The area of the figure bounded by y = e^x,y = | vedclass

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(C) The curves $y = e^x$ and $y = e^{-x}$ intersect at the point where $e^x = e^{-x}$,which implies $x = 0$.The region is bounded by $x = 0$ (intersec

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The curves $y = e^x$ and $y = e^{-x}$ intersect at the point where $e^x = e^{-x}$,which implies $x = 0$.The region is bounded by $x = 0$ (intersection point) and $x = 1$.In the interval $[0, 1]$,$e^x \ge e^{-x}$.The required area $A$ is given by the integral:$A = \int_{0}^{1} (e^x - e^{-x}) \, dx$Evaluating the integral:$A = [e^x - (-e^{-x})]_{0}^{1}$$A = [e^x + e^{-x}]_{0}^{1}$Applying the limits:$A = (e^1 + e^{-1}) - (e^0 + e^0)$$A = e + \frac{1}{e} - (1 + 1)$$A = e + \frac{1}{e} - 2$Thus,the correct option is $C$.

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

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