Evaluating a definite integral via triangle area

The region under $f(x) = x$ from $x=0$ to $x=4$ is bounded above by the line $y=x$, below by the $x$-axis, and on the right by the vertical line $x=4$. Since $f(x) \geq 0$ on this interval, this region is a right triangle sitting entirely above the $x$-axis.

The triangle has base $4$ (along the $x$-axis from $0$ to $4$) and height $4$ (the value of $f(4) = 4$). Using the area formula $A = \tfrac{1}{2} \cdot \text{base} \cdot \text{height}$, we get $A = \tfrac{1}{2}(4)(4) = 8$. Since the region lies entirely above the $x$-axis, the signed area equals the ordinary area, so $\int_0^4 x\,dx = 8$.