Question Number 89489 by cindiaulia last updated on 17/Apr/20
$$\int_{−\mathrm{1}} ^{\mathrm{4}} \mathrm{x}\sqrt{×+\mathrm{5}\:}\mathrm{dx} \\ $$
Commented by M±th+et£s last updated on 17/Apr/20
$${let}\:{x}+\mathrm{5}={u}\:\:{x}={u}−\mathrm{5}\:\:\:\:{du}={dx} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{4}} \left({u}−\mathrm{5}\right)\sqrt{{u}}\:\:{du} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{4}} {u}\sqrt{{u}}{du}\:−\mathrm{5}\int_{−\mathrm{1}} ^{\mathrm{4}} \sqrt{{u}}\:{du} \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}{u}^{\mathrm{2}} \:\sqrt{{u}}\mid_{−\mathrm{1}} ^{\mathrm{4}} \:−\:\frac{\mathrm{10}}{\mathrm{3}}{u}\sqrt{{u}}\:\mid_{−\mathrm{1}} ^{\mathrm{4}} \\ $$$${now}\:{put}\:{u}={x}+\mathrm{5}\:\:\: \\ $$$$=\frac{\mathrm{316}}{\mathrm{15}} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 18/Apr/20
![(√(x+5))=t ⇒x+5=t^2 ⇒ ∫_(−1) ^4 x(√(x+5))dx =∫_2 ^3 (t^2 −5)t(2t)dt =2 ∫_2 ^3 t^2 (t^2 −5)dt =2 ∫_2 ^3 (t^4 −5t^2 )dt =2[(t^5 /5)−(5/3)t^3 ]_2 ^3 =2{((3^5 /5)−(5/3)3^3 )−((2^5 /5)−(5/3)2^3 )}](https://www.tinkutara.com/question/Q89674.png)
$$\sqrt{{x}+\mathrm{5}}={t}\:\Rightarrow{x}+\mathrm{5}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{4}} {x}\sqrt{{x}+\mathrm{5}}{dx}\:=\int_{\mathrm{2}} ^{\mathrm{3}} \left({t}^{\mathrm{2}} −\mathrm{5}\right){t}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{2}} ^{\mathrm{3}} {t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{5}\right){dt}\:=\mathrm{2}\:\int_{\mathrm{2}} ^{\mathrm{3}} \left({t}^{\mathrm{4}} −\mathrm{5}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\mathrm{2}\left[\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{3}}{t}^{\mathrm{3}} \right]_{\mathrm{2}} ^{\mathrm{3}} =\mathrm{2}\left\{\left(\frac{\mathrm{3}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{3}}\mathrm{3}^{\mathrm{3}} \right)−\left(\frac{\mathrm{2}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{3}}\mathrm{2}^{\mathrm{3}} \right)\right\} \\ $$