Let f be a twice differentiable non-negative | vedclass

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(B) Given the equation: $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Differentiating both sides with respect to $x$ using the Leibniz rul

Vedclass · Accepted Answer

$1565$

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(B) Given the equation: $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $2 f(x) f'(x) = (f(x))^2 + (f'(x))^2$. Rearranging the terms: $(f(x))^2 - 2 f(x) f'(x) + (f'(x))^2 = 0$. This simplifies to: $(f(x) - f'(x))^2 = 0$, which implies $f'(x) = f(x)$. Solving this differential equation: $\frac{f'(x)}{f(x)} = 1 \Rightarrow \ln(f(x)) = x + C \Rightarrow f(x) = A e^x$. At $x = 0$, $(f(0))^2 = 25 + 0 \Rightarrow f(0) = 5$ (since $f$ is non-negative). Thus, $A e^0 = 5 \Rightarrow A = 5$, so $f(x) = 5 e^x$. We need the mean of $f(\ln 1), f(\ln 2), \ldots, f(\ln 625)$. Since $f(\ln n) = 5 e^{\ln n} = 5n$, the mean is: $	ext{Mean} = \frac{1}{625} \sum_{n=1}^{625} 5n = \frac{5}{625} 	imes \frac{625 	imes 626}{2} = \frac{5 	imes 626}{2} = 5 	imes 313 = 1565$.

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to:

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