∫_-2²|e^x-1|dx

=-∫_-2⁰（e^x-1）dx+∫₀²（e^x-1）dx

=-（e^x-x）|_-2⁰+（e^x-x）|₀²

=e^-2+e²-2．

故答案为：e^-2+e²-2．