I don't know whether you know any theorems about integration yet, but if you do, you can simplify this a good deal. Let
$$
g(x) = \begin{cases} 0 & |x| \ge 1 \\ 1 & Otherwise\end{cases}
$$
Then $f$ and $g$ sum to the constant function 1. So
$$
\int_0^5 f(x) + g(x) dx = 5
$$
because the integral of a constant function is the constant value times the length of the interval (since that's the exact value of all upper and lower sums).
I'll now show that $\int_0^5 g(x) dx$ exists (and compute its value), and hence so does $\int_0^5 -g(x) dx$; that'll show that
$$
\int_0^5 f(x) + g(x) dx + \int_0^5 -g(x) = \int_0^5 f(x) dx
$$
exists, since the sum of integrable functions is integrable. Simplifying, that says that
$$
5 - \int_0^5 g(x) = \int_0^5 f(x) dx
$$
To finish up, I need to integrate $g$.
Now
$$
\int_0^5 g(x) dx =
\int_0^1 g(x) dx +
\int_1^5 g(x) dx
$$
The rightmost integral is zero, because $g$ is zero on that interval; the middle integral is the integral of the constant function 1 on an interval of length 1...so it's 1. So we get
$$
\int_0^5 g(x) dx = 1
$$
and hence
$$
\int_0^5 f(x) dx = 5 - 1 = 4.
$$