| 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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