First set the equations equal to each other, and find the intersection points, then find the area between the curves.
color(red)(y=6x^2lnx)y=6x2lnx
color(blue)(y=24lnx)y=24lnx
Desmos
6x^2lnx=24lnx6x2lnx=24lnx
0=24lnx-6x^2lnx0=24lnx−6x2lnx
0=(-6)(lnx)(x^2-4)0=(−6)(lnx)(x2−4)
x=1,2x=1,2
By setting the equations equal to each other, I found that the functions intersect at x=1x=1 and x=2x=2:
"Area"=int_1^2(24lnx-6x^2lnx)dx Area=∫21(24lnx−6x2lnx)dx
Focus on the unbounded (indefinite) integral to solve:
int(24lnx-6x^2lnx)dx∫(24lnx−6x2lnx)dx
=-6 int (lnx)(x^2-4) dx=−6∫(lnx)(x2−4)dx
Use integration by parts: int color(blue)(u) color(green)(dv)= vu - int v du∫udv=vu−∫vdu
((u=color(blue)(lnx)),(du=1/x dx))((v=1/3x^3-4x),(dv=color(green)(x^2-4)))
=-6[(lnx)(1/3x^3-4x)- int(1/x)(1/3x^3-4x)dx]
=-6[(lnx)(1/3x^2-4x)-int(1/3x^2-4)dx]
=-6[(lnx)(1/3x^2-4x)-(1/9x^3-4x)]
=-6(lnx)(1/3x^3-4x)+2/3x^3-24x
Now go back to the definite integral:
int_1^2(24lnx-6x^2lnx)dx
=[-6(lnx)(1/3x^3-4x)+2/3x^3-24x]_1^2
=-6ln2(8/3-8) + 16/3 - 48 - 2/3 + 24
= 32ln2- 58/3
