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Question Number 99261 by Ar Brandon last updated on 19/Jun/20

Given F(x)=(1/2)∫((x+1)/(x−1))f(t)dt  Show that F is defined, continuous, and derivable.  And find its derivative

$$\mathrm{Given}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{F}\:\mathrm{is}\:\mathrm{defined},\:\mathrm{continuous},\:\mathrm{and}\:\mathrm{derivable}. \\ $$$$\mathrm{And}\:\mathrm{find}\:\mathrm{its}\:\mathrm{derivative} \\ $$

Answered by abdomathmax last updated on 19/Jun/20

F(x) =(1/2)((x+1)/(x−1)) ∫^x f(t)dt ⇒F^′ (x)=(1/2)(((x+1)/(x−1)))^′  ∫^x f(t)dt  +((x+1)/(2(x−1)))f^′ (x) =(1/2)×((x−1−(x+1))/((x−1)^2 )) ∫^x f(t)dt  +((x+1)/(2(x−1)))f^′ (x) =−(1/((x−1)^2 )) ∫^x f(t)dt+((x+1)/(2(x−1)))f^′ (x)

$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\int^{\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\:\Rightarrow\mathrm{F}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)^{'} \:\int^{\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)}\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{x}−\mathrm{1}−\left(\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:\int^{\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)}\mathrm{f}^{'} \left(\mathrm{x}\right)\:=−\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:\int^{\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)}\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$

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