Let f : [0 , ) R be a differentiable function such that f (x )=1-2x+ _0^x 0.05eme^x-tf (t ) dt for all x ∈ [0 , ). Then the area of the region bounded by y=f (x ) and the coordinate axes is

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Correct answer: D. (d) ∵f (x )=1-2x+ 0.05eme^x-tf (t ) dt or, f (x )=1-2x+e^x 0.05eme^-tf (t ) dt on differentiating both sides w.r.t. x we get f^ ' (x )=-2+e^x 0.05em 0.05eme^-tf (t ) dt+e^x e^-xf (x ) f^ ' (x )=-2+f (x )+2x-1+f (x )\ from eq . (i )\ ∴f (x )-2f (x )=2x-3 I .F .=e^ -2 dx=e^-2x ∴e^-2x f (x )= e^-2x (2x-3 )dx e^-2x f (x )= (2x-3 ) e^-2x-2- 2 e^-2x-2dx e^-2x f (x )= (2x-3 )e^-2x-2+ e^-2x-2+c f (x )=-x+1+c^ 'e^2x ∵f (x )=1 from eq . (i ) ∴1=0+1+c^ ' ⇒c^ '=0 ∴f (x )=-x+1 ⇒ Area= 12

$Let f : [0, \infty) \to R be a differentiable function such that f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) dt for all x \in [0, \infty) . Then the area of the region bounded by y = f (x) and the coordinate axes is$

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