Question

The area of the figure bounded by the curves $y=\ln x$ & $y={\left(\ln x\right)}^2$ is

• A. $e+1$
• B. $e-1$
• C. $3-e$
• D. $1$

Answer 1

Answer: (C)

$3-e$

Given curves, $y=\ln x$ & $y={\left(\ln x\right)}^2$

Points of intersection of given curves are, $(1,0)$ and $(e,1)$
Hence required area is,
$A={\int }_1^e(\ln x-(\ln x{)}^2)dx$
On solving it by parts we get

$A=[x\ln x-x{]}_1^e-\left[x\right(\ln x{)}^2-{\int }_1^{e}2\ln xdx]$

$A=[x\ln x-x{]}_1^e-{\left[x\right(\ln x{)}^2-2x\ln x+2x]}_1^e$

$A={[x\ln x-x-x(\ln x{)}^2+2x\ln x-2x]}_1^e$
$A={\left|3x(\ln x-1)\right|}_1^e-|x(\ln x{)}^2{|}_1^e=3-e$