Let a differentiable function f satisfy f(x)+_3^x(f(t))/(t),dt = x+1, x3.

Question

Let a differentiable function $f$ satisfy $f(x)+\displaystyle\int_3^x\frac{f(t)}{t}\,dt = \sqrt x+1$, $x\ge3$. Then $12f(8)$ is equal to:

Accepted Answer

Differentiating: $f'(x)+\dfrac{f(x)}x = \dfrac1{2\sqrt x}$. Integrating factor $=e^{\int dx/x}=x$:

$$xf(x) = \int\frac{x}{2\sqrt x}\,dx = \int\frac{\sqrt x}2\,dx$$

With $x+1=u^2$ substitution (as in source): $xf(x) = \dfrac{(x+1)^{3/2}}3-\sqrt{x+1}+C$.

Using $x=3$ in the original equation: $f(3)=\sqrt4=2$, giving $3(2) = \dfrac{8}3-2+C \implies C=\dfrac{16}3$.

$$f(x) = \frac{(x+1)^{3/2}/3-\sqrt{x+1}+16/3}{x}$$

$$f(8) = \frac{9-3+16/3}{8} = \frac{34/3}{8}=\frac{34}{24} \implies 12f(8) = 17$$

Solutions · 1