Get probability distribution function from density function and calculate the cumulative value For the given density function, how to find its distribution function and how to calculate the value of the distribution function?
Density function:
$$f(x) = \frac{1}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{1}{2}x}
\quad \quad \quad \quad \quad 
x \in (0, \infty), n \in \mathbb{N}.$$
I need to find distribution function and calculate its value, but don't really know how to start.

Firstly, your density function does not presently integrate to one --- to get the correct density you need to add an additional multiplicative term $2^{-n/2}$.  I will make this adjustment in my analysis.
For a continuous random variable, the distribution function is obtained by integrating the density function over the appropriate range (i.e., all values no greater than a stipulated input value).  In the present case the relevant integral leads you to the lower incomplete gamma function, which does not have a closed form.  To see this, we take the change-of-variable $t=r/2$ (giving $dt=dr/2)$ and we get the distribution function:
$$\begin{align}
F_X(x)
\equiv \mathbb{P}(X \leqslant x)
&= \int \limits_0^x f(r) \ dr \\[6pt]
&= \frac{2^{-n/2}}{\Gamma(\tfrac{n}{2})} \int \limits_0^x r^{n/2-1} e^{-r/2} \ dr \\[6pt]
&= \frac{1}{\Gamma(\tfrac{n}{2})} \int \limits_0^{x/2} t^{n/2-1} e^{-t} \ dt \\[6pt]
&= \frac{\gamma(\tfrac{n}{2},\tfrac{x}{2})}{\Gamma(\tfrac{n}{2})}, \\[6pt]
\end{align}$$
where $\gamma$ denotes the lower incomplete gamma function.  Unfortunately you can't "simplify" this integral except for special cases for $n$.  (For example, if $n$ is an odd integer then you can simplify to closed form using repeated application of integration by parts.)  In the general case, it is an integral that does not have a general closed form representation so we tend to just express it in its integral form.  As you gain more experience in calculus you will learn to spot integrals of this kind and save yourself the headache of trying unsuccessfully to put them into closed form.
The distribution you are dealing with is well-known, and is called the chi-squared distribution with $n$ degrees-of-freedom.  It is used widely in statistics and its properties are well-known.  It is also valid for any positive real value of $n$, and does not need to be restricted to an integer parameter.
