Question

求定积分，这么做为什么不对？答案应该是3π^2

Answer 1

如图所示：

Answer 2

∫(0->2π) t(1-cost)^2 dt
=∫(0->2π) [ t-2tcost + t( cost)^2 ] dt
=∫(0->2π) [ t-2tcost + (1/2)t(1+ cos2t) ] dt
=[(3/4)t^2 ] |(0->2π) + ∫(0->2π) [ -2tcost + (1/2)t.cos2t ] dt
=3π^2 -2∫(0->2π) tcost dx + (1/2)∫(0->2π) t.cos2t dt
=3π^2 + 0 +0
=3π^2

consider
∫(0->2π) tcost dx
=∫(0->2π) t dsint
=[tsint]|(0->2π) -∫(0->2π) sint dt
=0 +[cost]|(0->2π)
=0
∫(0->2π) t.cos2t dt
=(1/2)∫(0->2π) t dsin2t
=(1/2)[ tsin2t]|(0->2π) -(2/3)∫(0->2π) sin2t dt
=0 +(1/3) [cos2t]| (0->2π)
=0

Answer 3

3π^2 -2∫(0->2π) tcost dx + (1/2)∫(0...

Answer 4

解如下图所示