Question Number 66340 by mathmax by abdo last updated on 12/Aug/19
$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}\right)}{dx} \\ $$
Commented by mathmax by abdo last updated on 15/Aug/19
![let I =∫_0 ^2 (√(x^3 (2−x)))dx ⇒ I =∫_0 ^2 x(√(x(2−x)))dx =∫_0 ^2 x(√(−x^2 +2x))dx we have −x^2 +2x =−(x^2 −2x) =−(x^2 −2x+1−1) =−(x−1)^2 +1 ⇒I =∫_0 ^2 x(√(1−(x−1)^2 ))dx we use the changement x−1=sint ⇒ I = ∫_(−(π/2)) ^(π/2) (1+sint)cost(cost)dt =∫_(−(π/2)) ^(π/2) cos^2 t dt +∫_(−(π/2)) ^(π/2) sint cos^2 tdt =∫_(−(π/2)) ^(π/2) ((1+cos(2t))/2)dt +0 (t→sint cos^2 t is odd) =(π/2) +(1/4)[sin(2t)]_(−(π/2)) ^(π/2) =(π/2)+0 ⇒ I =(π/2)](https://www.tinkutara.com/question/Q66456.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}\right)}{dx}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} {x}\sqrt{{x}\left(\mathrm{2}−{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} {x}\sqrt{−{x}^{\mathrm{2}} +\mathrm{2}{x}}{dx}\:\:\:{we}\:{have}\:−{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:=−\left({x}^{\mathrm{2}} −\mathrm{2}{x}\right) \\ $$$$=−\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−\mathrm{1}\right)\:=−\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} {x}\sqrt{\mathrm{1}−\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${we}\:{use}\:{the}\:{changement}\:{x}−\mathrm{1}={sint}\:\Rightarrow \\ $$$${I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+{sint}\right){cost}\left({cost}\right){dt}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {t}\:{dt}\:+\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {sint}\:{cos}^{\mathrm{2}} {tdt} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:\:\:+\mathrm{0}\:\:\left({t}\rightarrow{sint}\:{cos}^{\mathrm{2}} {t}\:{is}\:{odd}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{2}}+\mathrm{0}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}} \\ $$