单选题 难度 3 定积分 ∫02∣1−x2∣ dx\int_0^2 |1-x^2|\,dx∫02∣1−x2∣dx 可化为（ ） A 选项 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 ∫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 ∫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 ∫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
