Let f(x) = int_{0}^{x} e^{t} f(t) dt + e^{x} | vedclass

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(A) Given $f(x) = \int_{0}^{x} e^{t} f(t) dt + e^{x}$.At $x=0$,$f(0) = \int_{0}^{0} e^{t} f(t) dt + e^{0} = 0 + 1 = 1$.Differentiating both sides with

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$2 e^{(e^{x}-1)}-1$

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(A) Given $f(x) = \int_{0}^{x} e^{t} f(t) dt + e^{x}$.At $x=0$,$f(0) = \int_{0}^{0} e^{t} f(t) dt + e^{0} = 0 + 1 = 1$.Differentiating both sides with respect to $x$ using the Leibniz rule:$f'(x) = e^{x} f(x) + e^{x} = e^{x}(f(x) + 1)$.Rearranging the terms,we get $\frac{f'(x)}{f(x) + 1} = e^{x}$.Integrating both sides with respect to $x$ from $0$ to $x$:$\int_{0}^{x} \frac{f'(t)}{f(t) + 1} dt = \int_{0}^{x} e^{t} dt$.$[\ln(f(t) + 1)]_{0}^{x} = [e^{t}]_{0}^{x}$.$\ln(f(x) + 1) - \ln(f(0) + 1) = e^{x} - e^{0}$.Since $f(0) = 1$,we have $\ln(f(x) + 1) - \ln(2) = e^{x} - 1$.$\ln\left(\frac{f(x) + 1}{2}\right) = e^{x} - 1$.Taking the exponential of both sides:$\frac{f(x) + 1}{2} = e^{(e^{x} - 1)}$.$f(x) = 2 e^{(e^{x} - 1)} - 1$.

Let $f(x) = \int_{0}^{x} e^{t} f(t) dt + e^{x}$ be a differentiable function for all $x \in R$. Then $f(x)$ equals ..... .

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