Let $$f : [0, \infty) \to \mathbb{R}$$ be differentiable function such that $$f(x) = 1 - 2x + \displaystyle\int_0^x e^{x-t} f(t) \, dt$$ for all $$x \in [0, \infty)$$. Then the area of the region bounded by $$y = f(x)$$ and the coordinate axes is

We start from the functional equation
$$f(x)=1-2x+\displaystyle\int_{0}^{x}e^{\,x-t}\,f(t)\,dt\qquad\bigl(x\in[0,\infty)\bigr).$$

Differentiate both sides with respect to $$x$$. The derivative of an integral with variable upper limit is obtained from Leibniz rule:

$$ \frac{d}{dx}\int_{0}^{x}e^{\,x-t}\,f(t)\,dt =\;e^{\,x-x}\,f(x)+\int_{0}^{x}\frac{\partial}{\partial x}\bigl[e^{\,x-t}\bigr]\,f(t)\,dt =f(x)+\int_{0}^{x}e^{\,x-t}\,f(t)\,dt. $$

Denote the integral in the given relation by $$I(x)$$, so that $$I(x)=\displaystyle\int_{0}^{x}e^{\,x-t}\,f(t)\,dt.$$ Using the original equation we have $$I(x)=f(x)-1+2x.$$ Hence

$$ f'(x)=-2+\bigl[f(x)+I(x)\bigr] =-2+\bigl[f(x)+f(x)-1+2x\bigr] =2f(x)+2x-3.\qquad-(1) $$

Equation $$(1)$$ is a linear first-order differential equation

$$ f'(x)-2f(x)=2x-3. $$

Its integrating factor is $$e^{-2x}$$. Multiplying throughout,

$$ e^{-2x}f'(x)-2e^{-2x}f(x)=\bigl(e^{-2x}f(x)\bigr)' =(2x-3)\,e^{-2x}. $$

Integrate from $$0$$ to $$x$$:

$$ e^{-2x}f(x)-e^{0}f(0)=\int_{0}^{x}(2t-3)\,e^{-2t}\,dt.\qquad-(2) $$

From the given relation at $$x=0$$, $$f(0)=1.$$

Next evaluate the integral on the right of $$(2)$$. First find an antiderivative:

$$ \int(2t-3)e^{-2t}\,dt =\int2t\,e^{-2t}\,dt-\int3e^{-2t}\,dt. $$

Using integration by parts for the first term (or a standard formula) and direct integration for the second,

$$ \int2t\,e^{-2t}\,dt=-t\,e^{-2t}-\tfrac12\,e^{-2t},\qquad \int3e^{-2t}\,dt=-\tfrac32\,e^{-2t}. $$

Therefore $$\int(2t-3)e^{-2t}\,dt=-t\,e^{-2t}+e^{-2t}=e^{-2t}(1-t)+C.$$

Applying limits $$0$$ to $$x$$:

$$ \int_{0}^{x}(2t-3)e^{-2t}\,dt =e^{-2x}(1-x)-1.\qquad-(3) $$

Substituting $$(3)$$ and $$f(0)=1$$ into $$(2)$$ gives

$$ e^{-2x}f(x)-1=e^{-2x}(1-x)-1. $$

Add $$1$$ to both sides and multiply by $$e^{2x}$$:

$$ f(x)=1-x\quad\text{for all }x\ge 0.\qquad-(4) $$

The curve $$y=f(x)$$ is the straight line $$y=1-x$$. Its intercepts with the coordinate axes are
• on the $$y$$-axis: $$(0,1)$$, • on the $$x$$-axis: $$(1,0)$$.

Thus the region bounded by the curve and the two coordinate axes is the right-triangle with vertices $$(0,0),(0,1),(1,0)$$. Its area is

$$ \text{Area}=\frac12\,(\,\text{base}\,)\times(\,\text{height}\,) =\frac12\,(1)\,(1)=\frac12. $$

Hence the required area equals $$\dfrac12$$, which matches Option B.