展开全部
I = ∫<0, 2>√(8-x^2)dx - (1/2)∫<0, 2>x^2dx
前者令 x = 2√2sint, x = 0 时, t = 0;x = 2 时, t = π/4
I = ∫<0, π/4>2√2cost 2√2cost dt - (1/6)[x^3]<0, 2>
= 8 ∫<0, π/4>(cost)^2 dt - 4/3 = 4 ∫<0, π/4>(1+cos2t)dt - 4/3
= 4 [t +(1/2)sin2t]<0, π/4> - 4/3 = 4 [π/4 +1/2] - 4/3 = π + 2/3
前者令 x = 2√2sint, x = 0 时, t = 0;x = 2 时, t = π/4
I = ∫<0, π/4>2√2cost 2√2cost dt - (1/6)[x^3]<0, 2>
= 8 ∫<0, π/4>(cost)^2 dt - 4/3 = 4 ∫<0, π/4>(1+cos2t)dt - 4/3
= 4 [t +(1/2)sin2t]<0, π/4> - 4/3 = 4 [π/4 +1/2] - 4/3 = π + 2/3
2017-06-17
展开全部
∫√(8-x²)dx
追答
不说上下限的话,被积函数是圆的上半部分,按照定积分的几何意义一一曲边梯形的面积,当然会出现π了。不信一试
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询