Question

The area of the region bounded by $x=0,y=0,x=2,y=2,y\ge e^x$ and $y\le \ell nx$

• A. $2e^2+(e^2-1)\log (e^2+1)+2+e+C$
• B. $2e^3-(e^3-1)\log (e^2-1)+2+e+C$
• C. $2e^2-(e^2-1)\log (e^2-1)+2+e+C$
• D. $2e^5-(e^2+1)\log (e^2-1)+2+e+C$

Answer 1

The correct option is C $2e^2-(e^2-1)\log (e^2-1)+2+e+C$

Consider the given equations, $y\le e^x,y\ge \log x,y=2,x=0,y=0$.

Let,

$y=e^x$ ………..(1)

$y={\log }_{e}x$ ……….(2)

$y=2$ ………(3)

From equation (1) and (3), we get

$2=e^x$

${\log }_{e}2={\log }_e{e}^x$

${\log }_{e}2=x.{\log }_{e}e(\because {\log }_{e}e=1)$

$x={\log }_{e}2$

So, point of intersection of equation (1) and (3)$(x,y)i.e.({\log }_{e}2,2)$ .

Now, from equation (2) and (3,we get .

$2={\log }_{e}x$

$\because ({\log }_{e}x=y\Rightarrow x=e^y)$

$\therefore e^2=x$

So, point of intersection of equation (2) and (3) $(x,y)i.e.(e^2,2)$

Area of rectangle OHBE is,$A_1=e^2\times 2=2e^2$

Area of region AHBA is,

$A_2={\int }_1^{e^2-1}\log xdx={[x\log x+x]}_1^{e^2-1}+C_1=[(e^2-1)\log (e^2-1)+e^2-1]-(1\log 1+1)+C_1$

$=(e^2-1)\log (e^2-1)+e^2-1-1+C_1=(e^2-1)\log (e^2-1)+e^2-2+C_1$

Area of region DCED is,

$A_3={\int }_1^2{e}^{x}dy=e^2-e+C_2$

Required area of region OABCDO is (A) =

Area of rectangle OHBE is($A_1$) - Area of region AHBA($A_2$) - Area of region DCED ($A_3$)

$=2e^2-((e^2-1)\log (e^2-1)+e^2-2+C_1)-(e^2-e+C_2)$

$=2e^2-(e^2-1)\log (e^2-1)+2+e-C_1-C_2$

$=2e^2-(e^2-1)\log (e^2-1)+2+e+C$

Hence, this is the answer.