Evaluating a piecewise definite integral using additivity

By additivity, $\int_0^5 f(x)\,dx = \int_0^2 x\,dx + \int_2^5 (4-x)\,dx$. On $[0,2]$, $f(x)=x$ traces a right triangle with base $2$ and height $2$, giving area $\frac{1}{2}(2)(2) = 2$, and since the region lies above the x-axis this integral equals $2$.

On $[2,5]$, $f(x) = 4-x$ is positive on $[2,4]$ (a triangle with base $2$, height $2$, area $2$) and negative on $[4,5]$ (a triangle with base $1$, height $1$, area $\frac{1}{2}$, counted as $-\frac{1}{2}$ since it's below the axis). So $\int_2^5(4-x)\,dx = 2 - \frac{1}{2} = \frac{3}{2}$. Adding both pieces: $\int_0^5 f(x)\,dx = 2 + \frac{3}{2} = \frac{7}{2}$.