# 007B Sample Midterm 2, Problem 4 Detailed Solution

Find the area of the region bounded by  $y=\ln x,~y=0,~x=1,$ and  $x=e.$ Background Information:
1. You can find the intersection points of two functions, say  $f(x),g(x),$ by setting  $f(x)=g(x)$ and solving for  $x.$ 2. The area between two functions,  $f(x)$ and  $g(x),$ is given by  $\int _{a}^{b}f(x)-g(x)~dx$ for  $a\leq x\leq b,$ where  $f(x)$ is the upper function and  $g(x)$ is the lower function.

3. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$ Solution:

Step 1:
We start by finding the intersection points of the functions  $y=\ln x$ and  $y=0.$ So, we consider the equation  $0=\ln x.$ The only solution to this equation is  $x=1.$ Also, for  $1 we have

$0<\ln(x).$ Step 2:
The area bounded by these functions is given by

$\int _{1}^{e}\ln x~dx.$ Now, we need to use integration by parts.
Let  $u=\ln x$ and  $dv=dx.$ Then,  $du={\frac {1}{x}}~dx$ and  $v=x.$ Therefore, we have
${\begin{array}{rcl}\displaystyle {\int _{1}^{e}\ln x~dx}&=&\displaystyle {x\ln(x){\bigg |}_{1}^{e}-\int _{1}^{e}dx}\\&&\\&=&\displaystyle {x\ln(x)-x{\bigg |}_{1}^{e}}\\&&\\&=&\displaystyle {e\ln(e)-e-(\ln(1)-1)}\\&&\\&=&\displaystyle {e-e-(0-1)}\\&&\\&=&\displaystyle {1.}\end{array}}$ $1$ 