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Question-108990




Question Number 108990 by shahria14 last updated on 20/Aug/20
Answered by bobhans last updated on 20/Aug/20
     ((♭o♭hans)/(ς−−−−ς))   set x−1= ♭ → { ((x=5 →♭=4)),((x=1→♭=0)) :}; dx = d♭  ∫_1 ^5 f(x−1)dx = ∫_0 ^4  f(♭) d♭ = 0
$$\:\:\:\:\:\frac{\flat{o}\flat{hans}}{\varsigma−−−−\varsigma} \\ $$$$\:{set}\:{x}−\mathrm{1}=\:\flat\:\rightarrow\begin{cases}{{x}=\mathrm{5}\:\rightarrow\flat=\mathrm{4}}\\{{x}=\mathrm{1}\rightarrow\flat=\mathrm{0}}\end{cases};\:{dx}\:=\:{d}\flat \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}{f}\left({x}−\mathrm{1}\right){dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:{f}\left(\flat\right)\:{d}\flat\:=\:\mathrm{0} \\ $$
Answered by Aziztisffola last updated on 20/Aug/20
 t=x−1 ⇒dt=dx   ∫_1 ^( 5) f(x−1)dx=∫_0 ^( 4) f(t)dt=0
$$\:\mathrm{t}=\mathrm{x}−\mathrm{1}\:\Rightarrow\mathrm{dt}=\mathrm{dx} \\ $$$$\:\int_{\mathrm{1}} ^{\:\mathrm{5}} \mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\:\mathrm{4}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{0} \\ $$

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