连接(0,0)及(1,1)的线段是y = x,dy/dx = 1
∫L (x + y) ds
= ∫(0→1) (x + x)√(1 + (dy/dx)²) dx
= ∫(0→1) 2x√(1 + 1) dx
= √2 * x²|(0→1)
= √2