Find the area under the curve $y=x+2$, bounded by the x-axis and the y-axis.

The curve is $y=x+2$.The intercepts are:When $y=0$, $x=-2$.When $x=0$, $y=2$.The region is a right-angled triangle formed by the vertices $(0,0)$, $(0,2)$, and $(-2,0)$.
$$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$$
$$\text{Area}=\frac{1}{2}\times |-2|\times 2=\frac{1}{2}\times 2\times 2=2$$
Alternatively, using integration:
$$\text{Area}=\left|{\int }_{-2}^0(x+2)dx\right|=\left|{\left[\frac{x^2}{2}+2x\right]}_{-2}^0\right|$$
$$=\left|0-\left(\frac{(-2{)}^2}{2}+2(-2)\right)\right|=|-(2-4)|=|-(-2)|=2$$