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Question Number 67963 by mhmd last updated on 02/Sep/19

Commented by mathmax by abdo last updated on 02/Sep/19

generally if F(x)=∫_(u(x)) ^(v(x)) f(t)dt ⇒F^′ (x)=v^′ (x)f(v(x))−u^′ (x)f(u(x))  we have F(x) =∫_0 ^x  tsin(t)dt ⇒F^′ (x)=(x)^′ (xsinx)−(0)^′ (0sin0)  =xsinx .

$${generally}\:{if}\:{F}\left({x}\right)=\int_{{u}\left({x}\right)} ^{{v}\left({x}\right)} {f}\left({t}\right){dt}\:\Rightarrow{F}^{'} \left({x}\right)={v}^{'} \left({x}\right){f}\left({v}\left({x}\right)\right)−{u}^{'} \left({x}\right){f}\left({u}\left({x}\right)\right) \\ $$$${we}\:{have}\:{F}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:{tsin}\left({t}\right){dt}\:\Rightarrow{F}\:^{'} \left({x}\right)=\left({x}\right)^{'} \left({xsinx}\right)−\left(\mathrm{0}\right)^{'} \left(\mathrm{0}{sin}\mathrm{0}\right) \\ $$$$={xsinx}\:. \\ $$

Answered by mr W last updated on 02/Sep/19

if G′(x)=x sin x, then  F(x)=∫_0 ^x t sin t dt=G(x)−G(0)  ⇒F ′(x)=G′(x)−0=x sin x

$${if}\:{G}'\left({x}\right)={x}\:\mathrm{sin}\:{x},\:{then} \\ $$$${F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {t}\:\mathrm{sin}\:{t}\:{dt}={G}\left({x}\right)−{G}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{F}\:'\left({x}\right)={G}'\left({x}\right)−\mathrm{0}={x}\:\mathrm{sin}\:{x} \\ $$

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