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

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 $e+8+\dfrac1e$, then the value of $a$ is:

Accepted Answer

For $-1\le x\le0$, $|x|=-x$, so $e^{|x|}-e^{-x}=0$; for $0\le x\le1$, $e^{|x|}-e^{-x}=e^x-e^{-x}$. The full area splits into a rectangle of height $a$ plus:

$$\text{Area} = a + \int_0^1\left(a+e^x-e^{-x}\right)dx = 2a + \left[e^x+e^{-x}\right]_0^1 - 2 = 2a + e + \frac1e - 1 - 1$$

Setting this equal to $e+8+\dfrac1e$: $2a - 2 = 8 \Rightarrow 2a = 10 \Rightarrow a = 5$.

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