Question

求定积分:∫(0→1)x^2√(1-x)dx

Answer 1

答：

设t=√(1-x),t^2=1-x,x=1-t^2

x=0,t=1

x=1,t=0

原式

=(1→0) ∫[(1-t^2)^2/t]d(1-t^2)

=(0→1) 2∫[(1-2t^2+t^4)/t] dt

=(0→1) 2∫(1-2t^2+t^4)dt

=(0→1) 2[t-(2/3)t^3+(1/5)t^5]

=2*(1-2/3+1/5)-0

=12/5-4/3

=16/15