Let for some function y=f(x), _0^x t f(t) d t=x^2 f(x), x0 and f(2)=3.

Question

Let for some function $\mathrm{y}=f(x), \int_0^x t f(t) d t=x^2 f(x), x>0$ and $f(2)=3$. Then $f(6)$ is equal to

Accepted Answer

$$\begin{aligned}
&\int_0^{\mathrm{x}} \mathrm{tf}(\mathrm{t}) \mathrm{dt}=\mathrm{x}^2+(\mathrm{x}), \mathrm{x}>0\
&\text { Diff. both side w.r. to } x\
&\begin{aligned}
& x f(x)=x^2 f^{\prime}(x)+2 x f(x) \
& -x f(x)=x^2 f^{\prime}(x) \
& \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-1}{x} d x \
& \operatorname{logdf}(x)=-\log x+\log c \
& f(x)=\frac{c}{x}
\end{aligned}\
\&\mathrm{f}(2)=3 \Rightarrow 3=\frac{\mathrm{c}}{2} \Rightarrow \mathrm{c}=6\
\&f(x)=\frac{6}{x}\
\&f(6)=1\quad\therefore (1)
\end{aligned}$$

Solutions · 1