Question

∫

2-2

|ex-1|dx=______．

Answer 1

∫_-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．