Question

求解一道高数题！！！！谢谢！！！！！！！！！！！！

Answer 1

里面进行配方，得到1-（x-1）^2，外面的x可以放到dx里面去，变成（x-1）^2,再进行计算

Answer 2

x-1=cost,dx=-sintdt
x=0,t=π
x=2,t=0
原式
=∫[π,0] cost*√(1-cos^2t)*(-sint)dt
=∫[π,0] cost*sint*(-sint)dt
=∫[0,π] sin^2tdsint
=sin^3t/3[0,π]
=0

Answer 3

∫(0->2)x√(2x-x^2) dx

2x-x^2 = 1-(x^2-2x+1) = 1-(x-1)^2
let
x-1= sina
dx=cosada
x=0, a=-π/2
x=2, a=π/2

∫(0->2)x√(2x-x^2) dx
=∫(-π/2->π/2) sina (cosa)^2 da
=(-1/3)[(cosa)^3](-π/2->π/2)
=0

Answer 4

结果我已经发了,楼下全错了.请查收

Answer 5

用换元法
令x-1=cost,dx=-sintdt
x=0,t=π
x=2,t=0
原式
=∫[π,0] cost*√(1-cos^2t)*(-sint)dt
=∫[π,0] cost*sint*(-sint)dt
=∫[0,π] sin^2tdsint
=sin^3t/3[0,π]
=0