5-t-1-t-4-dt-

Question Number 92888 by s.ayeni14@yahoo.com last updated on 09/May/20

$$\int\frac{\mathrm{5}−\mathrm{t}}{\mathrm{1}+\sqrt{\left(\mathrm{t}−\mathrm{4}\right)}}\mathrm{dt} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 09/May/20

I =∫((5−t)/(1+(√(t−4))))dt we do the changement (√(t−4))=x ⇒t−4=x^2 ⇒ I =∫((5−(x^2 +4))/(1+x))(2x)dx =2∫ ((1−x^2 )/(1+x))xdx =2∫ (1−x)xdx =2∫ (x−x^2 )dx =2((x^2 /2)−(1/3)x^3 ) +C =x^2 −(2/3)x^3 +C =t−4 −(2/3)(t−4)(√(t−4)) +C

$${I}\:=\int\frac{\mathrm{5}−{t}}{\mathrm{1}+\sqrt{{t}−\mathrm{4}}}{dt}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{t}−\mathrm{4}}={x}\:\Rightarrow{t}−\mathrm{4}={x}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\frac{\mathrm{5}−\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\mathrm{1}+{x}}\left(\mathrm{2}{x}\right){dx}\:=\mathrm{2}\int\:\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}}{xdx} \\ $$$$=\mathrm{2}\int\:\left(\mathrm{1}−{x}\right){xdx}\:=\mathrm{2}\int\:\left({x}−{x}^{\mathrm{2}} \right){dx}\:=\mathrm{2}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right)\:+{C} \\ $$$$={x}^{\mathrm{2}} \:−\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:+{C}\:={t}−\mathrm{4}\:−\frac{\mathrm{2}}{\mathrm{3}}\left({t}−\mathrm{4}\right)\sqrt{{t}−\mathrm{4}}\:+{C} \\ $$

Commented by john santu last updated on 09/May/20

(√(t−4)) = u ⇒dt = 2u du ∫ ((5−(u^2 +4) 2u du)/(1+u )) = ∫ (((1−u)(1+u)2u du)/(1+u)) = ∫ (2u−2u^2 ) du = u^2 −(2/3)u^3 +c = (2/3)u^2 ((3/2)−u) +c = (2/3)(t−4)((3/2)−(√(t−4))) +c

$$\sqrt{\mathrm{t}−\mathrm{4}}\:=\:\mathrm{u}\:\Rightarrow\mathrm{dt}\:=\:\mathrm{2u}\:\mathrm{du}\: \\ $$$$\int\:\frac{\mathrm{5}−\left(\mathrm{u}^{\mathrm{2}} +\mathrm{4}\right)\:\mathrm{2u}\:\mathrm{du}}{\mathrm{1}+\mathrm{u}\:}\:=\: \\ $$$$\int\:\frac{\left(\mathrm{1}−\mathrm{u}\right)\left(\mathrm{1}+\mathrm{u}\right)\mathrm{2u}\:\mathrm{du}}{\mathrm{1}+\mathrm{u}}\:=\: \\ $$$$\int\:\left(\mathrm{2u}−\mathrm{2u}^{\mathrm{2}} \right)\:\mathrm{du}\:=\:\mathrm{u}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\mathrm{3}} \:+\mathrm{c}\: \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{u}\right)\:+\mathrm{c}\: \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{t}−\mathrm{4}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{t}−\mathrm{4}}\right)\:+\mathrm{c}\: \\ $$

Commented by s.ayeni14@yahoo.com last updated on 09/May/20

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by maths mind last updated on 10/May/20

$$\mathrm{5}−{t}=\mathrm{1}−\left({t}−\mathrm{4}\right) \\ $$$$=\left(\mathrm{1}−\sqrt{{t}−\mathrm{4}}\right)\left(\mathrm{1}+\sqrt{{t}−\mathrm{4}}\right) \\ $$

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