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)

Given curves are $y=e^x$ and $y=e^{-x}$.
The point of intersection is $e^{-x}=e^x$
$X=0$
Then, $Y=1$

So, the point of intersection is $(0,1)$.
$\therefore$ Area of bounded region $ABC$
$=\underset{0}{\overset{1}{\int }}\left(y_2-y_1\right)dx=\underset{0}{\overset{1}{\int }}\left(e^x-e^{-x}\right)dx$
$={\left[e^x+e^{-x}\right]}_0^1=\left[e^1+e^{-1}-\left(e^0+e^{-0}\right)\right]$
$=\left[e+\frac{1}{e}-(1+1)\right]$
$=e+\frac{1}{e}-2$