Question

高等数学定积分问题

Answer 1

(1)
let dF(x) = (1/√(1+x^3))dx
f(a) = ∫(a->2a) dx/√(1+x^3)

= F(2a)- F(a)
f'(a) = 2F'(2a) - F'(a)
= 2/√(1+8a^3) - 1/√(1+a^3) =0
2√(1+a^3) -√(1+8a^3)=0
4(1+a^3) = 1+8a^3
4a^3 = 3
a= (3/4)^(1/3)
f''((3/4)^(1/3)) <0 ( max)
max ∫(a->2a) dx/√(1+x^3) at a= (3/4)^(1/3)
(2)
?????
(3)
let dF(x) = f(x) dx
∫(0->1) f(tx)dt =x
(1/x) [ F(x)-F(0)] = x

[ F(x)-F(0)] =x^2

f(x) = 2x

追问

（2）题连续函数是F(X)，且F(x)=.......

追答

F(x) =∫(2->t)dy∫(y->t)f(x) dx
∫(2->t)dy∫(y->t)f(x) dx 不是x的函数
∫(2->t)dy∫(y->t)f(x) dx 是t或者y的函数
如果是这样
F(x) = ∫(2->t)dy∫(y->t)f(x) dx
F'(x) = 0
F'(2) =0

追问

打错了F(t),不好意思

追答

let
dG(x) = f(x)dx
F(t) =∫(2->t)dy∫(y->t)f(x) dx

=∫(2->t) [G(t) -G(y)] dy
= G(t)(t-2) -∫(2->t)G(y) dy
F'(t) = G(t)+(t-2)f(x) - G(t)
= (t-2)f(x)
F'(2) =0