How do I find the median of this continous random variable?

Question

The question given was to find the median of $X$ given $f(x)=1.1e^{-1.1x}$
Random variable is $X≥0$

I have no issue working out the median of probability density functions with parameters like $1≤X≤5$ , I format them as:

$\int_1^m$(whatever the equation is)$dx=0.5$

So would I do the same for one with parameters $X≥0$?

Like, if would $X≥0$ using the original equation of $f(x)=1.1e^{-1.1x}$ turn into $\int_m^0 1.1e^{-1.1x} dx=0.5$

and $X≤0$ would turn into $\int_0^m 1.1e^{-1.1x} dx=0.5$ ?

I would then proceed to solve for m

Answer 1

$$\int_0^m 1.1e^{-1.1x}dx=0.5$$

integrate and you get:

$$1-e^{-1.1m}=0.5$$

$$e^{-1.1m}=0.5$$

$$m=\frac{\ln(0.5)}{-1.1}=\frac{\ln(2)}{1.1}$$

Answer 2

I'd like to add a remark on this specific exercise (in the general case you already have a detailed answer and so I won't dwell on it any longer).

We can observe that since $f(x) = 1.1 e^{-1.1 x} \mathbb{I}_{\{ x \ge 0 \}}$, we have $X \sim \text{Exp}(1.1)$. For this distribution the median is already known:

$$ m = \frac{\ln(2)}{\lambda} = \frac{\ln(2)}{1.1} $$