Question

高数题 有悬赏

Answer 1

f(x)= x^5 .(sinx)^2
f(-x) = -f(x)
∫(-π/2->π/2) (x^5+1)(sinx)^2 dx
=∫(-π/2->π/2) x^5.(sinx)^2 dx +∫(-π/2->π/2) (sinx)^2 dx
=0 +∫(-π/2->π/2) (sinx)^2 dx
=2∫(0->π/2) (sinx)^2 dx
=∫(0->π/2) (1-cos2x) dx
=[ u-(1/2)sin2u]|(0->π/2)
=π/2

追问

加号前边的为什么变成0了

追答

f(-x) = -f(x)
f(x)= x^5 .(sinx)^2 : 是奇函数
∫(-a->a) f(x) dx =0
let
u = -x
du = -dx
x=-a , u=a
x=a, u=-a
∫(-a->a) f(x) dx
=∫(a->-a) f(-u) ( -du)
=∫(a->-a) f(u) du
=-∫(-a->a) f(u) du
=-∫(-a->a) f(x) dx
2∫(-a->a) f(x) dx = 0
∫(-a->a) f(x) dx =0