Question

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

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

Answer 1

The correct option is C

$3-e$

$A={\int }_1^e(\ln x-(\ln x{)}^2)dx\{\because \int \ln xdx=x\ln x-x\text{and}\int (\ln x{)}^{2}dx=\left[\ln x\right(x\ln x-x)-\int (\ln x-1)dx]\}A=(x\ln x-x)-\left[\ln x\right(x\ln x-x)-\int (\ln x-1)dx]=[{x\ln x-x-x(\ln x{)}^2+x\ln x+x\ln x-x-x]}_1^e={[-x(\ln x{)}^2+3x\ln x-3x]}_1^e=-e+3e-3e+3=3-e$