Question

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

• 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

The correct option is C

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

Explanation for the correct option:

Draw the curves $y=e^x,y=e^{-x}$:

From the figure, we can see

Point of intersection of these curves is $\left(0,1\right)$

$\therefore$Area bounded by the curves $=$Area bounded by region $ABC$ $$\begin{gathered} ={\int }_0^1\left(e^x-e^{-x}\right)dx \\ ={\left[e^x+e^x\right]}_0^1 \\ =\left[e^1+e^{-1}-\left(e^0+e^{-0}\right)\right] \\ =e^1+\frac{1}{e}-2 \\ =e+\frac{1}{e}-2 \end{gathered}$$

Thus, Area bounded by the curves $y=e^x,y=e^{-x}$ is $\mathit{e}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{e}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}$

Hence, Option ‘C’ is Correct.