Q. The area between the curve $y = x\ e^{x}$ and $y = x\ e^{- x}$ and the line $x = 1$ in square unit, is
A) $2\ \left( \ e + \frac{1}{e} \right)$ sq unit
B) 0 sq unit
C) $2e$ sq unit
D) $\frac{2}{e}$ sq unit
Explanation
(d)
Required area $= \int_{0}^{1}{\left( xe^{x} - xe^{- x} \right)dx}$
$= \left\lbrack xe^{x} - e^{x} + xe^{- x} + e^{- x} \right\rbrack_{0}^{1}$
$= e - e + \frac{1}{e} + \frac{1}{e} = \frac{2}{e}$ sq unit
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.