用定积分的换元法求上限2,下限1,1⼀[x√(x^2-1)]的定积分

2024-11-08 00:30:38
推荐回答(2个)
回答1:

令x^2-x=x(x-1)>0
x>1或 x2)(x^2-x)dx
=(x^3/3-x^2/2)|(-1->0) -(x^3/3-x^2/2)|(0->1) +(x^3/3-x^2/2)|(1->2)
=[(0-0)-(-1/3-1/2)]-[(1/3-1/2)-(0-0)]+[(8/3-4/2)-(1/3-1/2)]
=5/6+1/6 +5/6
=11/6

回答2: