If the area of the region {(x, y): -1 leq x l | vedclass

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(D) The region is defined by $-1 \leq x \leq 1$ and $0 \leq y \leq a + e^{|x|} - e^{-x}$.Since $|x| = -x$ for $x \in [-1, 0]$ and $|x| = x$ for $x \in

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$5$

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(D) The region is defined by $-1 \leq x \leq 1$ and $0 \leq y \leq a + e^{|x|} - e^{-x}$.Since $|x| = -x$ for $x \in [-1, 0]$ and $|x| = x$ for $x \in [0, 1]$,we split the integral:Area $= \int_{-1}^0 (a + e^{-x} - e^{-x}) dx + \int_0^1 (a + e^x - e^{-x}) dx$Area $= \int_{-1}^0 a dx + \int_0^1 (a + e^x - e^{-x}) dx$Area $= a[x]_{-1}^0 + [ax + e^x + e^{-x}]_0^1$Area $= a(0 - (-1)) + (a(1) + e^1 + e^{-1}) - (a(0) + e^0 + e^0)$Area $= a + a + e + \frac{1}{e} - 2 = 2a + e + \frac{1}{e} - 2$Given Area $= \frac{e^2 + 8e + 1}{e} = e + 8 + \frac{1}{e}$Equating the two expressions:$2a + e + \frac{1}{e} - 2 = e + 8 + \frac{1}{e}$$2a - 2 = 8$$2a = 10 \Rightarrow a = 5$

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 $\frac{e^2 + 8e + 1}{e}$,then the value of $a$ is:

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