Evaluating a definite integral via trapezoid area

The region under $f(x) = x+2$ from $x=-2$ to $x=2$ is a trapezoid. At $x=-2$, $f(-2) = 0$, and at $x=2$, $f(2) = 4$; these are the lengths of the two parallel sides. The height of the trapezoid, measured horizontally, is the width of the interval: $2 - (-2) = 4$.

Since $f(x) \geq 0$ on $[-2,2]$, the definite integral equals the ordinary (positive) area of this region. Using the trapezoid area formula $A = \frac{1}{2}(b_1 + b_2)h$, we get $A = \frac{1}{2}(0+4)(4) = 8$. Therefore $\int_{-2}^{2}(x+2)\,dx = 8$.