Q. The area of the region enclosed by the curves $y=e^x$, $y=|e^x-1|$ and $y$-axis is:
A) $1-\log_e2$
B) $\log_e2$
C) $1+\log_e2$
D) $2\log_e2-1$
Explanation
[Solution by Dr. Abubacker] For $x<0$, $|e^x-1|=1-e^x$. The curves $y=e^x$ and $y=1-e^x$ meet at $e^x=\frac12$, i.e. $x=-\ln2$. Area $=\int_{-\ln2}^0\left(e^x-(1-e^x)\right)dx=\left[2e^x-x\right]_{-\ln2}^0=2-(1+\ln2)=1-\ln2$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.