计算瑕积分∫(1,2)x√x-1dx

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计算瑕积分∫(1,2)x√x-1dx计算瑕积分∫(1,2)x√x-1dx计算瑕积分∫(1,2)x√x-1dx∫(1→2)x√(x-1)dx=∫(1→2)(x-1+1)√(x-1)dx=∫(1→2)[(

计算瑕积分∫(1,2)x√x-1dx
计算瑕积分∫(1,2)x√x-1dx

计算瑕积分∫(1,2)x√x-1dx
∫(1→2) x√(x - 1) dx
= ∫(1→2) (x - 1 + 1)√(x - 1) dx
= ∫(1→2) [(x - 1)^(3/2) + √(x - 1)] d(x - 1)
= (2/5)(x - 1)^(5/2) + (2/3)(x - 1)^(3/2) |(1→2)
= 2/5 + 2/3
= 16/15

根(x-1)=t x=1时,t=0 x=2时,t=1 tE[0,1]
x-1=t^2
x=t^2+1
dx=2tdt
原式
=∫ (0,1) (t^2+1)t*2tdt
=2∫(0,1) [t^4+t^2)dt
2*[ 1/5t^5+1/3t^3] (0,1)
=2(1/5+1/3)
=2*8/15
=16/15