1-√x=tとおくとx=0→1はt=1→0となる。
両辺を微分すると{-1/(2√x)}dx=dt
dx=-2√xdt=-2(1-t)dt=2(t-1)dt
よって
π∫¹₀(1-√x)²dx=π∫⁰₁t²2(t-1)dt=π[(2/4)t⁴-(2/3)t³]⁰₁=π{(2/3)-(2/4)}=π/6
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