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Find the area enclosed by the curves y = e^x, y = e^(-x), and the vertical line x = 1, measured in square units.

1. e + 1/e
2. e + 1/e + 2
3. e + 1/e - 2
4. e - 1/e + 2

Correct answer: e + 1/e - 2

Solution

For 0<=x<=1, e^x >= e^-x, so area = integral_0^1 (e^x - e^-x) dx = [e^x + e^-x]_0^1 = (e + 1/e) - 2 = e + 1/e - 2.

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