9x²ln(x)=36ln(x) divide both sides by 9 x²ln(x)=4ln(x) [tex]ln(x^{x^2})=ln(x^4)[/tex] so [tex]x^{x^2}=x^4[/tex] so x=1 and and 2 (x can't be 0 or -2 because ln(0) and ln(-2) don't exist)
so intersect at x=1 and x=2 which is on top?
9(1.5)²ln(1.5)=20.25ln(1.5) 36ln(1.5)=36ln(1.5) 36ln(1.5) is on top so
that will be the area is [tex] \int\limits^2_1 {36ln(x)-9x^2ln(x)} \, dx= [/tex] [tex] [36x(ln(x)-1)-x^3(3ln(x)-1)]^2_1=[/tex] [tex] 48ln(2)-29[/tex]
