# 007B Sample Midterm 2, Problem 4 Detailed Solution

Find the area of the region bounded by  ${\displaystyle y=\ln x,~y=0,~x=1,}$  and  ${\displaystyle x=e.}$

Background Information:
1. You can find the intersection points of two functions, say  ${\displaystyle f(x),g(x),}$

by setting  ${\displaystyle f(x)=g(x)}$  and solving for  ${\displaystyle x.}$

2. The area between two functions,  ${\displaystyle f(x)}$  and  ${\displaystyle g(x),}$  is given by  ${\displaystyle \int _{a}^{b}f(x)-g(x)~dx}$

for  ${\displaystyle a\leq x\leq b,}$  where  ${\displaystyle f(x)}$  is the upper function and  ${\displaystyle g(x)}$  is the lower function.

3. Integration by parts tells us that
${\displaystyle \int u~dv=uv-\int v~du.}$

Solution:

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

${\displaystyle 0<\ln(x).}$

Step 2:
The area bounded by these functions is given by

${\displaystyle \int _{1}^{e}\ln x~dx.}$

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

${\displaystyle 1}$