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1-5-x-2-x-1-dx-

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Question Number 195515 by sulaymonnorboyev140 last updated on 04/Aug/23
∫_1 ^5 x^2 (√(x−1))dx=?
$$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}{x}^{\mathrm{2}} \sqrt{{x}−\mathrm{1}}{dx}=? \\ $$
Answered by tri26112004 last updated on 04/Aug/23
Say t=(√(x−1))⇒t^2 =x−1⇔x^2 =(t^2 +1)^2 →dx=2tdt •x=1 to 5 ⇔ t=0 to 2 = ∫^2 _0 (t^2 +1)^2 tdt = ∫^2 _0 (t^5 +2t^3 +t)dt = ((1/6)t^6 +(1/2)t^4 +(1/2)t^2 )∣_0 ^2 = ...
$${Say}\:{t}=\sqrt{{x}−\mathrm{1}}\Rightarrow{t}^{\mathrm{2}} ={x}−\mathrm{1}\Leftrightarrow{x}^{\mathrm{2}} =\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\bullet{x}=\mathrm{1}\:{to}\:\mathrm{5}\:\Leftrightarrow\:{t}=\mathrm{0}\:{to}\:\mathrm{2} \\ $$$$=\:\underset{\mathrm{0}} {\int}^{\mathrm{2}} \:\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {tdt} \\ $$$$=\:\underset{\mathrm{0}} {\int}^{\mathrm{2}} \:\left({t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}\right){dt} \\ $$$$=\:\left(\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{2}} \:=\:… \\ $$
Answered by Calculusboy last updated on 04/Aug/23
let u^2 =x−1=>u^2 +1=x(insert the limit) 2u(du/dx)=1 dx=2udu(continued sir)
$${let}\:{u}^{\mathrm{2}} ={x}−\mathrm{1}=>{u}^{\mathrm{2}} +\mathrm{1}={x}\left({insert}\:{the}\:{limit}\right) \\ $$$$\mathrm{2}{u}\frac{{du}}{{dx}}=\mathrm{1} \\ $$$${dx}=\mathrm{2}{udu}\left({continued}\:{sir}\right) \\ $$
Answered by Frix last updated on 04/Aug/23
∫x^2 (√(x−1))dx =^([t=(√(x−1))]) 2∫t^2 (t^2 +1)^2 dt= =((2t^3 (15t^4 +42t^2 +35))/(105))= =((2(x−1)^(3/2) (15x^2 +12x+8))/(105))+C ∫x^2 (√(x−1))dx =^([t=(1/( (√(x−1))))]) −2∫(((t^2 +1)^2 )/t^8 )dt= =((2(35t^4 +42t^2 +15))/(105t^7 ))= =((2(x−1)^(3/2) (15x^2 +12x+8))/(105))+C Answer is ((7088)/(105))
$$\int{x}^{\mathrm{2}} \sqrt{{x}−\mathrm{1}}{dx}\:\overset{\left[{t}=\sqrt{{x}−\mathrm{1}}\right]} {=}\:\mathrm{2}\int{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {dt}= \\ $$$$=\frac{\mathrm{2}{t}^{\mathrm{3}} \left(\mathrm{15}{t}^{\mathrm{4}} +\mathrm{42}{t}^{\mathrm{2}} +\mathrm{35}\right)}{\mathrm{105}}= \\ $$$$=\frac{\mathrm{2}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{15}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{8}\right)}{\mathrm{105}}+{C} \\ $$$$ \\ $$$$\int{x}^{\mathrm{2}} \sqrt{{x}−\mathrm{1}}{dx}\:\overset{\left[{t}=\frac{\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}\right]} {=}\:−\mathrm{2}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{8}} }{dt}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{35}{t}^{\mathrm{4}} +\mathrm{42}{t}^{\mathrm{2}} +\mathrm{15}\right)}{\mathrm{105}{t}^{\mathrm{7}} }= \\ $$$$=\frac{\mathrm{2}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{15}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{8}\right)}{\mathrm{105}}+{C} \\ $$$$ \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{7088}}{\mathrm{105}} \\ $$

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