2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ .
2. -1043 .

From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$

Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$