计算不定积分∫Ly^2(x+y)dx,其中L为从(0,1)到(1,0)点的直线段,

Question

计算不定积分 ∫L y^2(x+y)dx ,其中L为从(0,1)到(1,0)点的直线段,

Answer 1

连接(0,1)和(1,0)的直线段为(y - 0)/(x - 1) = (1 - 0)/(0 - 1) = - 1
==> y = 1 - x
∫L y²(x + y) dx
= ∫ (1 - x)²(x + 1 - x) dx
= ∫ (1 - 2x + x²) dx
= (x - x² + x³/3)：
= 1 - 1 + 1/3
= 1/3