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Question Number 31074 by abdo imad last updated on 02/Mar/18

$$find{\int }_a^b\sqrt{(b-x)(x-a)}dxwitha<b.thenfind$$$${\int }_1^{\sqrt{2}}\sqrt{(\sqrt{2}-x)(x-1)}dx.$$

Commented by abdo imad last updated on 09/Mar/18

$$letputI={\int }_a^b\sqrt{(b-x)(x-a)}dxthech.x=\frac{a-b}{2}t+\frac{a+b}{2}$$$$b-x=b-\frac{a+b}{2}-\frac{a-b}{2}t=\frac{b-a}{2}-\frac{a-b}{2}t=\frac{b-a}{2}(1+t)$$$$x-a=\frac{a-b}{2}t+\frac{a+b}{2}-a=\frac{a-b}{2}t+\frac{b-a}{2}=\frac{b-a}{2}(1-t)and$$$$wehave\frac{a-b}{2}t=x-\frac{a+b}{2}\Rightarrow t=\frac{2(x-\frac{a+b}{2})}{a-b}=\frac{2x-a-b}{a-b}$$$$x=a\Rightarrow t=1andx=b\Rightarrow t=-1\Rightarrow$$$$I=-{\int }_{-1}^1\sqrt{\frac{b-a}{2}(1+t)\frac{b-a}{2}(1-t)}\frac{a-b}{2}dt$$$$I=\frac{{(b-a)}^2}{4}{\int }_{-1}^1\sqrt{1-t^2}dt=\frac{{(b-a)}^2}{2}{\int }_0^1\sqrt{1-t^2}dtch.t=sin\theta$$$$give{\int }_0^1\sqrt{1-t^2}dt={\int }_0^{\frac{\pi }{2}}cos\theta cos\theta d\theta ={\int }_0^{\frac{\pi }{2}}{cos}^2\theta d\theta$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(1+cos\left(2\theta \right)d\theta =\frac{\pi }{4}\Rightarrow I=\frac{{(b-a)}^2}{8}.thenlettake$$$$a=1andb=\sqrt{2}weget$$$${\int }_1^{\sqrt{2}}\sqrt{(\sqrt{2}-x)(x-1)}dx=\frac{{(\sqrt{2}-1)}^2}{8}=\frac{3-2\sqrt{2}}{8}.$$$$$$

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