Question

The area bounded athe curves $y=e^x,y=e^{-x}$ and the straight line $x=1$ is given by

• A. $e+e^{-1}–2$
• B. $e-e^{-1}$
• C. $e+e^{-1}$
• D. $e+e^{-1}+2$

Answer 1

The correct option is A

$e+e^{-1}–2$

Explanation for the correct option:

Step:1 Finding coordinates of point A and B

Two curves are $y=e^x,y=e^{-x}$

Now, for finding the point of intersection of $e^x$ and $e^{-x}$

$$\begin{gathered} e^x=e^{-x} \\ \Rightarrow x=0 \\ \Rightarrow y=e^0 \\ \Rightarrow y=1 \end{gathered}$$

Therefore, the two curves meet at $A\left(0,1\right)$

Now, for finding the point of intersection of $e^x$ and $x=1$

$$\begin{gathered} y=e^x \\ =e^1 \\ =e \end{gathered}$$

Therefore, the two curves meet at $B\left(1,e\right)$.

Step:2 Finding coordinates of point B

ow, for finding the point of intersection of $e^{-x}$ and $x=1$

$$\begin{gathered} y=e^{-x} \\ =e^{-1} \end{gathered}$$

Therefore, the two curves meet at $C\left(1,e^{-1}\right)$

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

the area bounded is

Step:3 Finding Integral Area

$\begin{array}{rcl}A& =& {\int }_0^1\left(e^x-e^{-x}\right)dx\\ & =& {\left(e^x+e^{-x}\right)}_0^1\\ & =& \left[(e+e^{-1})-(e^0+e^{-0})\right]\\ & =& e+\frac{1}{e}-(1+1)\\ & =& e+e^{-1}-2\\ & & \end{array}$

Hence, the correct answer is option A.