calculus - Prove that f is integrable on [0,2]

Let

\begin{align}
f(x)=\left\{\begin{matrix}1,\:\: 0\leq x\leq 1,\\
0,\:\:1\end{matrix}\right.
\end{align}

Prove that $f$ is integrable on $\left[0,2\right]$, and find the value of

$$\begin{align} \int_0^2 f\left(x\right)\:dx. \end{align}$$

In order to show that $f$ is integrable I think I need to use the following theorem:

The bounded function $f$ is integrable on $\left[a,b\right]$ if and only if for
every positive number $\epsilon$ there exists a partition $P$ of $\left[a,b\right]$
such that $|U\left(f,P\right) - L\left(f,P\right)|<\epsilon$.

The problem is that I'm not sure how to actually use this theorem to show it, I dont understand how I can find the value of the integral either, any tips solution? thanks!

Answer

This function is integrable by definition because

$$\begin{align} \int_0^2 f\left(x\right)\:dx=\int_0^1\:dx+\int_{1+}^2 0\:dx=1. \end{align}$$