(1)-|f(x)|≤f(x)≤|f(x)|
所以-∫|f(x)|dx≤∫f(x)dx≤∫|f(x)|dx
从而|∫f(x)dx|≤∫|f(x)|dx
(2)直接用等比数列求和公式就可以得到
(3)1/(1-x²)-(1+x²+...+x^(2n-2))=x^(2n)/(1-x²)
令f(x)=1/(1-x²)-(1+x²+...+x^(2n-2)),a=0,b=1/2
则|∫[0,1/2]f(x)dx|=|∫[0,1/2]1/(1-x²)dx-∫[0,1/2](1+x²+...+x^(2n-2))dx|
=|1/2ln3-(1/2+(1/2)^3/3+..+(1/2)^(2n-1)/(2n-1))|
记S[n]=(1/2+(1/2)^3/3+..+(1/2)^(2n-1)/(2n-1)),则
|∫[0,1/2]f(x)dx|=|1/2ln(3)-S[n]|
由于1/(1-x²)在[0,1/2]上是增函数,所以x^(2n)/(1-x²)
