Question

根号下9-x2的原函数

Answer 1

在以下的论证中,\int_0^x表示从0到x的积分,\int_0^(x/3)表示从0到x/3的积分,
\sqrt(x)表示根号下x,并设其中一个原函数为F(x)(其他的原函数为F(x)+常数),则：
F(x)=\int_0^x[\sqrt(9-y^2)]dy
=x[\sqrt(9-x^2)]+\int_0^x{y^2/[\sqrt(9-y^2)]}dy
=x[\sqrt(9-x^2)]+\int_0^x{[(y^2-9)+9]/[\sqrt(9-y^2)]}dy
=x[\sqrt(9-x^2)]-\int_0^x[\sqrt(9-y^2)]dy+9\int_0^x{1/[\sqrt(1-(y/3)^2)]}d(y/3)
=x[\sqrt(9-x^2)]-\int_0^x[\sqrt(9-y^2)]dy+9\int_0^(x/3){1/[\sqrt(1-t^2)]}dt
=x[\sqrt(9-x^2)]-\int_0^x[\sqrt(9-y^2)]dy+9arcsin(x/3)
=x[\sqrt(9-x^2)]-F(x)+9arcsin(x/3)
从上边的方程中解出F(x)得：
F(x)=x[\sqrt(9-x^2)]/2+9[arcsin(x/3)]/2
注意：第二个等号来自于分部积分.