MathematicsHard210×since 2002Q3427If ∫02(2x−2x−x2)dx=∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }0∫2(2x−2x−x2)dx=0∫1(1−1−y2−2y2)dy+1∫2(2−2y2)dy+I, then I equalsA∫01(1+1−y2)dy\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}0∫1(1+1−y2)dyB∫01(y22−1−y2+1)dy\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy}0∫1(2y2−1−y2+1)dyC∫01(1−1−y2)dy\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy}0∫1(1−1−y2)dyD∫01(y22+1−y2+1)dy\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy}0∫1(2y2+1−y2+1)dyCheck answerSkip