If the area of the region (x,y):-1 x 1,0 y a+e^|x|-e^-x,a0 is e^2+8e+1e, then the value o…

Question

If the area of the region ${(x,y):-1\le x\le 1,0\le y\le a+{e}^{|x|}-{e}^{-x},a>0}$ is $\frac{{e}^{2}+8e+1}{e}$, then the value of $a$ is :

Accepted Answer

$\begin{gathered} a+\int _{0}^{1}(a+{e}^{x}-{e}^{-x})dx \ a+{[a+{e}^{x}+{e}^{-x}]}_{1}^{0} \ 2a+e-1+{e}^{-1}-1=e+8+\frac{1}{e} \ 2a=10\Rightarrow a=5 \end{gathered}$

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