Value of Nested Function for Integral Defined Function

To find the value of $f(f(1))$, we first determine the value of $f(1)$ and then derive the explicit functional form of $f(x)$.

Step 1: Find the value of $f(1)$

The given integral-defined function is: $f(x) = \int_{1}^{x} f(t) \, dt + (1-x)(\ln x - 1) + e$

Substituting $x = 1$ into the equation: $f(1) = \int_{1}^{1} f(t) \, dt + (1-1)(\ln 1 - 1) + e$ $f(1) = 0 + 0 + e = e$

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Step 2: Solve the differential equation to find $f(x)$

Differentiating both sides of the function $f(x)$ with respect to $x$ using Leibniz's Rule and the product rule: $f'(x) = \frac{d}{dx} \left( \int_{1}^{x} f(t) \, dt \right) + \frac{d}{dx} \left[ (1-x)(\ln x - 1) \right] + 0$ $f'(x) = f(x) + (-1)(\ln x - 1) + (1-x)\left(\frac{1}{x}\right)$ $f'(x) = f(x) - \ln x + 1 + \frac{1}{x} - 1$ $f'(x) - f(x) = \frac{1}{x} - \ln x$

This is a first-order linear differential equation of the form: $\frac{df}{dx} + P(x)f(x) = Q(x)$ where $P(x) = -1$ and $Q(x) = \frac{1}{x} - \ln x$.

The Integrating Factor ($\text{I.F.}$) is given by: $\text{I.F.} = e^{\int -1 \, dx} = e^{-x}$

Multiplying both sides by the integrating factor: $e^{-x} \left( f'(x) - f(x) \right) = e^{-x} \left( \frac{1}{x} - \ln x \right)$ $\frac{d}{dx} \left( f(x) e^{-x} \right) = e^{-x} \left( \frac{1}{x} - \ln x \right)$

Integrating both sides with respect to $x$: $f(x) e^{-x} = \int e^{-x} \left( \frac{1}{x} - \ln x \right) dx$

Notice that $\frac{d}{dx} \left( e^{-x} \ln x \right) = e^{-x} \cdot \frac{1}{x} - e^{-x} \ln x = e^{-x} \left( \frac{1}{x} - \ln x \right)$. Thus: $f(x) e^{-x} = e^{-x} \ln x + C$ $f(x) = \ln x + C e^x$

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Step 3: Determine the constant $C$ and calculate $f(f(1))$

Using the initial condition $f(1) = e$: $f(1) = \ln(1) + C e^1 = e$ $0 + C e = e \implies C = 1$

So, the function is: $f(x) = \ln x + e^x$

Now, calculating $f(f(1)) = f(e)$: $f(e) = \ln(e) + e^e = 1 + e^e$

Thus, the value of $f(f(1))$ is $(1 + e^e)$.