Question

Area bounded by the curves $y=e^x,y=e^{-x}$ and the straight line $x=1$ is (in sq units)

• A. $e+\frac{1}{e}$
• B. $e+\frac{1}{e}+2$
• C. $e+\frac{1}{e}-2$
• D. $e-\frac{1}{e}+2$
• E. $e-\frac{1}{e}$

Answer 1

Answer: (C)

$e+\frac{1}{e}-2$

The two curve meet at $x=0$

For $x>0,e^x>e^{-x}$

Area bounded by $e^x,e^{-x},x$-axis is equal to area bounded by $e^x,x$-axis - area bounded by $e^{-x},x$-axis

By above concept:

Area bounded is ${\int }_0^1{e}^{x}dx-{\int }_0^1{e}^{-x}dx$

So, area bounded by $e^x,e^{-x},x=1$ is ${\int }_0^1(e^x-e^{-x})dx$
Bounded Area $=e-1+e^{-1}-1=e+\frac{1}{e}-2$