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For-f-x-ax-n-b-when-f-o-f-is-continuous-Does-there-exist-a-solution-S-f-x-dx-lt-

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Question Number 4116 by Filup last updated on 29/Dec/15
For: f(x)=∣ax^n +b∣ when f(α) o f(β) is continuous, Does there exist a solution: S=∫_α ^( β) f(x)dx α<β
For:f(x)=∣axn+b∣whenf(α)of(β)iscontinuous,Doesthereexistasolution:S=∫αβf(x)dxα<β
Commented by Yozzii last updated on 29/Dec/15
f: R→R and α<β (α,β∈R) with f being defined at x=α,x=β. f(x)=∣ax^n +b∣ represents the modulus of a real polynomial,if n∈Z^+ +{0} and a,b∈R. Thus, f(x) is continuous ∀x∈R. ∴ f(x)=∣ax^n +b∣≥0⇒∫_α ^β f(x)dx≥0 ⇒∃S∈R^+ +{0} such that S=∫_α ^β f(x)dx. S≮0 since f(x)≮0 ∀x∈R. If n∈Z^− , and α<0<β then ∄S∈C such that S=∫_α ^β f(x)dx since the integral does not exist. ∫_α ^β f(x)dx=∫_α ^0 f(x)dx+∫_0 ^β f(x)dx =∫_α ^0 ∣(a/x^n )+b∣dx+∫_0 ^β ∣(a/x^n )+b∣dx f(0) is undefined so that each of the above integrals are undefined within the limit of x→0.
f:R→Randα<β(α,β∈R)withfbeingdefinedatx=α,x=β.f(x)=∣axn+b∣representsthemodulusofarealpolynomial,ifn∈Z++{0}anda,b∈R.Thus,f(x)iscontinuous∀x∈R.∴f(x)=∣axn+b∣⩾0⇒∫αβf(x)dx⩾0⇒∃S∈R++{0}suchthatS=∫αβf(x)dx.S≮0sincef(x)≮0∀x∈R.Ifn∈Z−,andα<0<βthen∄S∈CsuchthatS=∫αβf(x)dxsincetheintegraldoesnotexist.∫αβf(x)dx=∫α0f(x)dx+∫0βf(x)dx=∫α0∣axn+b∣dx+∫0β∣axn+b∣dxf(0)isundefinedsothateachoftheaboveintegralsareundefinedwithinthelimitofx→0.

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