Question

求一道高数计算定积分的题目

Answer 1

I=∫[0,1] f(x)dx
=∫[0,1] [1+(1/2)*∫[x,1]f(y)f(y-x)dy]dx
=∫[0,1] 1dx+(1/2)∫[0,1] dx∫[x,1]f(y)f(y-x)dy
=1+(1/2)∫[0,1] dx∫[x,1]f(y)f(y-x)dy
2I-2=∫[0,1] dx∫[x,1]f(y)f(y-x)dy
积分区域是y=1,x=0和y=x围成的三角形区域,交换顺序有 0<y<1,0<x<y
=∫[0,1] dy∫[0,y]f(y)f(y-x)dx
=∫[0,1] f(y)dy∫[0,y]f(y-x)dx
==令t=y-x==
=∫[0,1] f(y)dy(-∫[y,0]f(t)dt)
=∫[0,1] f(y)dy∫[0,y]f(t)dt
==设F(y)=∫[0,y]f(t)dt,则F'(y)=f(y)==
=∫[0,1] f(y)F(y)dy
=∫[0,1] F(y)dF(y)
=(F(y))^2/2 |[0,1]
={∫[0,y]f(t)dt }^2/2 |[0,1]
={∫[0,1]f(t)dt }^2/2 = I^2/2
所以I^2=4I-4,所以I=2