Q. $\int_{- 1}^{1}{\left( {e^{x}}^{3} + {e^{- x}}^{3} \right)\left( e^{x} - e^{- x} \right)dx}\ is\ equal\ to$
A) $\frac{e^{2}}{2} - 2e$
B) $e^{2} - 2e$
C) 2$(e^{2} - e)$
D) 0
Explanation
(d)
Let $f(x) = \left( {e^{x}}^{3} + {e^{- x}}^{3} \right)\left( e^{x} - e^{- x} \right)$
$\Rightarrow f( - x) = \left( {e^{- x}}^{3} + {e^{x}}^{3} \right)\left( e^{- x} - e^{x} \right)$
$= - \left( {e^{x}}^{3} + {e^{- x}}^{3} \right)\left( e^{x} - e^{- x} \right)$
$= - f(x)$
$\Rightarrow f(x)$is an odd function.
$\therefore\int_{- 1}^{1}{f(x)dx = 0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.