Hello, ahawk1!
I wish you'd shown us your work . . .
Find the intersections of the curves . . .Find the area of the region bounded by the given curves: .$\displaystyle \begin{array}{cccc}y &=& 35\ln x & {\color{blue}[1]}\\ y&=&x\ln x & {\color{blue}[2]}\end{array}$
$\displaystyle x\ln x \:=\:35\ln x \quad\Rightarrow\quad x\ln x - 35\ln x \:=\:0 \quad\Rightarrow\quad \ln x\bigg(x - 35\bigg) \:=\:0$
And we have: .$\displaystyle \begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & x \:=\:1 \\ x-35 \:=\:0 & \Rightarrow & x \:=\:35 \end{array}$
On the interval [1, 35], [1] is the upper curve.
The integral is: .$\displaystyle \int^{35}_1\bigg[35\ln x - x\ln x\bigg]\,dx \;=\;35\int^{35}_1\ln x\,dx - \int^{35}_1 x\ln x\,dx
$
Both integrals require integration-by-parts.
We get: .$\displaystyle 35x(\ln x - 1) - \tfrac{1}{4}x^2(2\ln x - 1)\,\bigg]^{35}_1$ . . . . etc.