∫(上限1,下限0)f(x)dx
= ∫1/(1+x²) + √(1-x²) dx x = 0 →1
其中：
J1 = ∫1/(1+x²) dx x = 0 →1
= arctan(x) arctan(1) = π / 4 arctan(0) = 0
= π / 4
J2 = ∫√(1-x²) dx = π / 4 x = 0 →1
= π / 4 正好是單位圓在第一象限的面積 = π / 4
原積分 = J1 + J2 = π / 2
補(bǔ)充：令x = sint dx = costdt x = 0 →1 時(shí) t= 0 →π / 2
J2 =∫√(1-x²) dx = ∫cos²t dt = 1/2∫(1+cos2t) dt = 1/2· [t + sin2t/2] =1/2·[π/2 + 0] = π/4