Question 1117609
{{{f(x)=(x*ln(x))^2}}}
Over the interval [1,3], the minimum value of the function occurs at {{{x=1}}},
{{{f(1)=(1*ln(1))^2=0}}}
and the maximum value occurs at {{{x=3}}},
{{{f(3)=(3*ln(3))^2=9*ln(3)^2}}}
So,
{{{f(1)*(3-1)<=int(f(x),dx,x=1,3)<=9*ln(3)^2*(3-1)}}}
{{{0<=int(f(x),dx,x=1,3)<=18*ln(3)^2}}}