Let f:[0, \infty) \rightarrow \mathbb{R} be a differentiable function such that f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t 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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Answer: D. \frac{1}{2} \because f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t or, f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t on differentiating both sides w.r.t. x we get \begin{aligned} & f^{\prime}(x)=-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x) \ & f^{\prime}(x)=-2+f(x)+2 x-1+f(x)\{	ext { from eq. (1) \}} \end{aligned} \begin{aligned} & 	herefore \quad f(x)-2 f(x)=2 x-3 \ & 	ext { I.F. }=e^{\int-2 d x}=e^{-2 x} \ & 	herefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x \ & e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x \ & e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c \ & f(x)=-x+1+c^{\prime} e^{2 x} \ & \because \quad f(x)=1 	ext { from eq. (1) } \ & 	herefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0 \ & 	herefore \quad f(x)=-x+1 \end{aligned} \Rightarrow \quad Area =\frac{1}{2}

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ 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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