Help calculating the following Integral?

Question

How can I calculate this? $\int_{0}^2 f(x)$ $\mathrm{dx}$ if $$f(x) = \begin{cases} x, & 0 \le x <1 \\ x-2 & 1 \le x \le 2 \end{cases} $$

Answer 1

Hint:
The integral you're working with is equivalent to $$\int_0^1 x\ dx +\int_1^2 (x-2)\ dx.$$

Can you take it from here?

Answer 2

$$\displaystyle \int\limits_{0}^{2}f(x)\mathrm{d}x = \int\limits_{0}^{1}f(x)\mathrm{d}x + \int\limits_{1}^{2}f(x)\mathrm{d}x$$