Let for some function y=f(x), _x^0tf(t)dt=x^2f(x),x0 and f(2)=3.

Question

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

Accepted Answer

$\begin{gathered} {\int }_{x}^{0}tf(t)dt={x}^{2}+(x),x>0 \ x \ xf(x)={x}^{2}{f}^{'}(x)+2xf(x) \ -xf(x)={x}^{2}{f}^{'}(x) \ \int \frac{{f}^{'}(x)}{f(x)}dx=\int \frac{-1}{x}dx \ logdf(x)=-\log x+\log c \ f(x)=\frac{c}{x} \ f(2)=3\Rightarrow 3=\frac{c}{2}\Rightarrow c=6 \ f(x)=\frac{6}{x} \ f(6)=1\, 	herefore (1) \end{gathered}$

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