A) \[e+\frac{1}{e}\]
B) \[e+\frac{1}{e}+2\]
C) \[e+\frac{1}{e}-2\]
D) \[e-\frac{1}{e}+2\]
Correct Answer: C
Solution :
[c] Given curves are \[y={{e}^{x}}\] and \[y={{e}^{-x}}\] Now, \[{{e}^{x}}={{e}^{-x}}\Rightarrow x=0\] \[\therefore \]Area \[=A=\int\limits_{0}^{1}{\left( {{e}^{x}}-{{e}^{-x}} \right)dx=\left( {{e}^{x}}+{{e}^{-x}} \right)_{0}^{1}}\] \[=\left[ \left( e+{{e}^{-1}} \right)-\left( {{e}^{0}}+{{e}^{-0}} \right) \right]=e+\frac{1}{e}-2\].
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