007B Sample Midterm 2, Problem 4 Detailed Solution

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

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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  $1<x<e,$  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
${\begin{array}{rcl}\displaystyle {\int _{1}^{e}\ln x~dx}&=&\displaystyle {x\ln(x){\bigg |}_{1}^{e}-\int _{1}^{e}dx}\\&&\\&=&\displaystyle {x\ln(x)-x{\bigg |}_{1}^{e}}\\&&\\&=&\displaystyle {e\ln(e)-e-(\ln(1)-1)}\\&&\\&=&\displaystyle {e-e-(0-1)}\\&&\\&=&\displaystyle {1.}\end{array}}$

Final Answer:
$1$

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