 | f(X)=(X＾２＋aX∫[０→１]f(t)dt−１)e＾X=(X＾２＋AX−１)e＾X
(但し、A=a∫[０→１]f(t)dt)
f'(1)={(X＾２＋(A+2)X+A−１)e＾X}[x=1]=0
∴A=-1
一方、A=a∫[０→１]f(t)dt=a∫[０→１](t＾２＋At−１)e＾tdt
={a(t＾２＋At−１)e＾t}[０→１]-a∫[０→１](２t＋A)e＾tdt
=a(Ae+1)-{a(２t＋A)e＾t}[０→１]+a{2e＾t}[０→１]
=a{A(e-e+1)+1-2+2e-2}=a{A+2e-3}
-1=a{-1+2e-3} ∴a=1/(4-2e)
f(x)=(X＾２＋AX−１)e＾X =(X＾２-X-1)e＾X
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