Question

这道题怎么解？

我算出来是-5Π/8

Answer 1

我认为不可能是负数，因为被积函数一定在x轴上方，所以积分值一定为非负。

以下是我的做法

正好和你相差一个符号

Answer 2

2x-x^2 = 1-(x-1)^2
let
x-1 =sinu
dx= cosu du
x=0, u=-π/2
x=2, u=π/2
∫(0->2) x^2.√(2x-x^2) dx
=∫(-π/2->π/2) (1+sinu)^2. (cosu)^2 du
=∫(-π/2->π/2) [1+2sinu + (sinu)^2 ]. (cosu)^2 du
= ∫(-π/2->π/2) [ (cosu)^2 + (sinu)^2.(cosu)^2 ] du
=2 ∫(0->π/2) [ (cosu)^2 + (sinu)^2.(cosu)^2 ] du
= ∫(0->π/2) [ (1+cos2u) + (1/2)(sin2u)^2 ] du
= ∫(0->π/2) [ (1+cos2u) + (1/4)(1-cos4u) ] du
=[ u +(1/2)sin2u + (1/4)u - (1/16)sin4u]|(0->π/2)
=(5/4)(π/2)
=5π/8