第1个回答 2014-08-17
请采纳哦。
本回答被提问者采纳第2个回答 2014-08-17
y的平均值 = [1/(√3/2-1/2)]∫<1/2,√3/2>x^2dx/√(1-x^2) 令 x=sint
= (1+√3)∫<π/6,π/3>（sint）^2dt
= (1+√3)∫<π/6,π/3>(1/2)(1-cos2t)dt
= [(1+√3)/2][t-(1/2)sin2t]<π/6,π/3>
= (1+√3)π/12.
= (1+√3)∫<π/6,π/3>（sint）^2dt
= (1+√3)∫<π/6,π/3>(1/2)(1-cos2t)dt
= [(1+√3)/2][t-(1/2)sin2t]<π/6,π/3>
= (1+√3)π/12.
第3个回答 2014-08-17
算出积分除以区间长度就可以了啊
第4个回答 2014-08-17
用积分中值定理