Suppose a continuous function f:(0, infty) ri | vedclass

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(A) Given the continuous function $f:[0, \infty) \rightarrow R$ satisfying $f(x) = 2 \int_0^x t f(t) d t + 1, \forall x \geq 0$. Applying the Leibniz

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$e$

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(A) Given the continuous function $f:[0, \infty) \rightarrow R$ satisfying $f(x) = 2 \int_0^x t f(t) d t + 1, \forall x \geq 0$. Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$: $f'(x) = 2x f(x)$. Rearranging the terms,we get $\frac{f'(x)}{f(x)} = 2x$. Integrating both sides with respect to $x$: $\int \frac{f'(x)}{f(x)} dx = \int 2x dx$. $\ln |f(x)| = x^2 + C$. $f(x) = K e^{x^2}$,where $K = e^C$. To find $K$,we evaluate the original equation at $x=0$: $f(0) = 2 \int_0^0 t f(t) dt + 1 = 0 + 1 = 1$. Substituting $x=0$ into $f(x) = K e^{x^2}$,we get $f(0) = K e^0 = K$. Thus,$K = 1$,which gives $f(x) = e^{x^2}$. Finally,$f(1) = e^{(1)^2} = e^1 = e$.

Suppose a continuous function $f:(0, \infty) \rightarrow R$ satisfies $f(x)=2 \int_0^x t f(t) d t+1, \forall x \geq 0$. Then,$f(1)$ equals

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