Question ID: #411
The area of the region enclosed by the curves $y=e^{x}$, $y=|e^{x}-1|$ and y-axis is:
- (1) $1+\log_{e}2$
- (2) $\log_{e}2$
- (3) $2~\log_{e}2-1$
- (4) $1-\log_{e}2$
Solution:
Area $= \int_{-\ln 2}^{0} [e^x – (1 – e^x)] dx = \int_{-\ln 2}^{0} (2e^x – 1) dx$
$$ = [2e^x – x]_{-\ln 2}^{0} $$
$$ = (2e^0 – 0) – (2e^{-\ln 2} – (-\ln 2)) $$
$$ = 2 – (2(\frac{1}{2}) + \ln 2) $$
$$ = 2 – (1 + \ln 2) = 1 – \ln 2 $$
Ans. (4)
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