Question

If $\left\{y\right\}$ denotes the fractional part of $y,$ then the value of the definite integral $\underset{-\infty }{\overset{\ln 3}{\int }}\left\{e^x\right\}dx$ equals

• A. $3+\ln 2-2\ln 3$
• B. $3-\ln 3$
• C. $2\ln 3+\ln 2$
• D. $1$

Answer 1

The correct option is A $3+\ln 2-2\ln 3$

$\underset{-\infty }{\overset{\ln 3}{\int }}\{e^x\}dx$
$=\underset{-\infty }{\overset{0}{\int }}{e}^{x}dx+\underset{0}{\overset{\ln 2}{\int }}(e^x-1)dx+\underset{\ln 2}{\overset{\ln 3}{\int }}(e^x-2)dx$
$=e^x{|}_{-\infty }^0+(e^x-x){|}_0^{\ln 2}+(e^x-2x){|}_{\ln 2}^{\ln 3}$
$=(1-0)+(2-\ln 2-1)+(3-2\ln 3-2+2\ln 2)$
$=1+1-\ln 2+1-2\ln 3+2\ln 2$
$=3+\ln 2-2\ln 3$