If $x \in [0,1)$, $f(x) = \int_0^x \lfloor t \rfloor \, dt= \int_0^x 0\, dt=0$.
If $x \in [1,2)$, $f(x) = \int_0^1 \lfloor t\rfloor \, dt + \int_1^x \lfloor t \rfloor \, dt= \int_0^1 0\, dt + \int_1^x 1\, dt=x-1$.
If $x \in [2, 3)$, $f(x) = \int_0^1 \lfloor t\rfloor \, dt + \int_1^2 \lfloor t\rfloor \, dt + \int_2^x \lfloor t \rfloor \, dt= \int_0^1 0\, dt + \int_1^2 1\, dt + \int_2^x 2\, dt=1+2(x-2)$.
In general, if $x \in [n, n+1), n \in \mathbb{N} \cup \{0\}$,
\begin{align}
f(x) &= \sum_{i=0}^{n-1} \int_i^{i+1} \lfloor t \rfloor \, dt + \int_n^x \lfloor t \rfloor \, dt \\
&= \sum_{i=0}^{n-1} \int_i^{i+1} i \, dt + \int_n^x n \, dt \\
&= \left(\sum_{i=0}^{n-1} i \right)+ n(x-n)
\end{align}
Try to simplify the last term.
If you sub $u=1-t$, you have
$$\int_{u=1}^{u=0}(-1)^{\lfloor{1994-1994u\rfloor}} (-1)^{\lfloor{1995-1995u\rfloor}} \binom{1993}{\lfloor{1994-1994u\rfloor}}\binom{1994}{\lfloor{1995-1995u\rfloor}} (-du)$$
Now note that there are values of $u$ where $\lfloor{1994-1994u\rfloor}$ is an integer, and at such places $\lfloor{1994-1994u\rfloor}\neq1993-\lfloor{1994u\rfloor}$. However those places have measure $0$ and contribute nothing to the integral. Everywhere else, where $\lfloor{1994-1994u\rfloor}$ is not an integer, then $\lfloor{1994-1994u\rfloor}=1993-\lfloor{1994u\rfloor}$.
And there is the analogous statement for $1995$. So you have:
$$-\int_{u=1}^{u=0}(-1)^{1993-\lfloor{1994u\rfloor}} (-1)^{1994-\lfloor{1995u\rfloor}} \binom{1993}{1993-\lfloor{1994u\rfloor}}\binom{1994}{1994-\lfloor{1995u\rfloor}} du$$
And for more basic reasons, this is equal to
$$-\int_{u=0}^{u=1}(-1)^{\lfloor{1994u\rfloor}} (-1)^{\lfloor{1995u\rfloor}} \binom{1993}{\lfloor{1994u\rfloor}}\binom{1994}{\lfloor{1995u\rfloor}} du$$
This is the negative of what we started with, using $u$ as the integration variable instead of $t$. So it all must be $0$.
Best Answer
Hint: try to prove (or find a prove in your textbook) that a function is Riemann integrable if it has only finitely many discontinuities in the interval of integration.
More generally, a function is Riemann integrable if it has countable many discontinuities. You might also want to try to find a prove for this.