B Impossible to find the quantile of any equation

[Addez123]

TL;DR Summary

Find the .5 quantile (which is same as median for continuous functions?) given the distribution function:
$$f(x) = 2xe^{-x^2}$$

Given the distribution function
$$f(x) = 2xe^{-x^2}$$
The probability density function would then be
$$F(x) = -e^{-x^2}$$

To find the .5 quantile I set F(x) = .5
$$.5 = -e^{-x^2}$$
$$ln(-.5) = -x^2$$

And already here we have the issue, you can't take ln of a negative number.

 

[Addez123]

Think I solved it. Issue was I wasn't solving the definitive integral, aka. I wasn't solving the integral with any specific range. Should've solved
$$F(x)= −e^{−x2} |_0^x \Leftrightarrow -e^{-x^2} + 1 = 0.5$$

 

[RPinPA]

Previous answer deleted. I was confused by your terminology.

$f(x)$ is a density, provided you include the restriction that $x \geq 0$, and $F(x)$ is the cumulative distribution, which is $\int_0^x f(t) dt$.

Sometimes the term "distribution" is used for the cumulative distribution function F(x), but it is never called a "probability density".

So you are correct, you need to do a definite integral starting at 0, which will give you the correct expression.