题目

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

提问时间：2020-11-02

答案

∫(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 / 2
J2 =∫√(1-x²) dx = ∫cos²t dt = 1/2∫(1+cos2t) dt = 1/2· [t + sin2t/2] =1/2·[π/2 + 0] = π/4