Question

1.函数f(x)在[0,1]连续,在（0,1）上可导,且f(0)=f(1)=0,f(1/2)=1,试证：

Answer 1

1、(2)设g(x)=e^(λx)f(x)-xe^(λx)

g(0)=0,g(η)=e^(λη)(f(η)-η)=0

由罗尔定理：存在ξ∈（0,η）,使得g'(ξ)=0

g'(x)=λe^(λx)f(x)+e^(λx)f '(x)-e^(λx)-λxe^(λx)

则：λe^(λξ)f(ξ)+e^(λξ)f '(ξ)-e^(λξ)-λxe^(λξ)=0

约去e^(λξ),得：λf(ξ)+f '(ξ)-1-λx=0,整理即得所证.

2、注意定积分的结果是一个常数

设a=∫(0,1)f(x)dx

则f(x)=arctanx+ax

a=∫(0,1)f(x)dx=∫(0,1) (arctanx+ax)dx

=∫(0,1) arctanxdx+1/2a

=xarctanx-∫x/(1+x^2)dx+1/2a 0-->1

=π/4-1/2∫1/(1+x^2)dx^2+1/2a 0-->1

=π/4-1/2ln(1+x^2)+1/2a 0-->1

=π/4-1/2ln2+1/2a

得a=π/4-1/2ln2+1/2a

解得：a=π/2-ln2

即：∫(0,1)f(x)dx=π/2-2ln2