Question

高数 设f（x）连续，f（x）f（-x）=1？

Answer 1

f(x)f(-x) = 1
f(-x) = 1/f(x)
let
u=-x
du=-dx
x=-π/2, u=π/2
x=π/2, u=-π/2
I
=∫(-π/2->π/2) (sinx)^4/[1+f(x)] dx
=∫(π/2->-π/2) { (sinu)^4/[1+f(-u)] } (-du)
=∫(-π/2->π/2) (sinu)^4/[1+f(-u)] du
=∫(-π/2->π/2) (sinx)^4/[1+f(-x)] dx
=∫(-π/2->π/2) (sinx)^4/[1+1/f(x)] dx
=∫(-π/2->π/2) f(x).(sinx)^4/[1+f(x)] dx
2I
=∫(-π/2->π/2) (sinx)^4/[1+f(x)] dx +∫(-π/2->π/2) f(x).(sinx)^4/[1+f(x)] dx
=∫(-π/2->π/2) (sinx)^4 dx
=2∫(0->π/2) (sinx)^4 dx
=(1/2)∫(0->π/2) (1-cos2x)^2 dx
=(1/2)∫(0->π/2) [1-2cos2x+ (cos2x)^2 ] dx
=(1/4)∫(0->π/2) [3-4cos2x+ cos4x ] dx
=(1/4) [3x-2sin2x+ (1/4)sin4x ]|(0->π/2)
=3π/8
I=3π/16
ie
∫(-π/2->π/2) (sinx)^4/[1+f(x)] dx =3π/16

Answer 2

我连续性能提的话，一般就是那种嗯，那个他如果能求出导数，必定连续。