Question Number 72337 by Rio Michael last updated on 27/Oct/19
$${Evaluate}\:\:\int_{−\mathrm{5}} ^{\mathrm{5}} \left(\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }\:\right)\:{dx}\:{using} \\ $$$$\Rightarrow\:{an}\:{algebraic}\:{method} \\ $$$$\Rightarrow\:{Geometrical}\:{mehod}\: \\ $$$${thanks}\:{in}\:{advanced}\:{great}\:{mathematicians} \\ $$
Commented by mathmax by abdo last updated on 27/Oct/19
![algebric method let A=∫_(−5) ^5 (√(25−x^2 ))dx ⇒A=2∫_0 ^5 (√(25−x^2 ))dx (even function) ⇒A =_(x=5sint) 2∫_0 ^(π/2) (√(25−25sin^2 t))5cost dt =50 ∫_0 ^(π/2) cos^2 t dt =25 ∫_0 ^(π/2) (1+cos(2t))dt =((25π)/2) +((25)/2)[sin(2t)]_0 ^(π/2) =((25π)/2)](Q72349.png)
$${algebric}\:{method}\:\:{let}\:{A}=\int_{−\mathrm{5}} ^{\mathrm{5}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }{dx}\:\Rightarrow{A}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{5}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }{dx} \\ $$$$\left({even}\:{function}\right)\:\Rightarrow{A}\:=_{{x}=\mathrm{5}{sint}} \:\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{25}−\mathrm{25}{sin}^{\mathrm{2}} {t}}\mathrm{5}{cost}\:{dt} \\ $$$$=\mathrm{50}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {t}\:{dt}\:=\mathrm{25}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{25}\pi}{\mathrm{2}}\:+\frac{\mathrm{25}}{\mathrm{2}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{25}\pi}{\mathrm{2}} \\ $$
Answered by MJS last updated on 27/Oct/19
![geometrical method y=(√(25−x^2 )) is the upper semicircle with center ((0),(0) ) and radius 5; integral = area ⇒ ∫_(−5) ^5 (√(25−x^2 ))dx=((5^2 π)/2)=((25)/2)π algebraic method ∫(√(25−x^2 ))dx= [t=arcsin (x/5) → dx=5cos t dt] =25∫cos^2 t dt=((25)/2)∫1+cos 2t dt= =((25)/2)x+((25)/4)sin 2t =((25)/2)arcsin (x/5) +(1/2)x(√(25−x^2 )) +C now use borders](Q72341.png)
$$\mathrm{geometrical}\:\mathrm{method} \\ $$$${y}=\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{semicircle}\:\mathrm{with}\:\mathrm{center} \\ $$$$\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{5};\:\mathrm{integral}\:=\:\mathrm{area}\:\Rightarrow \\ $$$$\underset{−\mathrm{5}} {\overset{\mathrm{5}} {\int}}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{5}^{\mathrm{2}} \pi}{\mathrm{2}}=\frac{\mathrm{25}}{\mathrm{2}}\pi \\ $$$$\mathrm{algebraic}\:\mathrm{method} \\ $$$$\int\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\frac{{x}}{\mathrm{5}}\:\rightarrow\:{dx}=\mathrm{5cos}\:{t}\:{dt}\right] \\ $$$$=\mathrm{25}\int\mathrm{cos}^{\mathrm{2}} \:{t}\:{dt}=\frac{\mathrm{25}}{\mathrm{2}}\int\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}\:{dt}= \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}}{x}+\frac{\mathrm{25}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{t}\:=\frac{\mathrm{25}}{\mathrm{2}}\mathrm{arcsin}\:\frac{{x}}{\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }\:+{C} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{borders} \\ $$
Commented by Rio Michael last updated on 27/Oct/19
$${thanks}\:{sir} \\ $$