问题
已知积分路径 $L$ 关于 $y$ 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 $x$ 轴右侧的部分。)
选项
[A]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(-x, y)=-f(x, y), \\2 \int_{L_{1}} f(x, y) \mathrm{d} s, & f(-x, y)=f(x, y). \end{array}\right.$[B]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(-x, y)=-f(x, y), \\\int_{2 L_{1}} f(x, y) \mathrm{d} s, & f(-x, y)=f(x, y). \end{array}\right.$
[C]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 0, & f(-x, y)=-f(x, y), \\\int_{L_{1}} f(x, y) \mathrm{d} s, & f(-x, y)=f(x, y). \end{array}\right.$
[D]. $\int_{L}$ $f(x, y)$ $\mathrm{d} s$ $=$ $\left\{\begin{array}{ll} 1, & f(-x, y)=-f(x, y), \\2 \int_{L_{1}} f(x, y) \mathrm{d} s, & f(-x, y)=f(x, y). \end{array}\right.$