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

May 14, 2017

${\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= 32 \ln 2 - \frac{58}{3}$

#### Explanation:

First set the equations equal to each other, and find the intersection points, then find the area between the curves.

$\textcolor{red}{y = 6 {x}^{2} \ln x}$
$\textcolor{b l u e}{y = 24 \ln x}$

$6 {x}^{2} \ln x = 24 \ln x$
$0 = 24 \ln x - 6 {x}^{2} \ln x$
$0 = \left(- 6\right) \left(\ln x\right) \left({x}^{2} - 4\right)$
$x = 1 , 2$

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

$\text{Area} = {\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$

Focus on the unbounded (indefinite) integral to solve:
$\int \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= - 6 \int \left(\ln x\right) \left({x}^{2} - 4\right) \mathrm{dx}$

Use integration by parts: $\int \textcolor{b l u e}{u} \textcolor{g r e e n}{\mathrm{dv}} = v u - \int v \mathrm{du}$
$\left(\begin{matrix}u = \textcolor{b l u e}{\ln x} \\ \mathrm{du} = \frac{1}{x} \mathrm{dx}\end{matrix}\right) \left(\begin{matrix}v = \frac{1}{3} {x}^{3} - 4 x \\ \mathrm{dv} = \textcolor{g r e e n}{{x}^{2} - 4}\end{matrix}\right)$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) - \int \left(\frac{1}{x}\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) \mathrm{dx}\right]$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{2} - 4 x\right) - \int \left(\frac{1}{3} {x}^{2} - 4\right) \mathrm{dx}\right]$

$= - 6 \left[\left(\ln x\right) \left(\frac{1}{3} {x}^{2} - 4 x\right) - \left(\frac{1}{9} {x}^{3} - 4 x\right)\right]$

$= - 6 \left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) + \frac{2}{3} {x}^{3} - 24 x$

Now go back to the definite integral:
${\int}_{1}^{2} \left(24 \ln x - 6 {x}^{2} \ln x\right) \mathrm{dx}$
$= {\left[- 6 \left(\ln x\right) \left(\frac{1}{3} {x}^{3} - 4 x\right) + \frac{2}{3} {x}^{3} - 24 x\right]}_{1}^{2}$

$= - 6 \ln 2 \left(\frac{8}{3} - 8\right) + \frac{16}{3} - 48 - \frac{2}{3} + 24$

$= 32 \ln 2 - \frac{58}{3}$