• # question_answer The area of figure bounded by $y={{e}^{x}},\,y={{e}^{-x}}$ and the straight line $x=1$ is                                            [Karnataka CET 1999] A)            $e+\frac{1}{e}$                 B)            $e-3$ C)            $e+\frac{1}{e}-2$             D)            $e+\frac{1}{e}+2$

Given equations of curves $y={{e}^{x}};y={{e}^{-x}}$ and straight line $x=1$ We know that area of the figure bounded by the curves and straight line                     $=\int_{0}^{1}{({{e}^{x}}-{{e}^{-x}})dx=[{{e}^{x}}+{{e}^{-x}}]\,_{0}^{1}\,=\,e+\frac{1}{e}-2.}$