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.

Solution:

Step 1:
We start by finding the intersection points of these two functions.
So, we consider the equation  $\sin x=\cos x.$
In the interval  $0\leq x\leq {\frac {\pi }{4}},$  the solutions to this equation are
$x=0$  and  $x={\frac {\pi }{4}}.$
Also, for  $0<x<{\frac {\pi }{4}},$  we have
$\sin(x)<\cos(x).$

Step 2:
The area bounded by these functions is given by
$\int _{0}^{\frac {\pi }{4}}\cos x-\sin x~dx.$
Then, we integrate to get
${\begin{array}{rcl}\displaystyle {\int _{0}^{\frac {\pi }{4}}\cos x-\sin x~dx}&=&\displaystyle {\sin x+\cos x{\bigg |}_{0}^{\frac {\pi }{4}}}\\&&\\&=&\displaystyle {\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}+\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}-(\sin(0)+\cos(0))}\\&&\\&=&\displaystyle {{\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}-1}\\&&\\&=&\displaystyle {{\sqrt {2}}-1.}\end{array}}$

Final Answer:
${\sqrt {2}}-1$

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