A) \[\left( e+\frac{1}{e} \right)\] sq. unit
B) \[\left( e-\frac{1}{e} \right)\] sq. unit
C) \[\left( e+\frac{1}{e}-2 \right)\] sq. unit
D) \[\left( e-\frac{1}{e}-2 \right)\] sq. unit
Correct Answer: C
Solution :
[c] Given equations of curves are \[y={{e}^{x}}\] and \[y={{e}^{-x}}\]. \[\Rightarrow {{e}^{x}}=\frac{1}{{{e}^{x}}}\Rightarrow {{e}^{2x}}={{e}^{0}}\Rightarrow x=0\] Also, equation of straight line gives \[x=1\] \[\therefore \]Required area \[=\int\limits_{0}^{1}{({{e}^{x}}-{{e}^{-x}})dx}\] \[=\left[ {{e}^{x}}+{{e}^{-x}} \right]_{0}^{1}=e+{{e}^{-1}}-{{e}^{0}}+{{e}^{-0}}\] \[=\left( e+\frac{1}{e}-2 \right)\] sq unit
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