Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the value of \(\dfrac{g(0)}{g(2)}\), we first need to understand and possibly simplify the given functions \(f(x)\) and \(g(x)\). The function \(f(x) = \int_0^1 e^{x+t} f(t)\, dt\) can be rewritten as \(f(x) = e^x \int_0^1 e^t f(t)\, dt\). Let's denote \(C = \int_0^1 e^t f(t)\, dt\), which is a constant since it does not depend on \(x\). Thus, \(f(x) = Ce^x\).</p> <p><strong>Step 2:</strong> Substituting \(f(x) = Ce^x\) back into its original equation, we get \(Ce^x = e^x \int_0^1 e^t Ce^t\, dt\), which simplifies to \(C = C \int_0^1 e^{2t}\, dt\). Solving the integral, \(\int_0^1 e^{2t}\, dt = \left[\frac{1}{2}e^{2t}\right]_0^1 = \frac{1}{2}(e^2 - 1)\). Thus, \(C = C \cdot \frac{1}{2}(e^2 - 1)\). For non-trivial solutions where \(C \neq 0\), we must have \(\frac{1}{2}(e^2 - 1) = 1\), which is not true, indicating \(C = 0\) for \(f(x)\) to satisfy the given condition, leading to \(f(x) = 0\).</p> <p><strong>Step 3:</strong> Now, let's analyze \(g(x) = x + \int_0^1 e^{x+t} g(t)\, dt\). Substituting \(f(x) = 0\) into the equation for \(g(x)\) does not directly affect \(g(x)\) since \(f(x)\) and \(g(x)\) are defined independently. However, the form of \(g(x)\) suggests we can find a similar representation. Let's denote \(D = \int_0^1 e^t g(t)\, dt\), so \(g(x) = x + e^x D\). Substituting this form into its integral, we get \(D = \int_0^1 e^t (t + e^t D)\, dt\), which simplifies to \(D = \int_0^1 te^t\, dt + D \int_0^1 e^{2t}\, dt\).</p> <p><strong>Step 4:</strong> Calculating the integrals, \(\int_0^1 te^t\, dt = \left[(t-1)e^t\right]_0^1 = 1\) (using integration by parts where \(u = t\), \(dv = e^t dt\), thus \(du = dt\), \(v = e^t\)), and \(\int_0^1 e^{2t}\, dt = \frac{1}{2}(e^2 - 1)\) as before. So, \(D = 1 + D \cdot \frac{1}{2}(e^2 - 1)\). Solving for \(D\), we get \(D = \frac{1}{1 - \frac{1}{2}(e^2 - 1)} = \frac{2}{2 - e^2 + 1} = \frac{2}{3 - e^2}\).</p> <p><strong>Step 5:</strong> Now, let's find \(g(0)\) and \(g(2)\). \(g(0) = 0 + e^0 D = D = \frac{2}{3 - e^2}\). For \(g(2)\), \(g(2) = 2 + e^2 D = 2 + e^2 \cdot \frac{2}{3 - e^2}\). Simplifying, \(g(2) = 2 + \frac{2e^2}{3 - e^2}\). To find \(\frac{g(0)}{g(2)}\), we calculate \(\frac{\frac{2}{3 - e^2}}{2 + \frac{2e^2}{3 - e^2}}\).</p> <p><strong>Step 6:</strong> Simplifying the expression for \(\frac{g(