设 $I_1=\iint_D(x+y) \operatorname{sgn}(x+y) \mathrm{d} x \mathrm{~d} y, I_2=\iint_D(x-y) \operatorname{sgn}(x-y) \mathrm{d} x \mathrm{~d} y$, 其中符号函数 $\operatorname{sgn} x=\left\{\begin{array}{l}1, x > 0, \\ 0, x=0, \\ -1, x < 0,\end{array}\right.$ 区域 $D=\{(x, y) \mid-1 \leqslant x \leqslant 1,-1 \leqslant y \leqslant 1\}$, 则
$ \text{A.} $ $I_1 > I_2$ $ \text{B.} $ $I_1 < I_2$ $ \text{C.} $ $I_1=I_2$ $ \text{D.} $ $I_1=-I_2$
【答案】 C

【解析】 $I_1=\iint_D|x+y| \mathrm{d} x \mathrm{~d} y, I_2=\iint_D|x-y| \mathrm{d} x \mathrm{~d} y$.
$$
\begin{aligned}
I_2 & =\iint_D|x-y| \mathrm{d} x \mathrm{~d} y=\int_{-1}^1 \mathrm{~d} x \int_{-1}^1|x-y| \mathrm{d} y \\
& \stackrel{y=-1}{=} \int_{-1}^1 \mathrm{~d} x \int_1^{-1}|x+t|(-\mathrm{d} t)=\int_{-1}^1 \mathrm{~d} x \int_{-1}^1|x+t| \mathrm{d} t \\
& =\int_{-1}^1 \mathrm{~d} x \int_{-1}^1|x+y| \mathrm{d} y=I_1 .
\end{aligned}
$$
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