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

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Answer 1 · 2017-11-30T01:24:03

The median is zero

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:

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

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