设f(x)在[0,1]上连续且满足f(x)=[1/(1+x^2)]-积分号(上限1,下限0)f(x)dx求f(x)在[0,1]的平均值

Question

设f(x)在[0,1]上连续 且满足f(x)=[1/(1+x^2)]-积分号(上限1,下限0)f(x)dx 求f(x)在[0,1]的平均值

Answer 1

令∫（0,1） f(x)dx=A,有f(x)=[1/(1+x^2)]-A,两边积分A=∫（0,1）[1/(1+x^2)]dx-Ax┃(上限1,下限0)
得A=π/8
f(x)在[0,1]的平均值=A/(1-0)=π/8