Question

一个定积分的问题

声明，由于定积分打不出来，我口述成如：0到2上的积分+表达式的形式

f(x)在[0,+∞)上二阶连续可导，且f(2)=1, f'(2)=0, 0到2上的积分f(x)dx=4.
求 0到1上的积分[x^2f''(2x)dx]

这里先谢谢解答了！！

Answer 1

let
g(x) = x^2f(x)
g'(x) = 2xf(x) + x^2f'(x)
g''(x) = 2f(x) + 2xf'(x) + x^2f''(x) + 2xf'(x)
= 2f(x) + 4xf'(x) + x^2f''(x)
∫x^2f''(x)dx = ∫g''(x)dx - 2∫f(x)dx - 4∫xf'(x)dx (0到2)

∫g''(x)dx (0到2)
= g'(2) - g'(0)
= 4f(2)+ 4f'(2) - 0 (g'(0)=0)
= 4 + 4(0) = 4

also
-2∫f(x)dx (0到2)
= -2(4) = -8

also
- 4∫xf'(x)dx (0到2)
let
h(x) = xf(x)
h'(x) = xf'(x) + f(x)
∫xf'(x)dx = ∫h'(x) dx -∫f(x)dx (0到2)
consider
∫h'(x) dx (0到2)
=h(2)- h(0)
=2f(2) - 0
=2(1) = 2
-∫f(x)dx (0到2)
= -4

- 4∫xf'(x)dx (0到2)
= -4(-2) = 8

∫x^2f''(x)dx = ∫g''(x)dx - 2∫f(x)dx - 4∫xf'(x)dx (0到2)
= 4 -8 + 8 = 4 (1)

I=∫x^2f''(x)dx (0到2)
let 2y = x
2dy = dx
x=0 , y=0
x=2, y=1

I=∫x^2f''(x)dx (0到2)
= ∫4y^2f''(2y)2dy (0到1)
= 8∫y^2f''(2y)dy (0到1)
= 8∫x^2f''(2x)dx (0到1)

from (1)
∫x^2f''(x)dx (0到2)
= 4
8∫x^2f''(2x)dx (0到1)
=4

∫x^2f''(2x)dx (0到1)
= 1/2 #

Answer 2

如图