How do you find the area of the region bounded by the given curves #y = 6x^2lnx # and #y = 24lnx#?

1 Answer
May 14, 2017

Answer:

#int_1^2(24lnx-6x^2lnx)dx#
#=32ln2 - 58/3#

Explanation:

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)#
#color(blue)(y=24lnx)#

Desmos

#6x^2lnx=24lnx#
#0=24lnx-6x^2lnx#
#0=(-6)(lnx)(x^2-4)#
#x=1,2#

By setting the equations equal to each other, I found that the functions intersect at #x=1# and #x=2#:

#"Area"=int_1^2(24lnx-6x^2lnx)dx #

Focus on the unbounded (indefinite) integral to solve:
#int(24lnx-6x^2lnx)dx#
#=-6 int (lnx)(x^2-4) dx#

Use integration by parts: #int color(blue)(u) color(green)(dv)= vu - int v du#
#((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#