定积分的求解

Question

第七题的详细过程谢谢

Answer 1

（7）I = ∫<0, π/2>x^2cosxdx + ∫<π/2, π>x^2(-cosx)dx
= ∫<0, π/2>x^2dsinx - ∫<π/2, π>x^2dsinx
因 ∫x^2dsinx = x^2sinx - ∫2xsinxdx = x^2sinx + ∫2xdcosx
= x^2sinx + 2xcosx - 2∫cosxdx = x^2sinx + 2xcosx - 2sinx
则 I = [x^2sinx + 2xcosx - 2sinx]<0, π/2> - [x^2sinx + 2xcosx - 2sinx]<π/2, π>

= π^2/4-2 - (-2π - π^2/4 + 2) = π^2/2 + 2π - 4追问

答案不是派

追答

题目抄错，已更改。

Answer 2

∫ x^2.cosx dx
=∫ x^2 dsinx
=x^2.sinx -2∫ xsinx dx
=x^2.sinx +2∫ x dcosx
=x^2.sinx +2x.cosx -2∫ cosx dx
=x^2.sinx +2x.cosx -2sinx +C
∫(0->π) x^2.|cosx| dx
=∫(0->π/2) x^2.cosx dx - ∫(π/2->π) x^2.cosx dx
=[x^2.sinx +2x.cosx -2sinx]|(0->π/2) -[x^2.sinx +2x.cosx -2sinx]|(π/2->π)
=[(π/2)^2 -2 ] +[ (π/2)^2 -2 ]
=(1/2)π^2 - 4

Answer 3

Answer 4

Answer 5

过程如图…………

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