请问40这道高数题目，答案划线部分是怎么推导的？
解:∵函数f(x)满足方程f(x)∫(0，x)f(x-t)dt=
(sinx)^4，设x-t=u，则
∫(0，x)f(x-t)dt=∫(x，0)f(u)d(x-u)=
∫(0，x)f(u)du(此时是关于u的积分，x可以看作常数) ∴设∫(0，x)f(u)du=g(x)，
则f(x)=g'(x)，原方程化为
g'(x)g(x)=(sinx)^4，
g(x)dg(x)/dx=(sinx)^4，
g(x)dg(x)=(sinx)^4 dx，
g(x)dg(x)=0.25(1-cos2x)²dx，
g(x)dg(x)=0.25[1-2cos2x+0.5(1+cos4x)]，g(x)dg(x)=0.25(1.5-2cos2x+0.5cos4x)，2g(x)dg(x)=0.75-cos2x+0.25cos4x，
g²(x)=0.75x-0.5sin2x+0.0625sin4x+c
∵∫(0，u)f(u)=g(x) ∴∫(0，0)f(u)du=g(0) ∴g(0)=0，则c=0，
g(x)=±√(0.75x-0.5sin2x+0.0625sin4x)
∴g(π/2)=±√0.75π，g(0)=0
∵f(x)在[0，π/2]上平均数=
∫(0，π/2) f(x)dx/(π/2-0)=[g(π/2)-g(0)]/(π/2)=±√3/√π
简单计算一下即可，答案如图所示
let
u=x-t
du = -dt
t=0, u=x
t=x, u=0
∫(0->x) f(x-t) dt
=∫(x->0) f(u) (-du)
=∫(0->x) f(u) du
=∫(0->x) f(t) dt
//
f(x).∫(0->x) f(x-t) dt = (sinx)^4
f(x).∫(0->x) f(t) dt = (sinx)^4
两边取定积分
∫(0->π/2) [f(x).∫(0->x) f(t) dt ] dx =∫(0->π/2) (sinx)^4 dx (1)
∫(0->π/2) (sinx)^4 dx
=(1/4) ∫(0->π/2) (1-cos2x)^2 dx
=(1/4) ∫(0->π/2) [1-2cos2x+ (cos2x)^2] dx
=(1/8) ∫(0->π/2) [3-4cos2x+ cos4x ] dx
=(1/8)[3x-2sin2x+ (1/4)sin4x ]|(0->π/2)
=3π/16
//
另外
∫(0->π/2) [f(x).∫(0->x) f(t) dt ] dx
=∫(0->π/2) [∫(0->x) f(t) dt ] [f(x) dx ]
=∫(0->π/2) [∫(0->x) f(t) dt ] d[∫(0->x) f(t) dt ]
=(1/2)[ {∫(0->x) f(t) dt }^2 ] | (0->π/2)
=(1/2)[ ∫(0->π/2) f(t) dt ]^2
from (1)
∫(0->π/2) [f(x).∫(0->x) f(t) dt ] dx =∫(0->π/2) (sinx)^4 dx
(1/2)[ ∫(0->π/2) f(t) dt ]^2 = 3π/16
[ ∫(0->π/2) f(t) dt ]^2 = 3π/8
∫(0->π/2) f(t) dt = √(3π/8)
在 (0,π/2) 的平均值
=∫(0->π/2) f(t) dt / (π/2)
= √(3π/8)/ (π/2)
=√[3/(2π)]
