Question
What is the area bounded by the curves $$y = {e^x},\,y = {e^{ - x}}$$ and the straight line $$x = 1\,?$$
A.
$$\left( {e + \frac{1}{e}} \right)\,{\text{square}}\,{\text{unit}}$$
B.
$$\left( {e - \frac{1}{e}} \right)\,{\text{square}}\,{\text{unit}}$$
C.
$$\left( {e + \frac{1}{e} - 2} \right)\,{\text{square}}\,{\text{unit}}$$
D.
$$\left( {e - \frac{1}{e} - 2} \right)\,{\text{square}}\,{\text{unit}}$$
Answer :
$$\left( {e + \frac{1}{e} - 2} \right)\,{\text{square}}\,{\text{unit}}$$
Solution :
Given equations of curves are $$y = {e^x}{\text{ and }}y = {e^{ - x}}$$
$$ \Rightarrow {e^x} = \frac{1}{{{e^x}}}\,\, \Rightarrow {e^{2x}} = {e^0}\,\, \Rightarrow x = 0$$
Also, equation of straight line gives $$x = 1$$
$$\therefore $$ Required area
$$\eqalign{
& = \int\limits_0^1 {\left( {{e^x} - {e^{ - x}}} \right)dx} \cr
& = \left[ {{e^x} + {e^{ - x}}} \right]_0^1 \cr
& = e + {e^{ - 1}} - {e^0} + {e^{ - 0}} \cr
& = \left( {e + \frac{1}{e} - 2} \right){\text{ sq}}{\text{. unit}} \cr} $$