Question

Area of the region bounded by $y=e^x,y=e^{-x},x=0$ and $x=1$ in sq. units is:

• A. ${(e+\frac{1}{e})}^2$
• B. ${(e-\frac{1}{e})}^2$
• C. ${(\sqrt{e}+\frac{1}{\sqrt{e}})}^2$
• D. ${(\sqrt{e}-\frac{1}{\sqrt{e}})}^2$

Answer 1

The correct option is D ${(\sqrt{e}-\frac{1}{\sqrt{e}})}^2$

$y=e^x,y=e^{-x}$

Then, area bounded by these two curves s given by
$A=\stackrel{1}{\underset{0}{\int }}(e^x-e^{-x})dx=\stackrel{1}{\underset{0}{\int }}{e}^{x}dx-\stackrel{1}{\underset{0}{\int }}{e}^{-x}dx$
$=(e^1-1)+1(e^{-1}-1)$
$=e+e^{-1}-2$
$={(\sqrt{e}-\frac{1}{\sqrt{e}})}^{2}squnits.$