2个回答
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题目是否 f(1) =0 ?
∫(0->1) f(x) dx
=[xf(x)]|(0->1) - ∫(0->1) xf'(x) dx
=f(1) -∫(0->1) xarcsin(1-x) dx
=0 -(1/2) ∫(0->卜搭答1) arcsin(1-x) dx^2
=-(1/2) [x^2. arcsin(1-x)]|(0->1) -(1/2) ∫(0->1) x^2/√[ 1-(1-x)^2] dx
=0-(1/2) ∫(0->1) x^2/√[ 1-(1-x)^2] dx
=(1/4) (4 - 3π/2)
//
let
1-x = sinu
dx = -cosu du
x=0, u=π/2
x=1, u=0
∫(0->1) x^2/√[ 1-(1-x)^2] dx
=-∫(π/2->型慧0) (1-sinu)^2 du
=∫(0->π/2) (1-sinu)^2 du
=∫(0->π/2) [1-2sinu + (sinu)^2 ] du
=(1/枝喊2)∫(0->π/2) [3-4sinu - cos2u ] du
=(1/2)[ 3u +4cosu - (1/2)sin2u]|(0->π/2)
=(1/2) ( 3π/2 - 4)
∫(0->1) f(x) dx
=[xf(x)]|(0->1) - ∫(0->1) xf'(x) dx
=f(1) -∫(0->1) xarcsin(1-x) dx
=0 -(1/2) ∫(0->卜搭答1) arcsin(1-x) dx^2
=-(1/2) [x^2. arcsin(1-x)]|(0->1) -(1/2) ∫(0->1) x^2/√[ 1-(1-x)^2] dx
=0-(1/2) ∫(0->1) x^2/√[ 1-(1-x)^2] dx
=(1/4) (4 - 3π/2)
//
let
1-x = sinu
dx = -cosu du
x=0, u=π/2
x=1, u=0
∫(0->1) x^2/√[ 1-(1-x)^2] dx
=-∫(π/2->型慧0) (1-sinu)^2 du
=∫(0->π/2) (1-sinu)^2 du
=∫(0->π/2) [1-2sinu + (sinu)^2 ] du
=(1/枝喊2)∫(0->π/2) [3-4sinu - cos2u ] du
=(1/2)[ 3u +4cosu - (1/2)sin2u]|(0->π/2)
=(1/2) ( 3π/2 - 4)
追问
这个f(1),题里没给,感觉用中值定理得0,谢谢回答,挺完整
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