Question Number 161294 by floor(10²Eta[1]) last updated on 15/Dec/21
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} \mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by puissant last updated on 20/Dec/21
![f(x)=x^3 cos((x/2))(√(4−x^2 )) + (1/2)(√(4−x^2 )) f(x)=g(x) + h(x) g(x)=x^3 cos((x/2))(√(4−x^2 )) et h(x)=(1/2)(√(4−x^2 )) ∫_(−2) ^2 x^3 cos((x/2))(√(4−x^2 )) dx = 0 ⇒ Ω=∫_(−2) ^2 (x^3 cos((x/2))+(1/2))(√(4−x^2 ))dx = (1/2)∫_(−2) ^2 (√(4−x^2 ))dx = ∫_0 ^2 (√(4−x^2 )) dx ; x=2sint → dx=2costdt ⇒ Ω = ∫_0 ^(π/2) (√(4(1−sin^2 t))) 2cost dt ⇒ Ω = 4∫_0 ^(π/2) cos^2 t dt = 2∫_0 ^(π/2) {1+cos2t}dt = 2[t+(1/2)sin2t]_0 ^(π/2) = 2[{(π/2)+(1/2)sinπ}−0] Ω = π....★ ..................Le puissant.....................](https://www.tinkutara.com/question/Q161304.png)
$${f}\left({x}\right)={x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)={g}\left({x}\right)\:+\:{h}\left({x}\right) \\ $$$${g}\left({x}\right)={x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:{et}\:{h}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\: \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} {x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\Omega=\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx}\:;\:{x}=\mathrm{2}{sint}\:\rightarrow\:{dx}=\mathrm{2}{costdt} \\ $$$$\Rightarrow\:\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{4}\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right)}\:\mathrm{2}{cost}\:{dt} \\ $$$$\Rightarrow\:\Omega\:=\:\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} {t}\:{dt}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{1}+{cos}\mathrm{2}{t}\right\}{dt} \\ $$$$=\:\mathrm{2}\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\mathrm{2}\left[\left\{\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\pi\right\}−\mathrm{0}\right] \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega\:\:\:=\:\:\:\pi….\bigstar \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………\mathscr{L}{e}\:{puissant}………………… \\ $$
Commented by Ar Brandon last updated on 15/Dec/21
$$\mathrm{g}\left({x}\right)\:\mathrm{is}\:\mathrm{even}. \\ $$
Commented by Ar Brandon last updated on 15/Dec/21
$$\mathrm{yep}\:\:\mathrm{odd},\:\mathrm{I}\:\mathrm{meant}. \\ $$