问题
已知积分路径 $L$ 关于 $x$ 轴对称, 则如何对第一类曲线积分 $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ 进行化简？（其中，积分路径 $L_{1}$ 是积分路径 $L$ 在 $y$ 轴上方的部分。）
选项
[A]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(x, -y)=-f(x, y), \\\int_{L_{1}} f(x, y) \mathrm{d} s, & f(x, -y)=f(x, y). \end{array}\right.$[B]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 1, & f(x, -y)=-f(x, y), \\2 \int_{L_{1}} f(x, y) \mathrm{d} s, & f(x, -y)=f(x, y). \end{array}\right.$
[C]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(x, -y)=-f(x, y), \\2 \int_{L_{1}} f(x, y) \mathrm{d} s, & f(x, -y)=f(x, y). \end{array}\right.$
[D]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(x, -y)=-f(x, y), \\\int_{2 L_{1}} f(x, y) \mathrm{d} s, & f(x, -y)=f(x, y). \end{array}\right.$
$\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(x, -y)=-f(x, y), \\2 \int_{L_{1}} f(x, y) \mathrm{d} s, & f(x, -y)=f(x, y). \end{array}\right.$
$f(-x, y)$ $=$ $-$ $f(x, y)$ $\Rightarrow$ $f(x, y)$ 是关于 $x$ 的奇函数.
$f(-x, y)$ $=$ $f(x, y)$ $\Rightarrow$ $f(x, y)$ 是关于 $x$ 的偶函数.