Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 139248 by ajfour last updated on 24/Apr/21

∫_a ^( b) f(t)g′(t)dt= ?

$$\int_{{a}} ^{\:{b}} {f}\left({t}\right){g}'\left({t}\right){dt}=\:? \\ $$

Answered by mathmax by abdo last updated on 25/Apr/21

∫_a ^b  f(t)g^′ (t)dt =[f(t)g(t)]_a ^b −∫_a ^b  f^′ (t)g(t)dt  =f(b)g(b)−f(a)g(a)−∫_a ^b  f′(t)g(t)dt

$$\int_{\mathrm{a}} ^{\mathrm{b}} \:\mathrm{f}\left(\mathrm{t}\right)\mathrm{g}^{'} \left(\mathrm{t}\right)\mathrm{dt}\:=\left[\mathrm{f}\left(\mathrm{t}\right)\mathrm{g}\left(\mathrm{t}\right)\right]_{\mathrm{a}} ^{\mathrm{b}} −\int_{\mathrm{a}} ^{\mathrm{b}} \:\mathrm{f}^{'} \left(\mathrm{t}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$=\mathrm{f}\left(\mathrm{b}\right)\mathrm{g}\left(\mathrm{b}\right)−\mathrm{f}\left(\mathrm{a}\right)\mathrm{g}\left(\mathrm{a}\right)−\int_{\mathrm{a}} ^{\mathrm{b}} \:\mathrm{f}'\left(\mathrm{t}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ $$

Answered by physicstutes last updated on 24/Apr/21

u = f(t) and dv = g′(t)dt  ⇒ du = f ′(t)dt and v = g(t)  ⇒ [g(t) f(t)]_a ^b −∫_a ^b g(t) f′(t)dt  I = ∫_a ^b g(t)f ′(t) dt = [g(t)f(t)]_a ^b −∫_a ^b f(t)g′(t)dt  ⇒ ∫_a ^b f(t)g′(t)dt = [g(t)f(t)]_a ^b −{[g(t)f(t)]_a ^b −∫_a ^b f(t)g′(t)dt}    determinant (((∫_a ^b f(t)g′(t)dt = 0))) determinant ((),())

$${u}\:=\:{f}\left({t}\right)\:\mathrm{and}\:{dv}\:=\:\mathrm{g}'\left({t}\right){dt} \\ $$$$\Rightarrow\:{du}\:=\:{f}\:'\left({t}\right){dt}\:\mathrm{and}\:{v}\:=\:\mathrm{g}\left({t}\right) \\ $$$$\Rightarrow\:\left[\mathrm{g}\left({t}\right)\:{f}\left({t}\right)\right]_{{a}} ^{{b}} −\int_{{a}} ^{{b}} \mathrm{g}\left({t}\right)\:{f}'\left({t}\right){dt} \\ $$$${I}\:=\:\int_{{a}} ^{{b}} \mathrm{g}\left({t}\right){f}\:'\left({t}\right)\:{dt}\:=\:\left[\mathrm{g}\left({t}\right){f}\left({t}\right)\right]_{{a}} ^{{b}} −\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt} \\ $$$$\Rightarrow\:\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt}\:=\:\left[\mathrm{g}\left({t}\right){f}\left({t}\right)\right]_{{a}} ^{{b}} −\left\{\left[\mathrm{g}\left({t}\right){f}\left({t}\right)\right]_{{a}} ^{{b}} −\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt}\right\} \\ $$$$\:\begin{array}{|c|}{\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt}\:=\:\mathrm{0}}\\\hline\end{array}\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$

Commented by mr W last updated on 24/Apr/21

wrong sir!  you showed only  ∫_a ^b f(t)g′(t)dt=∫_a ^b f(t)g′(t)dt

$${wrong}\:{sir}! \\ $$$${you}\:{showed}\:{only} \\ $$$$\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt}=\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt} \\ $$

Commented by mr W last updated on 24/Apr/21

if ∫_a ^b f(t)g′(t)dt=0, then it means  ∫_a ^b f(t)dt=0, because you can choose  g(t)=t and g′(t)=1.

$${if}\:\int_{{a}} ^{{b}} {f}\left({t}\right)\mathrm{g}'\left({t}\right){dt}=\mathrm{0},\:{then}\:{it}\:{means} \\ $$$$\int_{{a}} ^{{b}} {f}\left({t}\right){dt}=\mathrm{0},\:{because}\:{you}\:{can}\:{choose} \\ $$$${g}\left({t}\right)={t}\:{and}\:{g}'\left({t}\right)=\mathrm{1}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com