问题描述: f(x)=1/(1+x^2)+(1-x^2)^(1/2)∫(上限1,下限0)f(x)dx.求∫(上限1,下限0)f(x)dx 1个回答 分类:数学 2014-12-01 问题解答: 我来补答 ∫(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 正好是单位圆在第一象限的面积 = π / 4 原积分 = J1 + J2 = π / 2 补充:令x = sint dx = costdt x = 0 →1 时 t= 0 →π / 2J2 =∫√(1-x²) dx = ∫cos²t dt = 1/2∫(1+cos2t) dt = 1/2· [t + sin2t/2] =1/2·[π/2 + 0] = π/4 展开全文阅读