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

Step 1: Rewrite the integral equation

The given integral equation involves $e^{t-x}$. We can rewrite $e^{t-x}$ as $e^t e^{-x}$. Since $e^{-x}$ is independent of the integration variable $t$, we can take it out of the integral. This simplifies the expression and prepares it for differentiation.

Step 2: Differentiate the equation

We differentiate both sides of the rewritten equation with respect to $x$. For the integral term, we use the product rule for differentiation, treating $e^{-x}$ as one function and $\int_0^x e^t f(t) \, dt$ as another. The derivative of the integral part is found using the Leibniz integral rule, which states that $\frac{d}{dx} \int_{a(x)}^{b(x)} g(x,t) dt = g(x, b(x)) b'(x) - g(x, a(x)) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} g(x,t) dt$. In our case, $a(x)=0$, $b(x)=x$, and $g(x,t) = e^t f(t)$, so the derivative is $e^x f(x)$.

Step 3: Simplify and form a differential equation

We substitute the original expression for $e^{-x} \int_0^x e^t f(t) \, dt$ back into the differentiated equation. This allows us to eliminate the integral term and simplify the equation into a first-order linear differential equation. The terms involving $f(x)$ cancel out.

Step 4: Integrate to find f(x)

We integrate $f'(x)$ with respect to $x$ to find $f(x)$. This yields a general solution with an integration constant $C$. To find $C$, we use the initial condition $f(0)$. From the original equation, substituting $x=0$ gives $f(0) = 1 - 2(0) + \int_0^0 e^{t-0} f(t) \, dt = 1$. Therefore, $f(0) = 1$. Substituting $x=0$ into our integrated function gives $f(0) = -0 - 0^2 + C = C$. So, $C=1$.

Step 5: Determine the function f(x)

With the constant of integration $C=1$, the function $f(x)$ is determined as a quadratic polynomial. This function defines the curve whose area we need to find.

Step 6: Find x-intercepts and set up the integral for area

To find the area bounded by $y=f(x)$ and the coordinate axes, we first need to find the $x$ -intercepts of $f(x)$ by setting $f(x)=0$. Using the quadratic formula, we find two roots. Since the domain of $f(x)$ is $[0, \infty)$, we are interested in the positive root, which is $x = \frac{-1 + \sqrt{5}}{2}$. The $y$ -intercept is $f(0)=1$. The area will be the integral of $f(x)$ from $0$ to this positive $x$ -intercept.

Step 7: Calculate the area

We integrate $f(x)$ from $0$ to the positive $x$ -intercept $x_0 = \frac{\sqrt{5}-1}{2}$. We evaluate the definite integral and substitute the limits. After careful calculation, the area is found to be $\frac{5\sqrt{5}-1}{24}$.

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