Q. The area enclosed between the curve $y = \log_{e}(x + e)$and the coordinate axis is
A) 4 sq units
B) 3 sq units
C) 2 sq units
D) 1 sq unit
Explanation
(d)
$Required\ area,\ A = \int_{1 - e}^{0}{\log_{e}(x + e)dx}$
$Put\ \ x + e = t \Longrightarrow dx = dt$
$\therefore A = \int_{1}^{e}{\log_{e}\ t\ dt}$
$= \left\lbrack t\ \log_{e}\ t - t \right\rbrack_{1}^{e}$
$= (e - e - 0 + 1)$
$= 1\$sq unit
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.