definite-integral

Question: The area of figure bounded by $y={{e}^{x}},\,y={{e}^{-x}}$ and the straight line $x=1$ is



1) $e+\frac{1}{e}$
2) $e-3$
3) $e+\frac{1}{e}-2$
4) $e+\frac{1}{e}+2$
Solution: Explanation: No Explanation
Area bounded by region Volume and surface area of solids of revolution

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