Question Number 204902 by pticantor last updated on 01/Mar/24
$$\boldsymbol{{calculate}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\boldsymbol{{x}}\left(\mathrm{1}−\boldsymbol{{x}}\right)}\boldsymbol{{dx}} \\ $$
Answered by witcher3 last updated on 01/Mar/24
$$\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{x}=\mathrm{0}\Leftrightarrow\mathrm{y}^{\mathrm{2}} +\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{circle}\:\mathrm{center}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0}\right);\mathrm{R}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{S}=\pi.\frac{\mathrm{1}}{\mathrm{4}};\: \\ $$$$\mathrm{our}\:\mathrm{region}\:\mathrm{is}?\mathrm{y}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{S}}{\mathrm{2}}=\frac{\pi}{\mathrm{8}} \\ $$
Commented by pticantor last updated on 01/Mar/24
$${great}\: \\ $$$${thks}\:{too}\:{much} \\ $$
Answered by Mathspace last updated on 01/Mar/24
![I=∫_0 ^1 (√(x(1−x)))dx x=sin^2 t ⇒I=∫_0 ^(π/2) (√(sin^2 t.cos^2 t))2sint cost dt =2 ∫_0 ^(π/2) sin^2 t×cos^2 t dt =2∫_0 ^(π/2) ((1/2)sin(2t))^2 dt =(1/2)∫_0 ^(π/2) sin^2 (2t)dt =(1/4)∫_0 ^(π/2) (1−cos(4t))dt =(π/8)−(1/(16))[sin(4t)]_0 ^(π/2) =(π/8)−0=(π/8)](https://www.tinkutara.com/question/Q204915.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}\left(\mathrm{1}−{x}\right)}{dx} \\ $$$${x}={sin}^{\mathrm{2}} {t}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}^{\mathrm{2}} {t}.{cos}^{\mathrm{2}} {t}}\mathrm{2}{sint}\:{cost}\:{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} {t}×{cos}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{cos}\left(\mathrm{4}{t}\right)\right){dt} \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{16}}\left[{sin}\left(\mathrm{4}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{8}}−\mathrm{0}=\frac{\pi}{\mathrm{8}} \\ $$