1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Median of f(x) - quadratic

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that:

    f(x) = 2(1-x) for 0 < x < k
    ...........0 for otherwise

    sketch the graph of f(x) and find:

    (i) the value of k,

    (ii) the median of x

    2. Relevant equations

    3. The attempt at a solution

    (i) I sketched the graph and got a line from (0,2) to (1,0) which has an area of 1, so k = 1.

    (ii) [tex]^{}M[/tex][tex]_{}0[/tex][tex]\int[/tex](2 - 2x) = 1/2
    The code is messing up but intergation of (2 - 2x) with m (median) as the top limit and 0 being the bottom one and it = 1/2

    [2x - x^2]m0 = 1/2

    2m - m2 = 1/2

    I'm stuck on getting m separated as it's a negative square

    Help me please :(
  2. jcsd
  3. Sep 30, 2010 #2


    User Avatar
    Homework Helper

    Try arranging the equation so all terms are on one side.
  4. Sep 30, 2010 #3


    User Avatar
    Science Advisor

    You don't really need to integrate to get this. y= 2(1- x) is a straight line from (0, 2) to (k, 2(1- k)).

    If k= 1, the area under it is the area of the triangle with height 2 and base 1 and so area (1/2)(2)(1= 1. That is what we need to have in order to have a valid probability distribution.

    As a check, if k< 1, the area under the line is the area of the trapezoid with bases 2 and 2(1- k) and height k. It's area is k(2+ 2(1- k))/2= k(4- 2k)/2= k(2- k) and we want that equal to 1. [itex]k(2- k)= 2k- k^2= 1[/itex] is the same as [itex]k^2- 2k+ 1= (k- 1)^2= 0[/itex]. As before, it reduces to k= 1, the triangle case.
    (You say "I sketched the graph and got a line from (0,2) to (1,0)". Well, you couldn't know it would be a line from (0, 2) to (1, 0) until after you knew that k= 1.)

    Now the median is where the area both left and right is equal to 1/2. If we take that to be at "x", then we have a triangle on the right with base 1- x and height 2(1- x) which has area [itex](1/2)(1-x)(2(1-x))= (1- x)^2[/itex] and we want that equal to 1/2: [itex](1- x)^2= 2[/itex]. Solve that for x.

    If you really want to integrate, you are integrating
    [tex]\int_0^x 2(1- t)dt= 2 \int_0^x 1- t dt= 2\left(x- (1/2)x^2)= 1/2[/itex]] which gives the same result.

    You can simplify that integral by letting u= 1- t so that du= -dt, when t= 0, u= 1 and when t= x, u= 1- x. Now the integral is [itex]2\int_0^{1- x}u^2du= 1/2[/=itex], giving, again, the same result.
  5. Sep 30, 2010 #4
    Thanks for the input to both of you :)

    I understand now thanks, HallsOfIvy :smile: woo!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook