Question Number 69576 by Mr. K last updated on 25/Sep/19
$$\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 26/Sep/19
![let I =∫_(−2) ^2 (x^3 cos((x/2))+(1/2))(√(4−x^2 ))dx ⇒ I =∫_(−2) ^2 x^3 cos((x/2))(√(4−x^2 ))dx +(1/2) ∫_(−2) ^2 (√(4−x^2 ))dx but ∫_(−2) ^2 x^3 cos((x/2))(√(4−x^2 ))dx =0((function under ∫ is odd) ⇒ I =(1/2) ∫_(−2) ^2 (√(4−x^2 ))dx =∫_0 ^2 (√(4−x^2 ))dx =_(x=2sinθ) ∫_0 ^(π/2) 2 cosθ (2cosθ)dθ =4 ∫_0 ^(π/2) ((1+cos(2θ))/2)dθ = 2 ∫_0 ^(π/2) dθ +∫_0 ^(π/2) cos(2θ)dθ =π +(1/2)[sin(2θ)]_0 ^(π/2) =π +0 =π ⇒ I =π](https://www.tinkutara.com/question/Q69701.png)
$${let}\:{I}\:=\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${I}\:=\int_{−\mathrm{2}} ^{\mathrm{2}} {x}^{\mathrm{3}} \:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:\:{but} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} {x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:=\mathrm{0}\left(\left({function}\:{under}\:\int\:{is}\:{odd}\right)\:\Rightarrow\right. \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{2}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:=_{{x}=\mathrm{2}{sin}\theta} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}\:{cos}\theta\:\left(\mathrm{2}{cos}\theta\right){d}\theta \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\mathrm{2}\theta\right){d}\theta \\ $$$$=\pi\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\pi\:+\mathrm{0}\:=\pi\:\Rightarrow\:{I}\:=\pi \\ $$
Answered by mind is power last updated on 25/Sep/19
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}{x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{1}}{\mathrm{2}}\underset{−\mathrm{2}} {\overset{\mathrm{2}} {\int}}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{2}} {x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$$\Rightarrow\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }={A} \\ $$$${equation}\:{ofcircle}\:{of}\:{centerO}\left(\mathrm{0},\mathrm{0}\right)\:{Radius}\:\mathrm{2} \\ $$$${has}\:{equation}\:..{c}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\Rightarrow{y}=\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\:{if}\:{y}\geqslant\mathrm{0} \\ $$$${Area}\:{of}\:{S}=\left\{\left({x},{y}\right)\in{c}\:\mid\:{x},{y}\geqslant\mathrm{0}\right\}={Area}\:{S}'\left\{\left({x},{y}\right)\in{c}\mid\:{x},{y}\leqslant\mathrm{0}\right\} \\ $$$${cause}\:\left({x},{y}\right)\overset{{f}} {\rightarrow}\left(−{x},−{y}\right)\:\:\:{bijection}\:{S}\:{and}\:{S}' \\ $$$$\Rightarrow\int\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}.\pi.\mathrm{2}^{\mathrm{2}} =\mathrm{2}\pi \\ $$$$\Rightarrow\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{2}} ^{+\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\pi=\pi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Mr. K last updated on 26/Sep/19
$${Amazing}\:{work}! \\ $$$$ \\ $$$$ \\ $$
Commented by mind is power last updated on 26/Sep/19
$${thanx}\: \\ $$