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(B) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $f'(x) = -2 + e^x \int

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$\frac{1}{2}$

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(B) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $f'(x) = -2 + e^x \int_0^x e^{-t} f(t) dt + e^x (e^{-x} f(x)) = -2 + (f(x) - (1 - 2x)) + f(x) = 2f(x) + 2x - 3$. This is a linear differential equation: $\frac{dy}{dx} - 2y = 2x - 3$. The integrating factor is $IF = e^{\int -2 dx} = e^{-2x}$. Multiplying by $IF$: $\frac{d}{dx}(y e^{-2x}) = (2x - 3) e^{-2x}$. Integrating both sides: $y e^{-2x} = \int (2x - 3) e^{-2x} dx = (2x - 3) \frac{e^{-2x}}{-2} - \int 2 \cdot \frac{e^{-2x}}{-2} dx = -\frac{2x-3}{2} e^{-2x} - \frac{1}{2} e^{-2x} + C$. $y = -\frac{2x-3}{2} - \frac{1}{2} + C e^{2x} = -x + \frac{3}{2} - \frac{1}{2} + C e^{2x} = -x + 1 + C e^{2x}$. Since $f(0) = 1 - 0 + 0 = 1$,we have $1 = -0 + 1 + C e^0 \Rightarrow C = 0$. Thus,$f(x) = 1 - x$. The region bounded by $y = 1 - x$ and the coordinate axes is a triangle with vertices $(0,0), (1,0), (0,1)$. The area is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.

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