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Question Number 189825 by TUN last updated on 22/Mar/23

$${\underset{a}{\int }}^b\sqrt{(x-a)(b-x)}=¿$$

Answered by Frix last updated on 14/Apr/23

$$\mathrm{circle}:$$$${(x-\frac{a+b}{2})}^2+y^2=\frac{{(a-b)}^2}{4}=r^2\Rightarrow r=\frac{\mid a-b\mid }{2}$$$$\iff$$$$y=\pm \sqrt{(x-a)(b-x)}$$$$\Rightarrow$$$$\int \sqrt{(x-a)(b-x)}dx=\frac{r^2\pi }{2}=\frac{{(a-b)}^2\pi }{8}$$

Answered by Mathspace last updated on 24/Mar/23

$$\sqrt{x-a}=t\Rightarrow x-a=t^2\Rightarrow x=t^2+a$$$$I={\int }_0^{\sqrt{b-a}}t\sqrt{b-a-t^2}\left(2t\right)dt$$$$=2{\int }_0^{\sqrt{b-a}}{t}^2\sqrt{b-a-t^2}dt$$$${=}_{t=\sqrt{b-a}y}2{\int }_0^1(b-a)y^2\sqrt{b-a}\sqrt{1-y^2}\sqrt{b-a}dy$$$$=2{(b-a)}^2{\int }_0^1{y}^2\sqrt{1-y^2}dy$$$${\int }_0^1{y}^2\sqrt{1-y^2}dy{=}_{y=sint}{\int }_0^{\frac{\pi }{2}}{sin}^{2}tcostcostdt$$$$={\int }_0^{\frac{\pi }{2}}{\left(sintcost\right)}^{2}dt$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}{sin}^2\left(2t\right)dt$$$$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{1-cos\left(4t\right)}{2}dt$$$$=\frac{1}{8}.\frac{\pi }{2}-\frac{1}{8.4}{\left[sin\left(4t\right)\right]}_0^{\frac{\pi }{2}}(\to 0)$$$$=\frac{\pi }{16}\Rightarrow I=2{(b-a)}^2.\frac{\pi }{16}$$$$=\frac{\pi }{8}{(b-a)}^2$$$$herewehavesupposeda⩽b$$

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