发布网友 发布时间:2022-04-23 15:47
共2个回答
热心网友 时间:2023-10-08 21:03
因为∫(0,x)(x - t)f(t)dt =1-cosx
即:∫(0,x)x f(t)dt--∫(0,x) tf(t)dt =1-cosx
两边对x求导得:
∫(0,x) f(t)dt+xf(x)-- xf(x)=sinx
即:
∫(0,x) f(t)dt=sinx,将x=π/2代入得:
所以:∫(0,π/2) f(t)dt=sinx
改写积分变量得:∫(0,π/2) f(x)dx=sinx
热心网友 时间:2023-10-08 21:04
∫(0→x) (x - t)f(t) dt = 1 - cosx
x∫(0→x) f(t) dt - ∫(0→x) tf(t) dt = 1 - cosx
两边求导
∫(0→x) f(t) dt + xf(x) - xf(x) = sinx
再两边求导
f(x) = cosx
所以,
∫(0→π/2) f(x) dx
= ∫(0→π/2) cosx dx
= ∫(0→π/2) d(sinx)
= sin(π/2) - sin(0)
= 1