∫02(x−1)dx\int_{0}^{2}(x-1) d x∫02(x−1)dx
∫02∣x−1∣dx\int_{0}^{2}|x-1| d x∫02∣x−1∣dx
2∫12(1−x)dx2 \int_{1}^{2}(1-x) d x2∫12(1−x)dx
2∫01(x−1)dx2 \int_{0}^{1}(x-1) d x2∫01(x−1)dx