计算∫L(x2+y2)dx+(x2-y2)dy,其中L为连续O(0,0),A(1,1),
B(2,0)的折线OAB.
∫L(x2+y2)dx+(x2-y2)dy= ∫OA(x2+y2)dx+(x2-y2)dy+∫AB(x2+y2)dx+(x2-y2)dy 线段OA为y=x,0≤x≤1,线段AB为y=2-x,1≤x≤2, 所以∫L(x2+y2)dx+(x2-y2)dy =∫01[(x2+x2)+(x2-x2)•1]dx +∫12{[x2+(2-x)2]+[x2+(2-x)2]•(-1)}dx =2∫01x2dx+∫12(2x2-8x+8)dx =(2/3)x2∣01+(2/3)x2∣12 -4x2∣12+8x∣12=4/3