请问40这道高数题目，答案划线部分是怎么推导的？

Question

如题所述

Answer 1

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

Answer 2

简单计算一下即可，答案如图所示

Answer 3

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)]