What is the Median of a Quadratic Function f(x) with a Given Range and Area?

[Maatttt0]

Homework Statement

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

Homework Equations

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) $$^{}M$$$$_{}0$$$$\int$$(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 :(

 

[statdad]

Try arranging the equation so all terms are on one side.

 

[HallsofIvy]

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. $k(2- k)= 2k- k^2= 1$ is the same as $k^2- 2k+ 1= (k- 1)^2= 0$. 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 $(1/2)(1-x)(2(1-x))= (1- x)^2$ and we want that equal to 1/2: $(1- x)^2= 2$. 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.

 

[Maatttt0]

Thanks for the input to both of you :)

I understand now thanks, HallsOfIvy woo!