Question

What is the median of the function `x^2` over the interval #[-1,1]?

Answer 1

The median is zero

Explanation:

Consider:

`int_(-1)^(1) \ f(x) \ dx =int_(-1)^(1) \ x^2 \ dx`
`" " = [ x^3/3 ]_(-1)^(1)`
`" " = (1/3)-(-1/3)`
`" " = 2/3`

So the median value of the function is the value `M` st:

`int_(-1)^(M) \ f(x) \ dx =1/2 \ int_(-1)^(1) \ f(x) \ dx`

In this case (using symmetry) the solution `M=0` is trivial and we can explicitly demonstrate this using:

`int_(-1)^(M) \ x^2 \ dx =1/2 \ int_(-1)^(1) \ x^2 \ dx`
`:. [ x^3/3 ]_(-1)^(M) =1/2 * 2/3`
`:. (M^3/3) - (-1/3) = 1/3`
`:. M^3/3=0`
`:. M=0`