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math104-s21:s:antonthan [2021/05/07 05:20]
45.48.153.5 [Problem 4]
math104-s21:s:antonthan [2022/01/11 18:30] (current)
24.253.46.239 ↷ Links adapted because of a move operation
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 ==== Problem 17 ==== ==== Problem 17 ====
  
 +Consider f(x)eg(x)f(x)e^{g(x)}, which is continuous and differentiable on the same intervals.  We take its derivative and apply MVT, and we are done.
 ==== Problem 18 ==== ==== Problem 18 ====
  
 === (a) === === (a) ===
 +
 +Let partition $P_n = \{0, 1/n, 2/n, \ldots, n/n\}$.  Note that mesh(Pn)=1/nmesh(P_n) = 1/n and L(f,Pn)RnU(f,Pn)L(f, P_n) \leq R_n \leq U(f, P_n).  
 +
 +The idea now is to use Ross 32.7.  Let ϵ>0\epsilon > 0, and since ff is integrable, there exists a δ>0\delta > 0 that satisfies mesh(P)<δ    U(f,P)L(f,P)<ϵmesh(P) < \delta \implies U(f, P) - L(f, P) < \epsilon for all partitions PP.  We choose nn large enough such that mesh(Pn)<δmesh(P_n) < \delta, so that we get the inequalities: U(f,Pn)ϵ<RnU(f,PnU(f, P_n) - \epsilon < R_n \leq U(f, P_n and L(f,Pn)Rn<L(f,Pn)+ϵL(f, P_n) \leq R_n < L(f, P_n) + \epsilon This gives us:
 +
 +U(f)supn(U(f,Pn))=limRn=infn(L(f,Pn))L(f)U(f) \leq \sup_n(U(f, P_n)) = lim R_n = \inf_n(L(f, P_n)) \leq L(f)
 +
 +Since we also have L(f)=U(f)L(f) = U(f) by integrability, limRn=01f(x)dxlim R_n = \int_0^1 f(x)dx, as desired.
  
 === (b) === === (b) ===
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 ===== ඞ ඞ ඞ ඞ ඞ ===== ===== ඞ ඞ ඞ ඞ ඞ =====
-{{ :math104:s:amongus.mp4 |}}+{{ math104-s21:s:amongus.mp4 |}}
math104-s21/s/antonthan.1620390050.txt.gz · Last modified: 2021/05/07 05:20 by 45.48.153.5