# Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 112

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Determine the area of the region bounded by the curves $y = e^x, y = e^{-x}, x = -2$ and $x = 1$.

Notice that the orientation of the curves changes at its point of intersection at $x = 0$. We can divide the bounded region into two sub region. Let $A_1$ be the region bounded from $x = -2$ to $x = 0$ and $A_2$ e the region bounded from $x = 0$ to $x = 1$. Thus,

\begin{aligned} A_T = A_1 + A_2 =& \int^0_{-2} [e^{-x} - e^x] dx + \int^1_0 (e^x - e^{-x}) dx \\ \\ =& \left[ e^{-x} - e^x (1) \right]^0_{-2} + \left[ e^x (1) - e^{-x} (-1) \right]^1_0 \\ \\ =& \left[ -e^{-x} - e^x \right]^0_2 + \left[ e^x + e^{-x} \right]^1_0 \\ \\ =& [(e^{-0} + e^0) - (e^{-2} + e^2)] + [(e^1 - e^{-1}) - (e^0 + e^{-0})] \\ \\ =& -1 - 1 + \frac{1}{e^2} + e^2+ e^1 + \frac{1}{e^1} - 1 - 1 \\ \\ =& \frac{1}{e^2} + e^2 + \frac{1}{e^1} + e^1 - 4 \text{ Square units} \end{aligned}