Evaluate the integral

Question

Evaluate the integral

Answer 1

Between x = -1 and x = 0, e^x ≤ e⁰ = 1. So, e^x - 1 ≤ 0. Therefore, l e^x - 1 l = 1 - e^x.

Between x = 0 and x = 2, e^x ≥ e⁰ = 1, So, e^x - 1 ≥ 0. Therefore, l e^x - 1 l = e^x - 1

So, ∫_{(-1 to 2)} l e^x - 1 l dx = ∫_{(-1 to 0)} (1 - e^x)dx + ∫_{(0 to 2)} (e^x - 1)dx

= (x - e^x)_{(-1 to 0)} + (e^x - x)_{(0 to 2)} = (0 - 1) - (-1 - 1/e) + (e² - 2) - (1 - 0) = e² + 1/e - 3

Answer 2

Thinking about the graph of y = e^x, we realize that when x = 0, y = 1 (e⁰ = 1) and so the graph passes through (0, 1). Since y = e^x - 1 moves the graph down by 1, then y = e^x - 1 passes through the point (0, 0). Everything to the left of zero is below the x-axis and everything to the right of zero is above the x-axis.

Consequently we can think about the absolute value integral as follows:

_-1∫² | e^x - 1 | dx = _-1∫⁰ -(e^x - 1)dx + ₀∫² (e^x - 1) dx

_-1∫⁰ -(e^x - 1)dx = - _-1∫⁰ e^x - 1 dx = ₀∫^-1 e^x - 1 dx = (e^x - x) evaluated between zero and -1 = (e^-1 - (-1)) - (e⁰ - 0) = 1/e + 1 - 1 = 1/e

₀∫² (e^x - 1) dx = (e^x - x) evaluated between zero and two = (e² - 2) - (e⁰ - 0) = e² - 2 - 1 = e² - 3

So the entire integral _-1∫² | e^x - 1 | dx = 1/e + e² - 3