Answer
$$\int_{-1}^{2}\left|e^{x}-1\right|dx=\frac{1}{e}+e^{2}-3$$
Work Step by Step
$$\int_{-1}^{2}\left|e^{x}-1\right|dx=\int_{-1}^{0}(1-e^{x})dx+\int_{0}^{2}(e^{x}-1)dx$$
$$=\left [x -e ^{x}\right ]_{-1}^{0}+\left [ e^{x}-x \right ]_{0}^{2}$$
$$=\frac{1}{e}+e^{2}-3$$
