Φ﹙x﹚在[a,b]上连续,f﹙x﹚=(x﹣b﹚∫﹙a,x﹚Φ﹙t﹚dt,则由罗尔定理,必有ξ∈﹙a,b)使发f'(ξ)=

Question

Φ﹙x﹚在[a,b]上连续,f﹙x﹚=(x﹣b﹚∫﹙a,x﹚Φ﹙t﹚dt,则由罗尔定理,必有ξ∈﹙a,b)使发f'(ξ)=

Answer 1

f﹙x﹚=(x﹣b﹚∫﹙a,x﹚Φ﹙t﹚dt,
f(a)=f(b)=0
则由罗尔定理,必有ξ∈﹙a,b)使发f'(ξ)=0
f'﹙x﹚=∫﹙a,x﹚Φ﹙t﹚dt+(x-b)Φ﹙x﹚
即有∫﹙a,ξ﹚Φ﹙t﹚dt+(ξ-b)Φ﹙ξ﹚=0