Question

微积分题 求解

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Answer 1

1) F'(x) = 2xf(x^2+1) - 2f(2x-1)
2) integral [0,n] 1/(x+1) dx = ln(n+1) < 1 + 1/2 + 1/3 + ... + 1/n
and 1 + integral [1,n] 1/x dx = 1 + lnn >尺游 1 + 1/2 + 1/3 + ... + 1/n
Combining: ln(n+1) < 1 + 1/2 + 1/3 + ... + 1/n < 1 + lnn (具体过程画图看数值解可知）
3）i) Integration by parts:
I(n) = integral [0,pi/2] sin^nx dx
= -integral [0,pi/陵升销2] sin^(n-1)x dcosx
= -cosx sin^(n-1)x| [0,pi/2] + integral [0,pi/笑败2] cosx (n-1)sin^(n-2) x cosx dx
= integral [0,pi/2] (1-sin^2x) (n-1)sin^(n-2) x dx
= (n-1)I(n-2) - (n-1)I(n)
==> I(n) = [(n-1)/n] I(n-2)
ii) n = 2m
I(n) = [(n-1)(n-3)....1]/[n(n-2)(n-4)...2] ;

n = 2m+1
I(n) =(pi/2) [(n-1)(n-3)...(2)]/[n(n-2)(n-4)...1]