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Question-215386




Question Number 215386 by JasonHidd last updated on 04/Jan/25
Answered by mr W last updated on 04/Jan/25
∫_0 ^x f(x)dx=∫^(−x) _0 f(x)dx+∫_(−x) ^x f(x)dx  this is always true.   now the question is if f(x) exists s.t.  ∫_0 ^x f(t)dt=x^2 +2x+1  the answer is no.  say F(x)=∫_0 ^x f(t)dt  F(0)=∫_0 ^0 f(t)dt=0  but F(x)=x^2 +2x+1 ⇒F(0)=1≠0.
$$\int_{\mathrm{0}} ^{{x}} {f}\left({x}\right){dx}=\underset{\mathrm{0}} {\int}^{−{x}} {f}\left({x}\right){dx}+\int_{−{x}} ^{{x}} {f}\left({x}\right){dx} \\ $$$${this}\:{is}\:{always}\:{true}.\: \\ $$$${now}\:{the}\:{question}\:{is}\:{if}\:{f}\left({x}\right)\:{exists}\:{s}.{t}. \\ $$$$\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${the}\:{answer}\:{is}\:{no}. \\ $$$${say}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt} \\ $$$${F}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{0}} {f}\left({t}\right){dt}=\mathrm{0} \\ $$$${but}\:{F}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\:\Rightarrow{F}\left(\mathrm{0}\right)=\mathrm{1}\neq\mathrm{0}. \\ $$

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