Question

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

Answer 1

`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)`

`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`