solution
    Okay, so I know I'm attempting to solve ##∫_0^bf(x)dx=1## for ##b##, but I'm rather unsure of how to do that in this instance due to ##a## being a second unknown variable.

    I've attempted finding the x-intercept, ##b## of ##f(x)## but it is dependent on the variable ##a##. I then tried to find ##a## in terms of x and use that to somehow find ##b## but I soon realised that wasn't going anywhere. Normally for these types of questions, ##a## would just be a dilation factor, meaning it doesn't effect the values of the x-intercepts and can just be taken out as a common factor before integrating. I'm not sure why I'm supposed to do when it effects the location of the x-intercepts.
     
  2. jcsd
  3. Sep 8, 2016 #2

    Mark44

    Staff: Mentor

    With the integral you show in your attempt and the given information that f(b) = 0, you have two equations in the unknowns a and b. That should be enough information for you to solve for a and b.
     
  4. Sep 8, 2016 #3
    Oh thank you I think I understand now. Because ##b## is the x-intercept, ##f(x)=f(b)=0##, leaving us with just ##a## and ##b## as unknowns. As ##∫_0^bf(x)dx=1## also simplifies down to only containing ##a## and ##b##. Thus, I should be able to solve ##f(b)=0## and ##∫_0^bf(x)dx=1## simultaneously to find both ##a## and ##b##. Does this sound right?
     
  5. Sep 8, 2016 #4

    Mark44

    Staff: Mentor

    That's pretty much what I said.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted