Question Number 89238 by cindiaulia last updated on 16/Apr/20
$$\int_{\mathrm{7}} ^{\mathrm{12}} \mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}−\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 16/Apr/20
![I=∫ x^2 (√(x−3))dx we do the changement (√(x−3))=t ⇒x−3=t^2 ⇒ I =∫ (t^2 +3)^2 t (2t)dt =2∫ t^2 (t^4 +6t^2 +9)dt =2 ∫ ( t^6 +6t^4 +9t^2 )dt =2{(t^7 /7) +(6/5)t^5 +3t^3 } +c =(2/7)((√(x−3)))^7 +((12)/5)((√(x−3)))^5 +6((√(x−3)))^3 +c ⇒ ∫_7 ^(12) x^2 (√(x−3))dx =[(2/7)((√(x−3)))^7 +((12)/5)((√(x−3)))^5 +6((√(x−3)))^3 ]_7 ^(12) =((2/7)(3)^7 +((12)/5)(3)^5 +6(3)^3 )−((2/7)(2)^7 +((12)/5)(2)^5 +6(2)^3 )](https://www.tinkutara.com/question/Q89279.png)
$${I}=\int\:{x}^{\mathrm{2}} \sqrt{{x}−\mathrm{3}}{dx}\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{3}}={t}\:\Rightarrow{x}−\mathrm{3}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\left({t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} {t}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\int\:{t}^{\mathrm{2}} \left({t}^{\mathrm{4}} \:+\mathrm{6}{t}^{\mathrm{2}} \:+\mathrm{9}\right){dt} \\ $$$$=\mathrm{2}\:\int\:\left(\:{t}^{\mathrm{6}} \:+\mathrm{6}{t}^{\mathrm{4}} \:+\mathrm{9}{t}^{\mathrm{2}} \right){dt}\:=\mathrm{2}\left\{\frac{{t}^{\mathrm{7}} }{\mathrm{7}}\:+\frac{\mathrm{6}}{\mathrm{5}}{t}^{\mathrm{5}} \:+\mathrm{3}{t}^{\mathrm{3}} \right\}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{7}} \:+\frac{\mathrm{12}}{\mathrm{5}}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{5}} \:+\mathrm{6}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{3}} \:+{c}\:\Rightarrow \\ $$$$\int_{\mathrm{7}} ^{\mathrm{12}} {x}^{\mathrm{2}} \sqrt{{x}−\mathrm{3}}{dx}\:=\left[\frac{\mathrm{2}}{\mathrm{7}}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{7}} \:+\frac{\mathrm{12}}{\mathrm{5}}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{5}} \:+\mathrm{6}\left(\sqrt{{x}−\mathrm{3}}\right)^{\mathrm{3}} \right]_{\mathrm{7}} ^{\mathrm{12}} \\ $$$$=\left(\frac{\mathrm{2}}{\mathrm{7}}\left(\mathrm{3}\right)^{\mathrm{7}} \:+\frac{\mathrm{12}}{\mathrm{5}}\left(\mathrm{3}\right)^{\mathrm{5}} \:+\mathrm{6}\left(\mathrm{3}\right)^{\mathrm{3}} \right)−\left(\frac{\mathrm{2}}{\mathrm{7}}\left(\mathrm{2}\right)^{\mathrm{7}} \:+\frac{\mathrm{12}}{\mathrm{5}}\left(\mathrm{2}\right)^{\mathrm{5}} \:+\mathrm{6}\left(\mathrm{2}\right)^{\mathrm{3}} \right) \\ $$$$ \\ $$
Answered by 242242864 last updated on 16/Apr/20
![let p = (√(x−3)) p^2 = x−3 x=p^2 +3 x^2 =(p^2 +3) ⇒x^2 =p^4 +6p^2 +9 2pdp=dx changing limit for x=7,p=(√(7−3 )) =2 x=12, p=(√(12−3 )) =3 ∫_2^ ^3 (p^4 +6p^2 +9).p.2pdp ∫_2 ^3 (2p^6 +12p^4 +18p^2 )dp [((2p^7 )/7)+((12p^5 )/5)+6p^3 ]_2 ^3 =1208.69](https://www.tinkutara.com/question/Q89253.png)
$${let}\:\boldsymbol{{p}}\:=\:\sqrt{\boldsymbol{{x}}−\mathrm{3}} \\ $$$$\boldsymbol{{p}}^{\mathrm{2}} \:=\:\boldsymbol{{x}}−\mathrm{3} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{p}}^{\mathrm{2}} +\mathrm{3} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} =\left(\boldsymbol{{p}}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$\Rightarrow\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{p}}^{\mathrm{4}} +\mathrm{6}\boldsymbol{{p}}^{\mathrm{2}} +\mathrm{9} \\ $$$$\mathrm{2}\boldsymbol{{pdp}}=\boldsymbol{{dx}} \\ $$$$\boldsymbol{{changing}}\:\boldsymbol{{limit}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{x}}=\mathrm{7},\boldsymbol{{p}}=\sqrt{\mathrm{7}−\mathrm{3}\:\:\:}\:\:=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=\mathrm{12},\:\boldsymbol{{p}}=\sqrt{\mathrm{12}−\mathrm{3}\:}\:\:=\mathrm{3} \\ $$$$\int_{\mathrm{2}^{} } ^{\mathrm{3}} \left(\boldsymbol{{p}}^{\mathrm{4}} +\mathrm{6}\boldsymbol{{p}}^{\mathrm{2}} +\mathrm{9}\right).\boldsymbol{{p}}.\mathrm{2}\boldsymbol{{pdp}} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \left(\mathrm{2}\boldsymbol{{p}}^{\mathrm{6}} +\mathrm{12}\boldsymbol{{p}}^{\mathrm{4}} +\mathrm{18}\boldsymbol{{p}}^{\mathrm{2}} \right)\boldsymbol{{dp}} \\ $$$$\left[\frac{\mathrm{2}\boldsymbol{{p}}^{\mathrm{7}} }{\mathrm{7}}+\frac{\mathrm{12}\boldsymbol{{p}}^{\mathrm{5}} }{\mathrm{5}}+\mathrm{6}\boldsymbol{{p}}^{\mathrm{3}} \right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=\mathrm{1208}.\mathrm{69} \\ $$$$ \\ $$