question_answer

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

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 :

Required area\[=\int_{0}^{1}{({{e}^{x}}-{{e}^{-x}})\,\,dx}\]                 \[=[{{e}^{x}}+{{e}^{-x}}]_{0}^{1}\]                 \[=\left( e+\frac{1}{e}-2 \right)sq\,\,unit\]