Question Number 192024 by Shlock last updated on 05/May/23
Answered by a.lgnaoui last updated on 05/May/23
![The shaded Area betwen [y=0,y=(√(25−x^2 )) −3 ;y=(√(4−x^2 ))] ∫_0 ^4 ((√(25−x^2 )) −(√(4−x^2 )) −3)dx 5∫_0 ^4 (√(1−((x/5))^2 )) dx−2∣∫_0 ^2 (√(1−((x/2))^2 )) ∣−3∫_0 ^4 dx =(I_1 +I_3 )−I_2 Calcul de I_1 Posons (x/5)=cos t ⇒(√(1−cos ^2 t)) =sint x=5cost [ dx=−5sin tdt x=0 t=(π/2) x=4 t=arccos ((4/5)) I_1 =−5∫sin tcos tdt=−(5/2)∫_(𝛑/2) ^(srccos((4/5))) sin 2tdt =((−5)/4)[cos 2t]_(𝛑/2) ^(arccos ((4/5))) =((−5)/4)[ (0,4)−1]=(3/4) calcul de I_2 I_2 =2(√(1−((x/2))^2 )) dx ( (x/2))=cos t x=2cos t dx=−2sin tdt x=0 t=(𝛑/2) x=2 t=0 −2∫_(𝛑/2) ^0 sin2tdt=−[cos 2t]_(𝛑/2) ^0 =−(1+1)=−2 calcul de I_3 I_3 =3[x]_0 ^4 =12 I=∣(3/4)−12+2∣=((3−48+8)/4)∣=((37)/4)](https://www.tinkutara.com/question/Q192028.png)
$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{shaded}}\:\boldsymbol{\mathrm{Area}}\:\boldsymbol{\mathrm{betwen}} \\ $$$$\left[\boldsymbol{\mathrm{y}}=\mathrm{0},\boldsymbol{\mathrm{y}}=\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\mathrm{3}\:\:;\boldsymbol{\mathrm{y}}=\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \left(\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:\:−\mathrm{3}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\mathrm{5}\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{5}}\right)^{\mathrm{2}} \:}\:\mathrm{dx}−\mathrm{2}\mid\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\mid−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{dx} \\ $$$$=\left(\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{3}} \right)−\mathrm{I}_{\mathrm{2}} \:\:\: \\ $$$$\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{I}}_{\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{P}}\mathrm{osons}\:\:\frac{\mathrm{x}}{\mathrm{5}}=\mathrm{cos}\:\mathrm{t}\:\:\Rightarrow\sqrt{\mathrm{1}−\mathrm{cos}\:\:^{\mathrm{2}} \mathrm{t}}\:=\mathrm{sint} \\ $$$$\mathrm{x}=\mathrm{5cost}\:\:\left[\:\mathrm{dx}=−\mathrm{5sin}\:\:\mathrm{tdt}\:\:\:\right. \\ $$$$\mathrm{x}=\mathrm{0}\:\:\mathrm{t}=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\mathrm{x}=\mathrm{4}\:\:\:\:\mathrm{t}=\mathrm{arccos}\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\mathrm{I}_{\mathrm{1}} \:\:=−\mathrm{5}\int\mathrm{sin}\:\mathrm{tcos}\:\mathrm{tdt}=−\frac{\mathrm{5}}{\mathrm{2}}\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{s}\boldsymbol{\mathrm{rccos}}\left(\frac{\mathrm{4}}{\mathrm{5}}\right)} \mathrm{sin}\:\mathrm{2tdt} \\ $$$$=\frac{−\mathrm{5}}{\mathrm{4}}\left[\mathrm{cos}\:\mathrm{2t}\right]_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{arccos}\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)} =\frac{−\mathrm{5}}{\mathrm{4}}\left[\:\left(\mathrm{0},\mathrm{4}\right)−\mathrm{1}\right]=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\boldsymbol{\mathrm{calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{I}}_{\mathrm{2}} \\ $$$$\:\:\boldsymbol{\mathrm{I}}_{\mathrm{2}} \:\:\:=\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\boldsymbol{\mathrm{dx}}\:\:\:\left(\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)=\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{t}} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{2cos}\:\boldsymbol{\mathrm{t}}\:\:\:\boldsymbol{\mathrm{dx}}=−\mathrm{2}\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{tdt}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{0}\:\:\:\mathrm{t}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\:\:\mathrm{x}=\mathrm{2}\:\:\:\mathrm{t}=\mathrm{0} \\ $$$$−\mathrm{2}\int_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{tdt}}=−\left[\mathrm{cos}\:\mathrm{2t}\right]_{\frac{\boldsymbol{\pi}}{\mathrm{2}}} ^{\mathrm{0}} =−\left(\mathrm{1}+\mathrm{1}\right)=−\mathrm{2} \\ $$$$\boldsymbol{\mathrm{calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{I}}_{\mathrm{3}} \:\:\:\:\:\:\boldsymbol{\mathrm{I}}_{\mathrm{3}} =\mathrm{3}\left[\mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{4}} =\mathrm{12} \\ $$$$\boldsymbol{\mathrm{I}}=\mid\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{12}+\mathrm{2}\mid=\frac{\mathrm{3}−\mathrm{48}+\mathrm{8}}{\mathrm{4}}\mid=\frac{\mathrm{37}}{\mathrm{4}}\:\:\:\: \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 06/May/23
$${Ithink}\:{there}\:{is}\:{error}\:{in}\:{calcul}\:\left({integrals}\right)? \\ $$