alrighty
find where they intersect
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]
the area between the curves is 48ln(2)-29
