Question

Find the area of the region bounded by the given curves. y = 9x2 ln(x), y = 36 ln(x)

Answer 1

Alrighty
find where they intersect

9x²ln(x)=36ln(x)
divide both sides by 9
x²ln(x)=4ln(x)
$ln(x^{x^2})=ln(x^4)$
so
$x^{x^2}=x^4$
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
$\int\limits^2_1 {36ln(x)-9x^2ln(x)} \, dx=$
$[36x(ln(x)-1)-x^3(3ln(x)-1)]^2_1=$
$48ln(2)-29$

the area between the curves is 48ln(2)-29