If the area of the region $\{(x,y): -1\leq x\leq 1,\; 0\leq y\leq a+e^{|x|}-e^{-x},\; a>0\}$ is $\dfrac{e^2+8e+1}{e}$, then the value of $a$ is:
Step-by-Step Solution
Key Concept: Split into $x\in[-1,0]$ (where $e^{|x|}=e^{-x}$) and $x\in[0,1]$ (where $e^{|x|}=e^x$); the integral over $[-1,0]$ gives $a$ and the integral over $[0,1]$ gives $a+e+1/e-2$. Set the total equal to $(e^2+8e+1)/e$ and solve for $a$.
For $x\in[-1,0]$: $e^{|x|}-e^{-x} = e^{-x}-e^{-x}=0$, so integrand $= a$. Contribution: $a$.
For $x\in[0,1]$: integrand $= a+e^x-e^{-x}$. Contribution: $a + e+\tfrac{1}{e}-2$.
Total $= 2a+e+\tfrac{1}{e}-2 = \dfrac{e^2+8e+1}{e} = e+\dfrac{1}{e}+8$.
$2a-2=8 \Rightarrow a=5$.
Correct Answer: 3