6. B From the graph, we know that when $x \geqslant 0$, $f^{\prime}(x) \geqslant 0$, so $f(x)$ is increasing on $[0,+\infty)$. Since $a \geqslant 0, b \geqslant 0$, it follows that $2 a+b \geqslant 0$. Therefore,
$$
f(2 a+b)<f(4) \Leftrightarrow 2 a+b<4 .
$$

The given planar region is $\left\{\begin{array}{l}a \geqslant 0, \\ b \geqslant 0, \\ 2 a+b<4,\end{array}\right.$ with an area of $\frac{1}{2} \times 2 \times 4=4$. Choose B.
6. Given that the domain of the function $f(x)$ is $[-2,+\infty)$, and $f(4)=f(-2)=$ 1. $f^{\prime}(x)$ is the derivative of $f(x)$, and the graph of the function $y=f^{\prime}(x)$ is shown in the figure on the right. Then the area of the plane region $\left\{\begin{array}{l}a \geqslant 0, \\ b \geqslant 0, \\ f(2 a+b)<1\end{array}\right.$ is
A. 2
B. 4
C. 5
D. 8