Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 548: 30

$$\frac{1}{e}+e^2-3$$

Given \begin{aligned} \:\int _{-1}^2\:\left|e^x-1\right|dx \end{aligned} Since \begin{aligned} \:\int _{-1}^2\:\left|e^x-1\right|dx&= \int_{-1}^{0}(-(e^x-1))dx+ \int_{0}^{2}(e^x-1)dx\\ &= \int_{-1}^{0}(-e^x+1))dx+ \int_{0}^{2}(e^x-1)dx\\ &=-e^x+1\bigg|_{-1}^{0}+(e^x-1)\bigg|_{0}^{2}\\ &= \frac{1}{e}+e^2-3 \end{aligned}