Integrals · Question 8

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Let f : [1, ) R be a differentiable function defined as f(x)= 1x f(t) dt +(1-x) ( e x-1 )+e. Then the value of f(f(1)) is :

Answer

(1) (1+ee )

Put x=1: f(1)=0+0+e=e

Differentiating (Leibnitz rule): f'(x)=f(x)- ( e x-1 )+(1-x) 1x

f'(x)-f(x)= 1x- e x (linear, I.F. =e-x)

ddx (e-xf(x) )=e-x ( 1x- e x )= ddx (e-x e x )

e-xf(x)=e-x e x+C

x=1: e-1 e=0+C C=1

f(x)= e x+ex

f(f(1))=f(e)= e e+ee=1+ee

Answer: (1)

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