单选题 难度 3 5. 定积分 ∫01∣1−x2∣dx=\int_0^1 |1 - x^2| dx = ∫01∣1−x2∣dx= （ ）。 A 选项 A A. ∫01(1−x2)dx+∫12(1−x2)dx\int_0^1 (1 - x^2) dx + \int_1^2 (1 - x^2) dx∫01(1−x2)dx+∫12(1−x2)dx B 选项 B B. ∫01(x2−1)dx+∫12(x2−1)dx\int_0^1 (x^2 - 1) dx + \int_1^2 (x^2 - 1) dx∫01(x2−1)dx+∫12(x2−1)dx C 选项 C C. ∫01(x2−1)dx+∫12(1−x2)dx\int_0^1 (x^2 - 1) dx + \int_1^2 (1 - x^2) dx∫01(x2−1)dx+∫12(1−x2)dx D 选项 D D. ∫01(1−x2)dx+∫12(x2−1)dx\int_0^1 (1 - x^2) dx + \int_1^2 (x^2 - 1) dx∫01(1−x2)dx+∫12(x2−1)dx 答案与解析 微信扫描二维码 免费获取正确答案和解析
