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Question Number 4344 by Filup last updated on 12/Jan/16

for ∫f(x)dx=F(x)+c  and sgn(x)=(x/(∣x∣))=((∣x∣)/x)   ∀x≠0            let sgn(x)=0 for x=0    does   ∫sgn(f(x))f(x)dx=sgn(f(x))∫f(x)dx  ∵sgn(f(x)) is just a constant ±1 or 0.

$$\mathrm{for}\:\int{f}\left({x}\right){dx}={F}\left({x}\right)+{c} \\ $$$$\mathrm{and}\:{sgn}\left({x}\right)=\frac{{x}}{\mid{x}\mid}=\frac{\mid{x}\mid}{{x}}\:\:\:\forall{x}\neq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{let}\:\mathrm{sgn}\left({x}\right)=\mathrm{0}\:\mathrm{for}\:{x}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{does}\: \\ $$$$\int{sgn}\left({f}\left({x}\right)\right){f}\left({x}\right){dx}={sgn}\left({f}\left({x}\right)\right)\int{f}\left({x}\right){dx} \\ $$$$\because{sgn}\left({f}\left({x}\right)\right)\:\mathrm{is}\:\mathrm{just}\:\mathrm{a}\:\mathrm{constant}\:\pm\mathrm{1}\:\mathrm{or}\:\mathrm{0}. \\ $$

Commented by Filup last updated on 12/Jan/16

Let f(x)=x    sgn(x)x=x((∣x∣)/x)=x(x/(∣x∣))=∣x∣                     N.B.: (x^2 /(∣x∣))=∣x∣   ∵x^2 ∧∣x∣>0    S_1 =∫sgn(x)xdx  =∫∣x∣dx     (1)    S_2 =sgn(x)∫xdx  =(1/2)sgn(x)x^2 +c  =(1/2)∣x∣x+c  (2)      if true:  (d/dx)((1/2)∣x∣x+c)=∣x∣  (d/dx)(∣x∣)=(d/dx)((x^2 )^(1/2) )  =(1/2)(x^2 )^(−(1/2)) 2x  =(x/(∣x∣))  (d/dx)((1/2)∣x∣x+c)=(1/2)(u′v+uv′)+0  u=∣x∣     v=x  u′=(x/(∣x∣))    v′=1  =(1/2)((x^2 /(∣x∣))+∣x∣)  =(1/2)(∣x∣+∣x∣)  =∣x∣    ∴∫sgn(x)xdx=sgn(x)∫xdx    Can we prove that the above is true/false  for limits −1≤x≤1?

$$\mathrm{Let}\:{f}\left({x}\right)={x} \\ $$$$ \\ $$$$\mathrm{sgn}\left({x}\right){x}={x}\frac{\mid{x}\mid}{{x}}={x}\frac{{x}}{\mid{x}\mid}=\mid{x}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{N}.{B}.:\:\frac{{x}^{\mathrm{2}} }{\mid{x}\mid}=\mid{x}\mid\:\:\:\because{x}^{\mathrm{2}} \wedge\mid{x}\mid>\mathrm{0} \\ $$$$ \\ $$$${S}_{\mathrm{1}} =\int\mathrm{sgn}\left({x}\right){xdx} \\ $$$$=\int\mid{x}\mid{dx}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${S}_{\mathrm{2}} =\mathrm{sgn}\left({x}\right)\int{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sgn}\left({x}\right){x}^{\mathrm{2}} +{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}\mid{x}+{c}\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{true}: \\ $$$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mid{x}\mid{x}+{c}\right)=\mid{x}\mid \\ $$$$\frac{{d}}{{dx}}\left(\mid{x}\mid\right)=\frac{{d}}{{dx}}\left(\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}{x} \\ $$$$=\frac{{x}}{\mid{x}\mid} \\ $$$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mid{x}\mid{x}+{c}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({u}'{v}+{uv}'\right)+\mathrm{0} \\ $$$${u}=\mid{x}\mid\:\:\:\:\:{v}={x} \\ $$$${u}'=\frac{{x}}{\mid{x}\mid}\:\:\:\:{v}'=\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}^{\mathrm{2}} }{\mid{x}\mid}+\mid{x}\mid\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mid{x}\mid+\mid{x}\mid\right) \\ $$$$=\mid{x}\mid \\ $$$$ \\ $$$$\therefore\int\mathrm{sgn}\left({x}\right){xdx}=\mathrm{sgn}\left({x}\right)\int{xdx} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{above}\:\mathrm{is}\:\mathrm{true}/\mathrm{false} \\ $$$$\mathrm{for}\:\mathrm{limits}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}? \\ $$

Commented by prakash jain last updated on 11/Jan/16

If f(x) is such that it does not change sign  within the limits of integration. You  can take out.

$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{does}\:\mathrm{not}\:\mathrm{change}\:\mathrm{sign} \\ $$$$\mathrm{within}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{of}\:\mathrm{integration}.\:\mathrm{You} \\ $$$$\mathrm{can}\:\mathrm{take}\:\mathrm{out}. \\ $$

Commented by 123456 last updated on 11/Jan/16

i think its not im general

$$\mathrm{i}\:\mathrm{think}\:\mathrm{its}\:\mathrm{not}\:\mathrm{im}\:\mathrm{general} \\ $$

Commented by Filup last updated on 12/Jan/16

∫_(−1) ^( 1) f(x)dx=∫_(−1) ^( 0) f(x)dx+∫_0 ^( 1) f(x)dx    ∴if true:  ∫_(−1) ^( 0) sgn(x)xdx+∫_0 ^( 1) sgn(x)xdx=sgn(x)∫_(−1) ^( 0) xdx+sgn(x)∫_0 ^( 1) xdx  ⇒^(???) ∫_(−1) ^( 0) sgn(x)xdx+∫_0 ^( 1) sgn(x)xdx=sgn(x)(∫_(−1) ^( 0) xdx+∫_0 ^( 1) xdx)

$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}=\int_{−\mathrm{1}} ^{\:\mathrm{0}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$ \\ $$$$\therefore\mathrm{if}\:\mathrm{true}: \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{0}} \mathrm{sgn}\left({x}\right){xdx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{sgn}\left({x}\right){xdx}=\mathrm{sgn}\left({x}\right)\int_{−\mathrm{1}} ^{\:\mathrm{0}} {xdx}+\mathrm{sgn}\left({x}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} {xdx} \\ $$$$\overset{???} {\Rightarrow}\int_{−\mathrm{1}} ^{\:\mathrm{0}} \mathrm{sgn}\left({x}\right){xdx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{sgn}\left({x}\right){xdx}=\mathrm{sgn}\left({x}\right)\left(\int_{−\mathrm{1}} ^{\:\mathrm{0}} {xdx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} {xdx}\right) \\ $$

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