I4.1 In Figure 1 , let the area of the closed region bounded by the straight line $x+y=6$ and $y=4, x=0$ and $y=0$ be $A$ square units, find the value of $A$.
As shown in the figure, the intersection points
$$
\begin{array}{l}
P(6,0), Q(2,4), R(0,6) \\
O P=6, O R=4, Q R=2 \\
\text { Area }=A=\frac{1}{2}(6+2) \cdot 4=16
\end{array}
$$