Let y=f(x) be a twice differentiable non-negative function such that [y^2 = 25 + _0^x ((f(…

Question

Let $y=f(x)$ be a twice differentiable non-negative function such that
$$y^2 = 25 + \int_0^x \big((f(t))^2+(f'(t))^2\big)\,dt.$$
Then the mean of $f(\log_e 1), f(\log_e 2), \dots, f(\log_e 625)$ is equal to \_\_\_\_\_\_.

Answer: 1565

Accepted Answer

Differentiating $y^2 = 25+\int_0^x\big(y^2+(y')^2\big)dt$ with respect to $x$:
$$2y\,y' = y^2+(y')^2 \implies (y'-y)^2=0 \implies y'=y.$$
So $\dfrac{dy}{y}=dx \implies \ln|y|=x+c$.

At $x=0$: $y(0)^2 = 25+0 \implies y(0)=5$ (non-negative), so $c=\ln 5$, giving
$$y = 5e^x, \qquad f(\ln k) = 5e^{\ln k} = 5k.$$

We need the mean of $\{5\cdot1, 5\cdot2, \dots, 5\cdot625\}$:
$$\text{Mean} = \frac{5(1+2+\cdots+625)}{625} = \frac{5\cdot\frac{625\cdot626}{2}}{625} = 5\times313 = 1565.$$

Solutions · 1